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Math 211Math 211

Lecture #21

Determinants

October 16, 2002

Return Span

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Basis of a SubspaceBasis of a Subspace

Definition: A set of vectors v1, v2, . . . , and vk form a

basis of a subspace V if

1. V = span(v1, v2, . . . , vk)

2. v1, v2, . . . , and vk are linearly independent.

• The best way to describe a subspace is to give a basis.

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Examples of BasesExamples of Bases

• The vector v = (1, −1, 1)T is a basis for null(A).

� null(A) is the subspace of R3 with basis v.

• The vectors v = (1, −1, 1, 0)T and

w = (0, −2, 0, 1)T form a basis for null(B).

� null(B) is the subspace of R4 with basis {v,w}.

Return Examples

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Existence of a BasisExistence of a Basis

Proposition: Let V be a subspace of Rn.

1. If V �= {0}, then V has a basis.

2. Bases are not unique, but every basis of V has the

same number of elements.

Definition: The dimension of a subspace V is the

number of elements in a basis of V .

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Another Example of a NullspaceAnother Example of a Nullspace

A =

3 −3 1 −1−2 2 −1 11 −1 0 013 −13 5 −5

rref−→

1 −1 0 00 0 1 −10 0 0 00 0 0 0

• null(A) is the subspace of R4 with basis (1, 1, 0, 0)T

and (0, 0, 1, 1)T .

• null(A) has dimension 2.

• In MATLAB, use commands null(A) or null(A,’r’).

Return Theorem

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Nonsingular MatricesNonsingular Matrices

Let A be an n × n matrix. We know the following:

• A is nonsingular if the equation Ax = b has a solution

for any right hand side b. (This is the definition.)

• If A is nonsingular then Ax = b has a unique solution

for any right hand side b.

• A is singular if and only if the homogeneous equation

Ax = 0 has a non-zero solution.

� null(A) is non-trivial ⇔ A is singular.

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Determinants in 2DDeterminants in 2D

• How do we decide if a matrix A is nonsingular?

• A is nonsingular if and only if when put into row

echelon form, the matrix has nonzero entries along the

diagonal.

• Example: the general 2 × 2 matrix

A =(

a b

c d

)

is nonsingular if and only if ad − bc �= 0.

� We define ad − bc to be the determinant of A.

Return 2D

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Determinants in 3DDeterminants in 3D

A =

a11 a12 a13

a21 a22 a23

a31 a32 a33

• The same (but more difficult) argument shows that A

is nonsingular if and only if

a11a22a33 − a11a23a32 − a12a21a33

+ a12a23a31 − a13a22a31 + a13a21a32

�= 0.

• This will be the determinant of A.

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Main TheoremMain Theorem

We will define the determinant of a square matrix A so

that the next theorem is true.

Theorem: The n × n matrix A is nonsingular if and

only if det(A) �= 0.

Corollary: If A is an n × n matrix, then null(A)contains a nonzero vector if and only if det(A) = 0.

• The corollary contains the most important fact about

determinants for ODEs.

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Matrices and MinorsMatrices and Minors

The general n × n matrix has the form

A =

a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...

an1 an2 · · · ann

Definition: The ij-minor of an n × n matrix A is the

(n − 1) × (n − 1) matrix Aij obtained from A by deleting

the ith row and the jth column.

Return Matrix 3 × 3

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Definition of DeterminantDefinition of Determinant

Definition: The determinant of an n × n matrix A is

defined to be

det(A) =n∑

j=1

(−1)j+1a1j det(A1j).

• The definition is inductive.

� It assumes we know how to compute the

determinants of (n − 1) × (n − 1) matrices.

� We start with the 2 × 2 matrix.

Definition

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ExampleExample

det

2 1 03 −2 4−1 5 3

= (−1)2 × 2 × det(−2 4

5 3

)

+ (−1)3 × 1 × det(

3 4−1 3

)

= 2 × (−26) − 13

= −65

Return Definition

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Expansion by the ith RowExpansion by the ith Row

For any i, we have

det(A) =n∑

j=1

(−1)i+jaij det(Aij).

• This is called expansion by the ith row.

• Example:

det

5 −6 30 4 02 −16 9

= 4 · det

(5 32 9

)= 156.

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Properties of the DeterminantProperties of the Determinant

• The formula for the determinant of a matrix A is the

sum of n! products of the entries of A (sometimes

×− 1.)

� Each summand is the product of n entries, one from

each row, and one from each column.

• The determinant of a triangular matrix is the product

of the diagonal terms.

� We can use row operations to compute

determinants.

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Row Operations and DeterminantsRow Operations and Determinants

If B is obtained from A by

• adding a multiple of one row to another, then

det(B) = det(A).

• interchanging two rows, then

det(B) = −det(A).

• multiplying a row by c �= 0, then

det(B) = c det(A).

Return Row operations

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ExampleExample

A =

−5 2 325 −9 −1210 7 17

det(A) = 50

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More PropertiesMore Properties

• If A has two equal rows , then det(A) = 0.

• If A has a row of all zeros , then det(A) = 0.

• det(AT ) = det(A).

• If A has two equal columns, then det(A) = 0.

• If A has a column of all zeros, then det(A) = 0.

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Column Operations and DeterminantsColumn Operations and Determinants

If B is obtained from A by

• adding a multiple of one column to another, then

det(B) = det(A).

• interchanging two columns, then

det(B) = −det(A).

• multiplying a column by c �= 0, then

det(B) = c det(A).

Return det(AT ) = det(A) Expansion by row

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Expansion by a ColumnExpansion by a Column

We can also expand by a column.

det(A) =n∑

i=1

(−1)i+jaij det(Aij).

• This is called expansion by the jth column.

Return Expansion by row Expansion by column

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ExampleExample

A =

−5 −6 03 4 0−8 −16 9

det(A) = 9 · det(−5 −6

3 4

)

= 9 · (−2)

= −18

.

Theorem

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Determinants and BasesDeterminants and Bases

Proposition: A collection of n vectors v1, v2, . . . ,vn

in Rn is a basis for Rn if and only if

det([v1 v2 . . . vn]) �= 0.

Expansion by row Expansion by column Row operations Column operations

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ExamplesExamples

det

1 1 1 12 1 −1 −2−2 −1 1 12 2 1 1

= 1.

det

3 −1 0 112 −6 0 532 −15 −3 1318 −10 −1 8

= −1.

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The Span of a Set of VectorsThe Span of a Set of Vectors

Definition: The span of a set of vectors is the set of all

linear combinations of those vectors. The span of the

vectors v1, v2, . . . , and vk is denoted by

span(v1,v2, . . . ,vk).

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Linear IndependenceLinear Independence

Definition: The vectors v1, v2, . . . , and vk are linearly

independent if the only linear combination of them which

is equal to the zero vector is the one with all of the

coefficients equal to 0.

• In symbols,

c1v1 + c2v2 + · · · + ckvk = 0

⇒ c1 = c2 = · · · = ck = 0.

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Example of a NullspaceExample of a Nullspace

A =

4 3 −1−3 −2 11 2 1

rref−→

1 0 −10 1 10 0 0

The nullspace of A is the set

null(A) = {av | a ∈ R} ,

where v = (1, −1, 1)T .

• The nullspace of A consists of all multiples of v.

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Another Example of a NullspaceAnother Example of a Nullspace

B =

4 3 −1 6−3 −2 1 −41 2 1 4

rref−→

1 0 −1 00 1 1 20 0 0 0

The nullspace of B is the set

null(B) = {av + bw | a, b ∈ R} ,

where v = (1, −1, 1, 0)T and w = (0, −2, 0, 1)T .

• null(B) consists of all linear combinations of v and w.