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CS/ENGS 106 April 18, 2004 

1

Homework 4 

Problem 6.3 

Given nmC  A with nm , show that  A A is nonsingular if and only if A has full

rank. 

If  A A is nonsingular, its rank will be n and it has n nonzero eigenvalues. Then 

from Theorem 5.4, A has n nonzero singular values. So A has full rank. 

Inversely if A has full rank, the number of nonzero singular values is n. Then

 A A also has n nonzero eigenvalues and does not have zero as eigenvalue. So  A A

is nonsingular. 

Problem 6.4 

01

10

21

01

10

01

 B A 

a) The orthogonal projector P onto range(A) is: 

5.005.0

010

5.005.0

010

101)

10

02(

01

10

01

010

101)

01

1001

010

101(

01

1001

)(

1

11 A A A AP 

The image under P of the vector (1, 2, 3)* is:

2

2

2

3

2

1

5.005.0

010

5.005.0

Pv y 

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CS/ENGS 106 April 18, 2004 

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b) The orthogonal projector P onto range(B) is: 

6 / 53 / 16 / 1

3 / 13 / 13 / 1

6 / 13 / 16 / 5

012

101)

52

22(

01

10

21

012

101)

01

10

21

012

101(

01

10

21

)(

1

11 B B B BP 

The image under P of the vector (1, 2, 3)* is:

2

0

2

3

2

1

6 / 53 / 16 / 1

3 / 13 / 13 / 1

6 / 13 / 16 / 5

Pv y 

Problem 6.5 

mmC P is a nonzero projector.

First: if  12P , P will be an orthogonal projector.

Suppose 1,)(,)(2

vwithP NullvP Rangeu . Make an orthogonal

projection of u onto v, and then decompose u. We have: uvvur uvvr u , .

So 2

2

2

2

2

2  uvr u ,

uPuuvvuP )(Pr .

So that  1Pr

2

2

2

2

2

2

2

2

2

2

2

2

2

2

r 

uvr 

r 

u

r , since 1

2P .

Thus 02

2uv , this means that P is orthogonal.

Then if P is an orthogonal projector, 12

P .

Suppose vuwP NullvP Rangeu ,)(,)( . Then

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CS/ENGS 106 April 18, 2004 

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1)(

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

vu

u

uvvuvu

Pu

vu

vuP

w

Pw 

On the other hand,   wPPww .

Thus we can get 12

P .

Problem 7.1 

a) Reduced QR factorization for A: 

.10

02

02

210

02

2

0110

01

;

0

1

0

,1

,0

;10

1

2

2

,2

2

22

112222112222

2

*

112

11

112111

 A

ar 

qr aqqr ar 

aqr 

r 

a

qar  

Full QR factorization for A: (Assume a3=[0 0 1]*)

.10

02

2

20

2

2

0102

20

2

2

01

10

01

;

2

2

02

2

,2

2

,0,2

2

33

223113332223113333

3

*

2233

*

113

 A

r 

qr qr aqqr qr ar 

aqr aqr  

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CS/ENGS 106 April 18, 2004 

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b) Reduced QR factorization for B: 

.30

22

3

3

2

2

3

30

3

3

2

2

01

10

21

;

1

1

1

3

3,3

,2

;

1

0

1

2

2,2

22

112222112222

2

*

112

11

112111

 B

r 

qr aqqr ar 

aqr 

r 

aqar  

Full QR factorization for A: (Assume b3=[0 0 1]*)

.30

22

6

6

3

3

2

2

3

6

3

30

3

6

3

3

2

2

01

10

21

;

6

6

3

6

6

6

,6

6

,3

3,

2

2

33

223113332223113333

3

*

2233

*

113

 B

r 

qr qr bqqr qr br 

bqr bqr  

Problem 7.3 

Give an algebraic proof of Hadamard s inequality:m

 j

 ja A1

2det .

Make a QR factorization of A. We get: 

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