Homework1 of multivariate data analysis
7
Math 537 Homework1 2.5 Check that Q= [ 5 13 12 13 −12 13 5 13 ] Is an orthogonal matrix. Solution: Since Q= [ 5 13 12 13 −12 13 5 13 ] ∴Q T =[ 5 13 −12 13 12 13 5 13 ] Q T Q= [ 5 13 −12 13 12 13 5 13 ] × [ 5 13 12 13 −12 13 5 13 ] = [ 5 13 × 5 13 + ( −12 13 ) × ( −12 13 ) 5 13 × 12 13 + ( −12 13 ) × 5 13 12 13 × 5 13 + 5 13 × ( −12 13 ) 12 13 × 12 13 + 5 13 × 5 13 ] = [ 1 0 0 1 ] Thus, Q Is an orthogonal matrix. 2.6 Let A = [ 9 −2 −2 6 ] (a) Is A symmetric? (b) Show that A is positive definite. Solution: A is symmetric because numbers are symmetric of the diagonal line. Since |A| =9 × 6−(−2 ) × (−2 )=58 > 0 A is positive definite.
description
Homework1 of multivariate data analysis
Transcript of Homework1 of multivariate data analysis
Math 537 Homework1
2.5 Check that
Q=[ 513 1213
−1213
513
]Is an orthogonal matrix.
Solution:
Since Q=[ 513 1213
−1213
513
]∴QT=[
513
−1213
1213
513
]
QTQ=[ 513 −1213
1213
513
]×[ 513
1213
−1213
513
]=[ 513× 513+(−1213 )×(−1213 ) 513×1213
+(−1213 )× 5131213×513
+513×(−1213 ) 12
13×1213
+513×513
]=[1 00 1]
Thus, Q Is an orthogonal matrix.
2.6 Let
A=[ 9 −2−2 6 ]
(a) Is A symmetric?(b) Show that A is positive definite.Solution:A is symmetric because numbers are symmetric of the diagonal line.Since |A|=9×6−(−2 )× (−2 )=58>0A is positive definite.
2.8 Given the matrix
A=[1 22 −2
]
Find the eigenvalues λ1∧λ2 and the associated normalized eigenvectors e1∧e2. Determine the spectral decomposition (2-16) of A.
Solution:Since |A−λI|=0
|1−λ 22 −2−λ|=0
∴ (1−λ ) (−2− λ )−4=0So λ1=2 , λ2=−3For λ1=2
(1 22 −2)(x1x2)=2(
x1x2)
∴ e1=(−0.8944−0.4472
)
For λ2=−3
(1 22 −2)(x1x2)=−3 (x1x2)∴ e1=(−0.4472
0.8944)
the spectral decomposition of A is
A=2(−0.8944−0.4472)(−0.8944−0.4472)'
−3 (−0.44720.8944
)(−0.44720.8944
)'
.
2.9 Let A be as in Exercise 2.8.
(a) Find A−1 .Solution:
A−1=(0.3333 0.33330.3333 −0.1667
)
(b) Compute the eigenvalues and eigenvectors of A−1 .Solution:Since |A−1−λI|=0
|0.3333−λ 0.33330.3333 −0.1667−λ|=0∴ (1−λ ) (−2− λ )−4=0
So λ1=0.5 , λ2=−0.3333For λ1=0.5
(0.3333 0.33330.3333 −0.1667)(x1x2)=0.5 (
x1x2)
∴ e1=(−0.8944−0.4472
)
For λ2=−0.3333
(0.3333 0.33330.3333 −0.1667)(x1x2)=−0.3333 (x1x2)
∴ e1=( 0.4472−0.8944
)
(c) Write the spectral decomposition of A−1 , and compare it with that of A from Exercise 2.8.Solution: the spectral decomposition of A−1 is
A−1=0.5(−0.8944−0.4472)(−0.8944−0.4472)'
−0.3333( 0.4472−0.8944
)(0.44720.8944
)'
Compare to Exercise 2.8, two pairs of eigenvalues are each other’s reciprocals, one of the eigenvectors of A−1 has opposite sign with the eigenvectors of A.
2.10 Consider the matrices
A=[ 4 4.0014.001 4.002
] and B¿ [ 4 4.0014.001 4.002001
]
These matrices are identical except for a small difference in the (2,2) position. Moreover, the columns of A (and B) are nearly linearly dependent. Show that
A−1≐(−3)B−1. Consequently,
small changes---perhaps caused by rounding---can give substantially different inverses.
Solution:
Since A=[ 4 4.0014.001 4.002
] and B¿ [ 4 4.0014.001 4.002001
]
Then A−1=[−4002000 40010004001000 −4000000]
B−1=[ 1334000 −1333667−1333667 1333333
]
Thus, A−1≐ (−3 ) B−1 ,
which means small changes can give substantially different inverses.
2.11 Show that the determinant of the p× p diagonal matrix A={a ij} with a ij=0 ,i ≠ j , is given by the product of the diagonal elements; thus, |A|=a11a22…app
.
Hint: By Definition 2A.24, |A|=a11A11+0+…+0. Repeat for the submatrix A11
obtained by
deleting the first row and first column of A.
Proof:
Since |A|=∑j=1
k
a1 j∨A1 j∨(−1)1+ j if k>1,
∴|A|=a11A11+0+…+0 because a ij=0 ,i ≠ j ,
Then |A|=a11A11=a11a22 A22, which
A22 is the matrix deleting the first and second rows and
columns of A. If we keep doing this, we get the final matrix as a scaler a pp .Thus |A|=a11A11=a11a22 A22=…=a11a22…app
.
2.12 Show that the determinant of a square symmetric p× p matrix A can be expressed as the
product of its eigenvalues λ1 , λ2 ,…, λp; that is, |A|=∏
i=1
p
λi.
Hint: From (2-16) and (2-20), A=PΛP' with P' P=I . From Result 2A.11(e),
|A|=|PΛ P'|=|P||ΛP '|=|P||Λ||P'|=|Λ|∨I∨¿, since |I|=|P' P|=|P'|∨P∨¿
. Apply
Exercise 2.11.
Proof:Since A is symmetric matrix, we can write A=PΛP' with P' P=I .And |A|=|PΛ P'|=|P||ΛP '|=|P||Λ||P'|=|Λ|∨I∨¿
, since |I|=|P' P|=|P'|∨P∨¿.
Apply Exercise 2.11, if A is a diagonal matrix, |A|=a11a22…app,
Thus |A|=|Λ||I|=∏i=1
p
λi∏i=1
p
1=∏i=1
p
λ i.
2.13 Show that |Q|=+1∨−1 if Q is a p× p orthogonal matrix.
Hint: |QQ'|=¿ I∨¿. Also, from Result 2A.11, |Q||Q'|=¿Q∨¿2 .¿
Thus, ¿Q∨¿2=¿ I∨¿¿
.
Now use Exercise 2.11.
Proof:Since Q is a p× p orthogonal matrix.
So |QQ'|=|I|,be
And because |Q|=|Q'| according to Result 2A.11 (a),
Then |Q||Q'|=¿Q∨¿2=|I|=1×1…×1=1 ,¿ which has 1 multiply 1 p times.
Thus, |Q|=+1∨−1.
2.14 Show that Q' AQ and A have the same eigenvalues if Q is orthogonal.Hint: Let λ be an eigenvalue of A. Then 0=¿ A− λI∨¿. By Exercise 2.13 and Result 2A.11(e), we
can write 0=|Q'||A− λ I||Q|=|Q' AQ− λ I|, since
Q'Q=I.
Proof:Since let λ be an eigenvalue of A.
∴|A−λI|=0According to Exercise 2.13 and Result 2A.11(e), and Q is orthogonal, Q'Q=I ,ijwe have
|Q' AQ− λI|=|Q'||A−λI||Q|=0
Thus, Q' AQ and A have the same eigenvalues if Q is orthogonal.
2.15 A quadratic from x ' A x is said to be positive definite if the matrix A is positive definite. Is
the quadratic from 3 x1
2+3 x22−2 x1 x2
positive definite?
Solution:
Since a quadratic from x ' A x=∑
i=1
p
∑j=1
p
x iaij x j
So 3 x12+3 x2
2−2 x1 x2=(x1 x2 ) A (x1x2)Let A=(a b
c d)∴ax1
2+(b+c ) x1 x2+d x22=3 x1
2+3 x22−2 x1 x2
