Linear Algebra. Session 9 - Texas A&M Universityroquesol/Math_304_Fall_2019_Session_9.pdf ·...

Abstract Linear Algebra ISingular Value Decomposition (SVD)

Linear Algebra. Session 9

Dr. Marco A Roque Sol

10 / 25 / 2018

Dr. Marco A Roque Sol Linear Algebra. Session 9


Complex EigenvaluesRepeated EigenvaluesDiagonalization

Complex Eigenvalues

In this section we consider again a system of n linear homogeneousequations with constant coefficients

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues

In this section

we consider again a system of n linear homogeneousequations with constant coefficients

X′ = AX



λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues

In this section we consider again

a system of n linear homogeneousequations with constant coefficients

X′ = AX



λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues

In this section we consider again a system of n

linear homogeneousequations with constant coefficients

X′ = AX



λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues

In this section we consider again a system of n linear homogeneousequations

with constant coefficients

X′ = AX



λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX



λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where

the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix

A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A

is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued.

If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek

solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek solutions

of the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form

x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt ,

then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then

it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that

λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be

an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand

v a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v

a corresponding eigenvector of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector

of the coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the

coefficient matrix A.


λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX



λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX


In the case,

λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX


In the case, λ

is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX


In the case, λ is complex,

we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX


In the case, λ is complex, we have

complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX


In the case, λ is complex, we have complex eigenvalues and

eigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX


In the case, λ is complex, we have complex eigenvalues andeigenvectors

always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX


In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear

in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX


In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate.

Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX


In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus,

if wehave that

λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX



λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues


X′ = AX



λ±k = µ± i ν;

v±k = a± i b




Complex Eigenvalues


X′ = AX



λ±k = µ± i ν; v±k = a± i b




Complex Eigenvalues

are two complex-conjugate eigenvalues and eigenvectors of thematrix A, then

X±(t) = e(µ±i ν)t (a± i b)

are complex-valued solutions, but taking in account that

e(µ±i ν)t = eµt (cos(νt)± i sin(νt))

and the principle of superposition, then we have that

X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues

are two complex-conjugate

eigenvalues and eigenvectors of thematrix A, then

X±(t) = e(µ±i ν)t (a± i b)




X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues

are two complex-conjugate eigenvalues and eigenvectors

of thematrix A, then

X±(t) = e(µ±i ν)t (a± i b)




X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues

are two complex-conjugate eigenvalues and eigenvectors of thematrix A,

then

X±(t) = e(µ±i ν)t (a± i b)




X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)




X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)

are complex-valued

solutions, but taking in account that



X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)

are complex-valued solutions,

but taking in account that



X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)

are complex-valued solutions, but

taking in account that



X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)




X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)


e(µ±i ν)t =

eµt (cos(νt)± i sin(νt))


X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)




X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)



and the principle of superposition,

then we have that

X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)




X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)




X1(t) =1

2

(X+(t) + X−(t)

)

X2(t) =1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues


X±(t) = e(µ±i ν)t (a± i b)




X1(t) =1

2

(X+(t) + X−(t)

)X2(t) =

1

2i

(X+(t)− X−(t)

)




Complex Eigenvalues

are two (real) solutions !!!

X1(t) = eµt (acos(νt)− bsin(νt))

X2(t) = eµt (acos(νt) + bsin(νt))




Complex Eigenvalues

are two

(real) solutions !!!






Complex Eigenvalues

are two (real) solutions !!!






Complex Eigenvalues

Example 9.6

Solve the following ODE

x′ = Ax =

3 1 10 2 10 −1 2

x

Solution

Let’s find the eigenvalues of the matrix A




Complex Eigenvalues

Example 9.6

Solve

the following ODE

x′ = Ax =

3 1 10 2 10 −1 2

x

Solution





Complex Eigenvalues

Example 9.6


x′ = Ax =

3 1 10 2 10 −1 2

x

Solution





Complex Eigenvalues

Example 9.6


x′ = Ax =

3 1 10 2 10 −1 2

x

Solution

Let’s find

the eigenvalues of the matrix A




Complex Eigenvalues

Example 9.6


x′ = Ax =

3 1 10 2 10 −1 2

x

Solution

Let’s find the eigenvalues

of the matrix A




Complex Eigenvalues

Example 9.6


x′ = Ax =

3 1 10 2 10 −1 2

x

Solution

Let’s find the eigenvalues of the matrix

A




Complex Eigenvalues

Example 9.6


x′ = Ax =

3 1 10 2 10 −1 2

x

Solution





Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣ =

(3− λ)(λ2 − 4λ+ 5) = 0 =⇒




Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=

0 1 10 −1 10 −1 −1

v1v2v3

=

0 1 10 0 20 0 0

v1v2v3

=

000




Complex Eigenvalues

λ1 = 2,

λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=

0 1 10 −1 10 −1 −1

v1v2v3

=

0 1 10 0 20 0 0

v1v2v3

=

000




Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2=

2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=

0 1 10 −1 10 −1 −1

v1v2v3

=

0 1 10 0 20 0 0

v1v2v3

=

000




Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=

0 1 10 −1 10 −1 −1

v1v2v3

=

0 1 10 0 20 0 0

v1v2v3

=

000




Complex Eigenvalues

and a corresponding eigenvector is

v(1) =

100

If λ2 = 2 + i , then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=




Complex Eigenvalues

and

a corresponding eigenvector is

v(1) =

100


(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=




Complex Eigenvalues

and a corresponding

eigenvector is

v(1) =

100


(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=




Complex Eigenvalues


v(1) =

100


(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=




Complex Eigenvalues


v(1) =

100

If

λ2 = 2 + i , then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=




Complex Eigenvalues


v(1) =

100

If λ2 = 2 + i ,

then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=




Complex Eigenvalues


v(1) =

100


(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

=




Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

=

1− i 1 10 −i 10 −1 −i

v1v2v3

=

1− i 1 10 −i 10 0 0

v1v2v3

=

1− i 0 1− i0 −i 10 0 0

v1v2v3

=

000




Complex Eigenvalues


v(2) =

10−1

+ i

010

The corresponding solutions of the differential equation are

x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues

and


v(2) =

10−1

+ i

010


x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues

and a corresponding

eigenvector is

v(2) =

10−1

+ i

010


x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues


v(2) =

10−1

+ i

010


x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues


v(2) =

10−1

+

i

010


x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues


v(2) =

10−1

+ i

010


x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues


v(2) =

10−1

+ i

010

The corresponding

solutions of the differential equation are

x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues


v(2) =

10−1

+ i

010

The corresponding solutions

of the differential equation are

x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues


v(2) =

10−1

+ i

010

The corresponding solutions of the differential equation

are

x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues


v(2) =

10−1

+ i

010


x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues


v(2) =

10−1

+ i

010


x(1) =

100

e3t ;

x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues


v(2) =

10−1

+ i

010


x(1) =

100

e3t ; x(2) = e2t

10−1

cos(t)−

010

sin(t)

x(3) = e2t

10−1

cos(t) +

010

sin(t)




Complex Eigenvalues

The Wronskian of these solutions is

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣e3t e2tcos(t) e2tsin(t)0 −e2tsin(t) e2tcos(t)0 −e2tcos(t) −e2tsin(t)

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣1 cos(t) sin(t)0 −sin(t) cos(t)0 −cos(t) −sin(t)

∣∣∣∣∣∣ =

e3te2te2t(sin2(t) + cos2(t)

)= e7t 6= 0




Complex Eigenvalues

The Wronskian

of these solutions is

W [x(1), x(2), x(3)](t) =


∣∣∣∣∣∣ =

e3te2te2t


∣∣∣∣∣∣ =


)= e7t 6= 0




Complex Eigenvalues

The Wronskian of these solutions

is

W [x(1), x(2), x(3)](t) =


∣∣∣∣∣∣ =

e3te2te2t


∣∣∣∣∣∣ =


)= e7t 6= 0




Complex Eigenvalues


W [x(1), x(2), x(3)](t) =


∣∣∣∣∣∣ =

e3te2te2t


∣∣∣∣∣∣ =


)= e7t 6= 0




Complex Eigenvalues


W [x(1), x(2), x(3)](t) =


∣∣∣∣∣∣ =

e3te2te2t


∣∣∣∣∣∣ =


)=

e7t 6= 0




Complex Eigenvalues


W [x(1), x(2), x(3)](t) =


∣∣∣∣∣∣ =

e3te2te2t


∣∣∣∣∣∣ =


)= e7t

6= 0




Complex Eigenvalues


W [x(1), x(2), x(3)](t) =


∣∣∣∣∣∣ =

e3te2te2t


∣∣∣∣∣∣ =


)= e7t 6= 0




Complex Eigenvalues

Hence, the solutions x(1), x(2) and x(3) form a fundamental set,and the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues

Hence,

the solutions x(1), x(2) and x(3) form a fundamental set,and the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues

Hence, the solutions

x(1), x(2) and x(3) form a fundamental set,and the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues

Hence, the solutions x(1), x(2) and

x(3) form a fundamental set,and the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues

Hence, the solutions x(1), x(2) and x(3)

form a fundamental set,and the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues

Hence, the solutions x(1), x(2) and x(3) form a fundamental set,and

the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues

Hence, the solutions x(1), x(2) and x(3) form a fundamental set,and the general solution

of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues

Hence, the solutions x(1), x(2) and x(3) form a fundamental set,and the general solution of the system

is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues


X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues


X =

c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues


X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues


X = c1x(1) + c2x(2) + c3x(3) =⇒

X =

c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues


X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t +

c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues


X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

100

e3t + c2

10−1

e2tcos(t)−

010

e2tsin(t)

+

c3

10−1

e2tcos(t) +

010

e2tsin(t)




Complex Eigenvalues

X =

x1x2x3

=

c1e3t + e2t(c2cos(t) + c3sin(t))

0 e2t(−c2sin(t) + c3cos(t))0 −e2t(c2cos(t) + c3sin(t))

Here is the direction field associated with the system

x ′1x ′2x ′3

=

3 1 10 2 10 −1 2

x1x2x3




Complex Eigenvalues

X =

x1x2x3

=



Here is

the direction field associated with the system

x ′1x ′2x ′3

=

3 1 10 2 10 −1 2

x1x2x3




Complex Eigenvalues

X =

x1x2x3

=



Here is the direction field

associated with the system

x ′1x ′2x ′3

=

3 1 10 2 10 −1 2

x1x2x3




Complex Eigenvalues

X =

x1x2x3

=



Here is the direction field associated with

the system

x ′1x ′2x ′3

=

3 1 10 2 10 −1 2

x1x2x3




Complex Eigenvalues

X =

x1x2x3

=



Here is the direction field associated with the system

x ′1x ′2x ′3

=

3 1 10 2 10 −1 2

x1x2x3




Complex Eigenvalues




Complex Eigenvalues

Example 9.7


X′ = AX =

(−1/2 1−1 −1/2

)X

Solution


|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

4= 0




Complex Eigenvalues

Example 9.7

Solve

the following ODE

X′ = AX =

(−1/2 1−1 −1/2

)X

Solution


|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

4= 0




Complex Eigenvalues

Example 9.7


X′ = AX =

(−1/2 1−1 −1/2

)X

Solution


|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

4= 0




Complex Eigenvalues

Example 9.7


X′ = AX =

(−1/2 1−1 −1/2

)X

Solution

Let’s find


|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

4= 0




Complex Eigenvalues

Example 9.7


X′ = AX =

(−1/2 1−1 −1/2

)X

Solution


of the matrix A

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

4= 0




Complex Eigenvalues

Example 9.7


X′ = AX =

(−1/2 1−1 −1/2

)X

Solution


A

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

4= 0




Complex Eigenvalues

Example 9.7


X′ = AX =

(−1/2 1−1 −1/2

)X

Solution


|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

4= 0




Complex Eigenvalues

Example 9.7


X′ = AX =

(−1/2 1−1 −1/2

)X

Solution


|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 =

(λ)2 + λ+5

4= 0




Complex Eigenvalues

Example 9.7


X′ = AX =

(−1/2 1−1 −1/2

)X

Solution


|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

4= 0




Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

)=

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

)=(

−i 1−1 −i

)(v1v2

)=

(00

)




Complex Eigenvalues

λ1 = −1

2+ i ,

λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

)=

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

)=(

−i 1−1 −i

)(v1v2

)=

(00

)




Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

)=

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

)=(

−i 1−1 −i

)(v1v2

)=

(00

)




Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If

λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

)=

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

)=(

−i 1−1 −i

)(v1v2

)=

(00

)




Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i ,

then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

)=

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

)=(

−i 1−1 −i

)(v1v2

)=

(00

)




Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

)=

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

)=(

−i 1−1 −i

)(v1v2

)=

(00

)




Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

)=

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

)=

(−i 1−1 −i

)(v1v2

)=

(00

)




Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

)=

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

)=(

−i 1−1 −i

)(v1v2

)=

(00

)




Complex Eigenvalues


v(1) =

(1i

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

)=




Complex Eigenvalues

and


v(1) =

(1i

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

)=




Complex Eigenvalues

and a corresponding

eigenvector is

v(1) =

(1i

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

)=




Complex Eigenvalues


v(1) =

(1i

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

)=




Complex Eigenvalues


v(1) =

(1i

)

If λ2 = −12 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

)=




Complex Eigenvalues


v(1) =

(1i

)If

λ2 = −12 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

)=




Complex Eigenvalues


v(1) =

(1i

)If λ2 = −1

2 − i ,

then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

)=




Complex Eigenvalues


v(1) =

(1i

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

)=




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=


v(2) =

(1−i

)The corresponding solutions of the differential equation are

x(1) =

(1i

)e(−1/2+i)t ; x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=

and


v(2) =

(1−i


x(1) =

(1i

)e(−1/2+i)t ; x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=

and a corresponding

eigenvector is

v(2) =

(1−i


x(1) =

(1i

)e(−1/2+i)t ; x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=


v(2) =

(1−i


x(1) =

(1i

)e(−1/2+i)t ; x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=


v(2) =

(1−i

)


x(1) =

(1i

)e(−1/2+i)t ; x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=


v(2) =

(1−i

)The corresponding solutions

of the differential equation are

x(1) =

(1i

)e(−1/2+i)t ; x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=


v(2) =

(1−i

)The corresponding solutions of the differential equation

are

x(1) =

(1i

)e(−1/2+i)t ; x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=


v(2) =

(1−i


x(1) =

(1i

)e(−1/2+i)t ; x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=


v(2) =

(1−i


x(1) =

(1i

)e(−1/2+i)t ;

x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

(i 1−1 i

)(v1v2

)=


v(2) =

(1−i


x(1) =

(1i

)e(−1/2+i)t ; x(2) =

(1

− i

)e(−1/2−i)t




Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)

)Hence, a set of real-valued solutions are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues

To obtain

a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues

To obtain a set of

real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues

To obtain a set of real-valued solutions,

we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose

the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand

imaginary parts of either x (1) or x (2). In fact,

x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts

of either x (1) or x (2). In fact,

x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or

x (2). In fact,

x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2).

In fact,

x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues


x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues


x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+

i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues


x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)

)

Hence, a set of real-valued solutions are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues


x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)

)Hence,

a set of real-valued solutions are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues


x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)

)Hence, a set of

real-valued solutions are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues


x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)

)Hence, a set of real-valued

solutions are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues


x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)

)Hence, a set of real-valued solutions

are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues


x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues


x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)

v(t) = e−t/2(sin(t)cos(t)

)




Complex Eigenvalues


x(1) =

(1i

)e(−1/2+i)t =

(1i

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

)+ i

(e−t/2sin(t)

e−t/2cos(t)


u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

)




Complex Eigenvalues

The Wronskian of these two real-valued solutions is

W [x(1), x(2)](t) =

∣∣∣∣ e−t/2cos(t) e−t/2sin(t)

−e−t/2sin(t) e−t/2cos(t)

∣∣∣∣ =

e−t/2e−t/2∣∣∣∣ cos(t) sin(t)−sin(t) cos(t)

∣∣∣∣ = e−t 6= 0

Hence, the solutions x(1), x(2) form a fundamental set,




Complex Eigenvalues

The Wronskian

of these two real-valued solutions is

W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0





Complex Eigenvalues

The Wronskian of these two

real-valued solutions is

W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0





Complex Eigenvalues

The Wronskian of these two real-valued solutions

is

W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0





Complex Eigenvalues


W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0





Complex Eigenvalues


W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t

6= 0





Complex Eigenvalues


W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0





Complex Eigenvalues


W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0

Hence,

the solutions x(1), x(2) form a fundamental set,




Complex Eigenvalues


W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0

Hence, the solutions

x(1), x(2) form a fundamental set,




Complex Eigenvalues


W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0

Hence, the solutions x(1),

x(2) form a fundamental set,




Complex Eigenvalues


W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0

Hence, the solutions x(1), x(2)

form a fundamental set,




Complex Eigenvalues


W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0

Hence, the solutions x(1), x(2) form a

fundamental set,




Complex Eigenvalues


W [x(1), x(2)](t) =



∣∣∣∣ =


∣∣∣∣ = e−t 6= 0





Complex Eigenvalues

and the general solution of the system is

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field associated with the system(

x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues

and

the general solution of the system is

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)


x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues

and the general solution

of the system is

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)


x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues

and the general solution of the system

is

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)


x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues


X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)


x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues


X =

c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)


x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues


X = c1x(1) + c2x(2) =

c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)


x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues


X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+

c2e−t/2

(sin(t)cos(t)


x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues


X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)

Here is the direction field associated with the system(x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues


X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is

the direction field associated with the system(x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues


X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field

associated with the system(x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues


X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field associated with

the system(x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues


X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field associated with the system

(x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues


X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)


x ′1x ′2

)=

(−1/2 1−1 −1/2

)(x1x2

)




Complex Eigenvalues




Repeated Eigenvalues

We conclude our consideration of the linear homogeneous systemwith constant coefficients

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.





We conclude

our consideration of the linear homogeneous systemwith constant coefficients

x′ = Ax


∣∣∣ = 0


{veλt ,weλt

}.





We conclude our consideration

of the linear homogeneous systemwith constant coefficients

x′ = Ax


∣∣∣ = 0


{veλt ,weλt

}.





We conclude our consideration of the linear homogeneous system

with constant coefficients

x′ = Ax


∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax

with a discussion

of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax

with a discussion of the case

in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax

with a discussion of the case in which the matrix

A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax

with a discussion of the case in which the matrix A

has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues.

suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that

λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ

is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root

of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the

characteristicequation ∣∣∣A− λI

∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation

∣∣∣A− λI∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then,

λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ

is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue

of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity

2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2

of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA.

In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event,

there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are

two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities:

The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A

isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and

there is still a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still

a fundamental set of solutions ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions

ofthe form

{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0


{veλt ,weλt

}.






x′ = Ax


∣∣∣ = 0


{veλt ,

weλt}

.






x′ = Ax


∣∣∣ = 0


{veλt ,weλt

}.





However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

In this way, it may be natural to attempt to find a secondindependent solution of the form

x = wteλt





However,

if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A

is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective,

there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is

just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is just one solution

ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is just one solution ofthe form

veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is just one solution ofthe form veλt

associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is just one solution ofthe form veλt associated with

this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue.

Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore,

toconstruct the general solution, it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution,

it is necessary to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary

to find other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find

other solutionof a different form.



x = wteλt





However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solution

of a different form.



x = wteλt








x = wteλt






Recall that

a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation

occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for

the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0

when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation

had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r .

In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case

we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found

one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution

y1(t) = ert ,but a second independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,

but a second independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second

independent solution had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution

had the form y2(t) = tert


x = wteλt






Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form

y2(t) = tert


x = wteλt








x = wteλt







In this way,

it may be natural to attempt to find a secondindependent solution of the form

x = wteλt







In this way, it may be natural

to attempt to find a secondindependent solution of the form

x = wteλt







In this way, it may be natural to attempt to find

a secondindependent solution of the form

x = wteλt







In this way, it may be natural to attempt to find a second

independent solution of the form

x = wteλt







In this way, it may be natural to attempt to find a secondindependent solution

of the form

x = wteλt








x = wteλt





but, doing this and substituting x in the system we find thatw = 0. Thus, we propose

x = wteλt + ueλt

and substituting this new x in the system we find the system

(A− λI) w = 0

(A− λI) u = w

The first equation is already solved with w = v and only thesecond one is remaining to be solved.





but,

doing this and substituting x in the system we find thatw = 0. Thus, we propose

x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w






but, doing this and

substituting x in the system we find thatw = 0. Thus, we propose

x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w






but, doing this and substituting x in the system

we find thatw = 0. Thus, we propose

x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w






but, doing this and substituting x in the system we find that

w = 0. Thus, we propose

x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w






but, doing this and substituting x in the system we find thatw = 0. Thus,

we propose

x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w







x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w







x = wteλt + ueλt

and

substituting this new x in the system we find the system

(A− λI) w = 0

(A− λI) u = w







x = wteλt + ueλt

and substituting

this new x in the system we find the system

(A− λI) w = 0

(A− λI) u = w







x = wteλt + ueλt

and substituting this new x

in the system we find the system

(A− λI) w = 0

(A− λI) u = w







x = wteλt + ueλt

and substituting this new x in the system

we find the system

(A− λI) w = 0

(A− λI) u = w







x = wteλt + ueλt

and substituting this new x in the system we find

the system

(A− λI) w = 0

(A− λI) u = w







x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w







x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w

The first equation

is already solved with w = v and only thesecond one is remaining to be solved.






x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w

The first equation is already solved

with w = v and only thesecond one is remaining to be solved.






x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w

The first equation is already solved with w = v and

only thesecond one is remaining to be solved.






x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w

The first equation is already solved with w = v and only thesecond one

is remaining to be solved.






x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w

The first equation is already solved with w = v and only thesecond one is remaining

to be solved.






x = wteλt + ueλt


(A− λI) w = 0

(A− λI) u = w






Example 9.8

Find the solution of the system

x′ = Ax =

(1 −11 3

)x

Solution


|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0





Example 9.8

Find

the solution of the system

x′ = Ax =

(1 −11 3

)x

Solution


|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0





Example 9.8

Find the solution

of the system

x′ = Ax =

(1 −11 3

)x

Solution


|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0





Example 9.8


x′ = Ax =

(1 −11 3

)x

Solution


|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0





Example 9.8


x′ = Ax =

(1 −11 3

)x

Solution

Let’s find


|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0





Example 9.8


x′ = Ax =

(1 −11 3

)x

Solution


of the matrix A

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0





Example 9.8


x′ = Ax =

(1 −11 3

)x

Solution


A

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0





Example 9.8


x′ = Ax =

(1 −11 3

)x

Solution


|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0





λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

)=

(−1 −11 1

)(v1v2

)=

(00

)and a corresponding eigenvector is

v(1) =

(1

− 1

)





λ1 = 2,

λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

)=

(−1 −11 1

)(v1v2

)=

(00


v(1) =

(1

− 1

)





λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

)=

(−1 −11 1

)(v1v2

)=

(00


v(1) =

(1

− 1

)





λ1 = 2, λ2 = 2,

If

λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

)=

(−1 −11 1

)(v1v2

)=

(00


v(1) =

(1

− 1

)





λ1 = 2, λ2 = 2,

If λ1,2 = 2,

then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

)=

(−1 −11 1

)(v1v2

)=

(00


v(1) =

(1

− 1

)





λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

)=

(−1 −11 1

)(v1v2

)=

(00


v(1) =

(1

− 1

)





λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

)=

(−1 −11 1

)(v1v2

)=

(00

)


v(1) =

(1

− 1

)





λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

)=

(−1 −11 1

)(v1v2

)=

(00

)and


v(1) =

(1

− 1

)





λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

)=

(−1 −11 1

)(v1v2

)=

(00

)and a corresponding

eigenvector is

v(1) =

(1

− 1

)





λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

)=

(−1 −11 1

)(v1v2

)=

(00


v(1) =

(1

− 1

)





and the solution is

x(1) =

(1

− 1

)e2t

Now, for the second solution we propose

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and

the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution

is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t

Now,

for the second solution we propose

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t

Now, for the second

solution we propose

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t

Now, for the second solution

we propose

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t +

ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where

u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where u

satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where u satisfies

(A− λI) u =

(A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u =

v

(−1 −11 1

)(u1u2

)=

(v1v2

)





and the solution is

x(1) =

(1

− 1

)e2t


x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

)=

(v1v2

)





we have

−u1 − u2 = 1

so if u1 = k , where k is arbitrary, then u2 = −k − 1. If we write

u =

(k

−1− k

)=

(0−1

)+ k

(1−1

)then by substituting for w and u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1

so

if u1 = k , where k is arbitrary, then u2 = −k − 1. If we write

u =

(k

−1− k

)=

(0−1

)+ k

(1−1


x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1

so if u1 = k ,

where k is arbitrary, then u2 = −k − 1. If we write

u =

(k

−1− k

)=

(0−1

)+ k

(1−1


x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1

so if u1 = k , where k

is arbitrary, then u2 = −k − 1. If we write

u =

(k

−1− k

)=

(0−1

)+ k

(1−1


x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1

so if u1 = k , where k is arbitrary,

then u2 = −k − 1. If we write

u =

(k

−1− k

)=

(0−1

)+ k

(1−1


x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1

so if u1 = k , where k is arbitrary, then u2 = −k − 1.

If we write

u =

(k

−1− k

)=

(0−1

)+ k

(1−1


x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1


u =

(k

−1− k

)=

(0−1

)+ k

(1−1


x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1


u =

(k

−1− k

)=

(0−1

)+ k

(1−1

)

then by substituting for w and u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1


u =

(k

−1− k

)=

(0−1

)+ k

(1−1

)then

by substituting for w and u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1


u =

(k

−1− k

)=

(0−1

)+ k

(1−1

)then by substituting

for w and u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1


u =

(k

−1− k

)=

(0−1

)+ k

(1−1

)then by substituting for w and

u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1


u =

(k

−1− k

)=

(0−1

)+ k

(1−1

)then by substituting for w and u,

we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1


u =

(k

−1− k

)=

(0−1

)+ k

(1−1


x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





we have

−u1 − u2 = 1


u =

(k

−1− k

)=

(0−1

)+ k

(1−1


x(2) =

(1−1

)te2t +

(0−1

)e2t +

ke2t(

1−1

)





we have

−u1 − u2 = 1


u =

(k

−1− k

)=

(0−1

)+ k

(1−1


x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

)





The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t

An elementary calculation shows that W [x (1), x (2)](t) = − e4t 6= 0and therefore

{x (1), x (2)

}form a fundamental set of solutions of

the system. The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





The last term above

is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





The last term above is merely

a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





The last term above is merely a multiple

of the first solutionx (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





The last term above is merely a multiple of the first solution

x (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





The last term above is merely a multiple of the first solutionx (1)(t) and

may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored,

but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but

the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first

two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms

constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms constitute

anew solution:

x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t

An elementary calculation

shows that W [x (1), x (2)](t) = − e4t 6= 0and therefore

{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t

An elementary calculation shows that W [x (1), x (2)](t) =

− e4t 6= 0and therefore

{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t

An elementary calculation shows that W [x (1), x (2)](t) = − e4t 6= 0and

therefore{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)

}

form a fundamental set of solutions ofthe system. The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)

}form a fundamental set

of solutions ofthe system. The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)

}form a fundamental set of solutions

ofthe system. The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)


the system.

The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)


the system. The general solution

is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x =

c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) +

c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)






x(2) =

(1−1

)te2t +

(0−1

)e2t


{x (1), x (2)



x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

)e2t)





Consider again the system

x′ = Ax

and suppose that r = λ is a double eigenvalue of A, but, there isonly one corresponding eigenvector v. Then one solution is

x(1)(t) = veλtwhere v satisfies

(A− λI) v = 0

and a second solution is given by

x(2)(t) = vteλt + ueλt

where u satisfies

(A− λI) u = v





Consider again

the system

x′ = Ax



(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax



(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax

and

suppose that r = λ is a double eigenvalue of A, but, there isonly one corresponding eigenvector v. Then one solution is


(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax

and suppose that

r = λ is a double eigenvalue of A, but, there isonly one corresponding eigenvector v. Then one solution is


(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax

and suppose that r = λ

is a double eigenvalue of A, but, there isonly one corresponding eigenvector v. Then one solution is


(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax

and suppose that r = λ is a double eigenvalue

of A, but, there isonly one corresponding eigenvector v. Then one solution is


(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax

and suppose that r = λ is a double eigenvalue of A,

but, there isonly one corresponding eigenvector v. Then one solution is


(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax

and suppose that r = λ is a double eigenvalue of A, but, there isonly one

corresponding eigenvector v. Then one solution is


(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax

and suppose that r = λ is a double eigenvalue of A, but, there isonly one corresponding eigenvector v.

Then one solution is


(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax

and suppose that r = λ is a double eigenvalue of A, but, there isonly one corresponding eigenvector v. Then

one solution is


(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax



(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax


x(1)(t) = veλt

where v satisfies

(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax


x(1)(t) = veλtwhere

v satisfies

(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax


x(1)(t) = veλtwhere v

satisfies

(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax



(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax



(A− λI) v = 0

and

a second solution is given by


where u satisfies

(A− λI) u = v






x′ = Ax



(A− λI) v = 0

and a second solution

is given by


where u satisfies

(A− λI) u = v






x′ = Ax



(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax



(A− λI) v = 0



where

u satisfies

(A− λI) u = v






x′ = Ax



(A− λI) v = 0



where u

satisfies

(A− λI) u = v






x′ = Ax



(A− λI) v = 0



where u satisfies

(A− λI) u = v






x′ = Ax



(A− λI) v = 0



where u satisfies

(A− λI) u = vDr. Marco A Roque Sol Linear Algebra. Session 9




Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

The vector u is known as a generalized eigenvector.





Even though

|A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0,

it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that

it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible

to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it

for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u

( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually,

there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are

infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .

Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now,

Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation,

together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with

the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation

for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0






Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,

we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0







(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0







(A− λI)

[(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0







(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0







(A− λI) [(A− λI) u = v]

(A− λI)2 u =

(A− λI) v

(A− λI)2 u = 0







(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0







(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u =

0







(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0







(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

The vector

u is known as a generalized eigenvector.






(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

The vector u

is known as a generalized eigenvector.






(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

The vector u is known as

a generalized eigenvector.






(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0





Diagonalization.

Diagonalizable Matrices.

The basic reason why a system of linear (algebraic or differential)equations presents some difficulty is that the equations are usuallycoupled.

Hence, the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.


The basic reason

why a system of linear (algebraic or differential)equations presents some difficulty is that the equations are usuallycoupled.






Diagonalization.


The basic reason why a system

of linear (algebraic or differential)equations presents some difficulty is that the equations are usuallycoupled.






Diagonalization.


The basic reason why a system of linear

(algebraic or differential)equations presents some difficulty is that the equations are usuallycoupled.






Diagonalization.


The basic reason why a system of linear (algebraic or differential)

equations presents some difficulty is that the equations are usuallycoupled.






Diagonalization.


The basic reason why a system of linear (algebraic or differential)equations

presents some difficulty is that the equations are usuallycoupled.






Diagonalization.


The basic reason why a system of linear (algebraic or differential)equations presents some difficulty

is that the equations are usuallycoupled.






Diagonalization.


The basic reason why a system of linear (algebraic or differential)equations presents some difficulty is that

the equations are usuallycoupled.






Diagonalization.


The basic reason why a system of linear (algebraic or differential)equations presents some difficulty is that the equations

are usuallycoupled.






Diagonalization.


The basic reason why a system of linear (algebraic or differential)equations presents some difficulty is that the equations are usually

coupled.






Diagonalization.








Diagonalization.



Hence,

the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.





Diagonalization.



Hence, the equations

in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.





Diagonalization.



Hence, the equations in the system

must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.





Diagonalization.



Hence, the equations in the system must be solved

simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.





Diagonalization.



Hence, the equations in the system must be solved simultaneously.

On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.





Diagonalization.



Hence, the equations in the system must be solved simultaneously.On the contrary,

if the system is uncoupled, then each equationcan be solved independently of all the others.





Diagonalization.



Hence, the equations in the system must be solved simultaneously.On the contrary, if the system

is uncoupled, then each equationcan be solved independently of all the others.





Diagonalization.



Hence, the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled,

then each equationcan be solved independently of all the others.





Diagonalization.



Hence, the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equation

can be solved independently of all the others.





Diagonalization.



Hence, the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved

independently of all the others.





Diagonalization.



Hence, the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently

of all the others.





Diagonalization.








Diagonalization.




Transforming

the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system

into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent

uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem

( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which

each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation

contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only

one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable )

corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds

to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming

the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix

A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A

intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa

diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix.

Eigenvectors are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors

are useful in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful

in accomplishing such atransformation.




Diagonalization.




Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing

such atransformation.




Diagonalization.








Diagonalization.

Let’s assume that the matrix A has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then

Ax(1) = λ1x(1); Ax(2) = λ2x(2); ...Ax(n) = λnx(n)

and considering the matrix

U =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n




Diagonalization.

Let’s assume

that the matrix A has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then



U =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n




Diagonalization.

Let’s assume that the matrix

A has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then



U =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n




Diagonalization.

Let’s assume that the matrix A

has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then



U =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n




Diagonalization.

Let’s assume that the matrix A has n eigenvectors x(1), x(2), ...,x(n)

linearly indepedent, then



U =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n




Diagonalization.




U =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n




Diagonalization.



and

considering the matrix

U =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n




Diagonalization.




U =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n




Diagonalization.

we have

AU =

Ax(1) · · · Ax

(n)1

......

... · · ·...

=

λ1x

(1)1 · · · λnx

(n)1

λ1x(1)2

...

λ1x(1)n λnx

(n)n

= UD




Diagonalization.

we have

AU =

Ax(1) · · · Ax

(n)1

......

... · · ·...

=

λ1x

(1)1 · · · λnx

(n)1

λ1x(1)2

...

λ1x(1)n λnx

(n)n

=

UD




Diagonalization.

we have

AU =

Ax(1) · · · Ax

(n)1

......

... · · ·...

=

λ1x

(1)1 · · · λnx

(n)1

λ1x(1)2

...

λ1x(1)n λnx

(n)n

= UD




Diagonalization.

where D is the diagonal matrix

D =

λ1

λ2. . .

λn

whose diagonal elements are the eigenvalues of A. From the lastequations we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.




Diagonalization.

where D

is the diagonal matrix

D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1





Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1





Diagonalization.


D =

λ1

λ2. . .

λn

whose

diagonal elements are the eigenvalues of A. From the lastequations we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1





Diagonalization.


D =

λ1

λ2. . .

λn

whose diagonal elements

are the eigenvalues of A. From the lastequations we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1





Diagonalization.


D =

λ1

λ2. . .

λn

whose diagonal elements are the eigenvalues

of A. From the lastequations we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1





Diagonalization.


D =

λ1

λ2. . .

λn

whose diagonal elements are the eigenvalues of A.

From the lastequations we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1





Diagonalization.


D =

λ1

λ2. . .

λn

whose diagonal elements are the eigenvalues of A. From the lastequations

we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1





Diagonalization.


D =

λ1

λ2. . .

λn

whose diagonal elements are the eigenvalues of A. From the lastequations we have that

it follows that

U−1AU = D ⇐⇒ A = UDU−1





Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1





Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1

Thus,

if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.




Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and

eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.




Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors

of A are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.




Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A

are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.




Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known,

A canbe transformed into a diagonal matrix by the process shown in theabove equation.




Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed

into a diagonal matrix by the process shown in theabove equation.




Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a

diagonal matrix by the process shown in theabove equation.




Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix

by the process shown in theabove equation.




Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process

shown in theabove equation.




Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process shown in

theabove equation.




Diagonalization.


D =

λ1

λ2. . .

λn


U−1AU = D ⇐⇒ A = UDU−1





Diagonalization.

This process is known as a similarity transformation.Alternatively, we may say that A is diagonalizable.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.




Diagonalization.

This process

is known as a similarity transformation.Alternatively, we may say that A is diagonalizable.






Diagonalization.

This process is known as

a similarity transformation.Alternatively, we may say that A is diagonalizable.






Diagonalization.

This process is known as a similarity transformation.

Alternatively, we may say that A is diagonalizable.






Diagonalization.

This process is known as a similarity transformation.Alternatively,

we may say that A is diagonalizable.






Diagonalization.

This process is known as a similarity transformation.Alternatively, we may say

that A is diagonalizable.






Diagonalization.

This process is known as a similarity transformation.Alternatively, we may say that A

is diagonalizable.






Diagonalization.







Diagonalization.


If

A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A

is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian

( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ),

then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination

of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1

isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple.

The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n)

of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A

are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known

to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to be

mutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal,

so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us

choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them

so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that

they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are

alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized

by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1

for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i .

It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy

verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify that

U−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗.

In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words,

the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse

of U is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U

is the same as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same

as itsadjoint (the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint

(the transpose of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose

of its complex conjugate).





Diagonalization.


If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its

complex conjugate).





Diagonalization.







Diagonalization.



Finally,

we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.




Diagonalization.



Finally, we note

that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.




Diagonalization.



Finally, we note that if A

has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.




Diagonalization.



Finally, we note that if A has fewer

than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.




Diagonalization.



Finally, we note that if A has fewer than n

linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.




Diagonalization.



Finally, we note that if A has fewer than n linearly independenteigenvectors,

then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.




Diagonalization.



Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no

matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.




Diagonalization.



Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U

such that U−1AU = D. Inthis case, A is not diagonalizable.




Diagonalization.



Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that

U−1AU = D. Inthis case, A is not diagonalizable.




Diagonalization.



Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D.

Inthis case, A is not diagonalizable.




Diagonalization.



Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case,

A is not diagonalizable.




Diagonalization.



Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A

is not diagonalizable.




Diagonalization.



Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not

diagonalizable.




Diagonalization.







Diagonalization.

Fundamental Matrices

Let’s start with the system

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.




Diagonalization.



x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t)

form a fundamental set ofsolutions on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n





Diagonalization.



x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set

ofsolutions on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n





Diagonalization.



x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions

on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n





Diagonalization.



x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions on some interval

α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n





Diagonalization.



x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions on some interval α < t < β.

Then the matrix

Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n





Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n





Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns

are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.




Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns are the vectors

x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.




Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns are the vectors x(1)(t), ..., x(n)(t),

is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.




Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be

afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.




Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix

for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.




Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system.

Since the set ofsolutions is linearly independent the matrix is nonsingular.




Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions

is linearly independent the matrix is nonsingular.




Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent

the matrix is nonsingular.




Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix

is nonsingular.




Diagonalization.



x′ = P(t)x


Ψ(t) =

x(1)1 · · · x

(n)1

......

x(1)n · · · x

(n)n





Diagonalization.

Thus, for example, a fundamental matrix for the system

x′ = Ax =

(1 −11 3

)x

can be formed from the solutions x (1)(t) and x (2)(t):

x(1) =

(1

− 1

)e2t ; x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.

Thus,

for example, a fundamental matrix for the system

x′ = Ax =

(1 −11 3

)x


x(1) =

(1

− 1

)e2t ; x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.

Thus, for example,

a fundamental matrix for the system

x′ = Ax =

(1 −11 3

)x


x(1) =

(1

− 1

)e2t ; x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.

Thus, for example, a fundamental matrix

for the system

x′ = Ax =

(1 −11 3

)x


x(1) =

(1

− 1

)e2t ; x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.


x′ = Ax =

(1 −11 3

)x


x(1) =

(1

− 1

)e2t ; x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.


x′ = Ax =

(1 −11 3

)x

can be formed

from the solutions x (1)(t) and x (2)(t):

x(1) =

(1

− 1

)e2t ; x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.


x′ = Ax =

(1 −11 3

)x

can be formed from the solutions

x (1)(t) and x (2)(t):

x(1) =

(1

− 1

)e2t ; x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.


x′ = Ax =

(1 −11 3

)x

can be formed from the solutions x (1)(t) and

x (2)(t):

x(1) =

(1

− 1

)e2t ; x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.


x′ = Ax =

(1 −11 3

)x


x(1) =

(1

− 1

)e2t ; x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.


x′ = Ax =

(1 −11 3

)x


x(1) =

(1

− 1

)e2t ;

x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.


x′ = Ax =

(1 −11 3

)x


x(1) =

(1

− 1

)e2t ; x(2) =

(1−1

)te2t +

(0−1

)e2t

then




Diagonalization.

then

Ψ(t) =

(e2t te2t

−e2t −te2t − e2t

)= e2t

(1 t−1 −1− t

)Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be foundfrom the relation Φ(t) = Ψ(t)Ψ−1(0).




Diagonalization.

then

Ψ(t) =

(e2t te2t


)=

e2t(

1 t−1 −1− t


Ψ′ = P(t)Ψ





Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t


Ψ′ = P(t)Ψ





Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t

)

Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ





Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t

)Recall that each column

of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ





Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t

)Recall that each column of the fundamental matrix Ψ(t)

is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ





Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t

)Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE.

It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ





Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t

)Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t)

satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ





Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t


Ψ′ = P(t)Ψ





Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t


Ψ′ = P(t)Ψ

In particular,

the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be foundfrom the relation Φ(t) = Ψ(t)Ψ−1(0).




Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t


Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = I

is called the special fundamental matrix and can also be foundfrom the relation Φ(t) = Ψ(t)Ψ−1(0).




Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t


Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and

can also be foundfrom the relation Φ(t) = Ψ(t)Ψ−1(0).




Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t


Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be found

from the relation Φ(t) = Ψ(t)Ψ−1(0).




Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t


Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be foundfrom the relation

Φ(t) = Ψ(t)Ψ−1(0).




Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t


Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be foundfrom the relation Φ(t) =

Ψ(t)Ψ−1(0).




Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t


Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be foundfrom the relation Φ(t) = Ψ(t)

Ψ−1(0).




Diagonalization.

then

Ψ(t) =

(e2t te2t


)= e2t

(1 t−1 −1− t


Ψ′ = P(t)Ψ





Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix is also known as the exponential matrix eAt .




Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)

=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t





Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t





Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) =

Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t





Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) =

e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t





Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

)

(1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t





Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t





Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) =

e2t(

1− t −tt 1 + t





Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t

)

The latter matrix is also known as the exponential matrix eAt .




Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix

is also known as the exponential matrix eAt .




Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix is also known

as the exponential matrix eAt .




Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix is also known as the

exponential matrix eAt .




Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

)

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

)

Φ(t) = e2t(

1− t −tt 1 + t





Diagonalization.

The Matrix eAt

Recall that the solution of the initial value problem

x ′ = ax , x(0) = x0, a = constant

is given by

x(t) = x0eat

Now, consider the corresponding initial value problem for an n × nsystem

x′ = Ax, x(0) = x0

where A is a constant matrix.




Diagonalization.

The Matrix eAt



is given by

x(t) = x0eat

Now, consider the corresponding initial value problem

for an n × nsystem

x′ = Ax, x(0) = x0





Diagonalization.

The Matrix eAt



is given by

x(t) = x0eat

Now, consider the corresponding initial value problem for an n × nsystem

x′ = Ax, x(0) = x0





Diagonalization.

Applying the results of already obtained, we can write its solutionas

x = Φ(t)x0

where Φ(0) = I. Thus, Φ(t), is playing the roll of eat . let’s seethis with more detail.

The scalar exponential function eat can be represented by thepower series

eat = 1 +∞∑n=1

antn

n!

which converges for all t. Let us now replace the scalar a by then × n constant matrix A and consider the corresponding series




Diagonalization.

Applying the results of already obtained,

we can write its solutionas

x = Φ(t)x0



eat = 1 +∞∑n=1

antn

n!





Diagonalization.


x = Φ(t)x0



eat = 1 +∞∑n=1

antn

n!





Diagonalization.


x = Φ(t)x0

where Φ(0) = I.

Thus, Φ(t), is playing the roll of eat . let’s seethis with more detail.


eat = 1 +∞∑n=1

antn

n!





Diagonalization.


x = Φ(t)x0

where Φ(0) = I. Thus, Φ(t), is playing the roll of eat .

let’s seethis with more detail.


eat = 1 +∞∑n=1

antn

n!





Diagonalization.


x = Φ(t)x0



eat = 1 +∞∑n=1

antn

n!





Diagonalization.


x = Φ(t)x0


The scalar exponential function eat

can be represented by thepower series

eat = 1 +∞∑n=1

antn

n!





Diagonalization.


x = Φ(t)x0



eat = 1 +∞∑n=1

antn

n!





Diagonalization.


x = Φ(t)x0



eat = 1 +∞∑n=1

antn

n!

which converges for all t.

Let us now replace the scalar a by then × n constant matrix A and consider the corresponding series




Diagonalization.


x = Φ(t)x0



eat = 1 +∞∑n=1

antn

n!

which converges for all t. Let us now replace the scalar a by then × n constant matrix A and

consider the corresponding series




Diagonalization.


x = Φ(t)x0



eat = 1 +∞∑n=1

antn

n!





Diagonalization.

I +∞∑n=1

Antn

n!= I + At +

A2t2

2!+ ...+

Ant2

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat each element of this matrix sum converges for all t asn→∞. Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

Antn

n!

By differentiating the above series term by term, we obtain

d

dt

[eAt]

=∞∑n=1

Antn−1

(n − 1)!= A

[I +

∞∑n=1

Antn

n!

]= AeAt




Diagonalization.

I +∞∑n=1

Antn

n!=

I + At +A2t2

2!+ ...+

Ant2

n!+ ...


eAt = I +∞∑n=1

Antn

n!


d

dt

[eAt]

=∞∑n=1

Antn−1

(n − 1)!= A

[I +

∞∑n=1

Antn

n!

]= AeAt




Diagonalization.

I +∞∑n=1

Antn

n!= I +

At +A2t2

2!+ ...+

Ant2

n!+ ...


eAt = I +∞∑n=1

Antn

n!


d

dt

[eAt]

=∞∑n=1

Antn−1

(n − 1)!= A

[I +

∞∑n=1

Antn

n!

]= AeAt
