Solution to problem set 10Solution to problem set 10Solution to problem set 10Solution to problem set 10

�

×λ

⇒ λ

⇒ λ

⇒ λ λ

⇒ λ λ

⇒ λ λ⇒ ×

2*

2* *

2* * * *

2 2

#1 (20pts)(a)let A is a n n skew-Hermitian matrixAx= x

x Ax= x

(x Ax) = x

(x Ax) = -x Ax = -x ( x) = - x

x =- x

=-

the eigenvalues of any n n skew-Hermitian matrix are pure imaginary num

−

−

− − + − −

− −

− −

= =

× = ×− × = × −

∴ × = = ⇒ =

⇒ = =

⇒

−

= = =

⇒

∵

T

*

A T A A

A T A A A

A A ( A) A A A 1

A T A A 1

A

A * A A A 1

A

ber.

(b)1.

(e ) e e

(e ) e e e( A) (A) (A) ( A)

e e e I e (e )

(e ) e (e )

e is an orthogonal matrix

2.

if A is skew Hermitian

(e ) e e (e )

e is unitary matrix

#2 (15pts)

step1 : fi

λ = ⇒ = σ =

λ = ⇒ = σ =

T

1 1 1

2 2 2

nd the eigenvalues and eigenvectors of A A, and the singular value of A

00

2 v , 210

111

2 v , 2020

− λ = ⇒ = σ =

3 3 3

111

0 v , 0020

λ = ⇒ = σ =

4 4 4

00

0 v , 001

−

=

=

= = σ

= = = σ

= ⇒ −

∑

∑

1 11

2 22

2

step2 : Set up V and

1 10 0

2 21 1

0 0V2 2

1 0 0 00 0 0 1

2 0 0 020 0 0

0 0 0 0

stept3 : Build U

01 1

u Av 12

1

2 11 1

u Av 0 02

0 0

0let e 1 by Gram Schmidt orthogon

0

< > < > ⇒ = − − = < > < >

−

2 1 2 23 2 1 2

1 1 2 2

alization

0e ,u e ,u 1

u e u uu ,u u ,u 2

12

=

−

0 1 01 1

U 02 2

1 10

2 2

− = = =

−

= =

∑

T

T

1 2

1 10 0

0 1 0 2 220 0 0 0 01 11 11 1 0 02A 0 0 0 U V 0 0 0 012 22 2

0 0 0 0 0 0 01 1 0 0 01 10

0 0 0 12 20 1

1the orthonormal basis of C(A) {u ,u } { 1 , 0 }

21 0

the orthonor

= =

− = =

= = −

T1 2

43

T3

0 10 11

mal basis of C(A ) {v ,v } { , }1 020 0

1 01 01

the orthonormal basis of N(A) {v ,v } { , }0 020 1

01

the orthonormal basis of N(A ) {u } { 1 }2

1

− − − = ≠ ⇒ = = − − −

= ≠ ⇒ =

2

#3 (20pts)(a)True

(b)False

1 1 1 1 1 1 0 0counter example : let A 0 A

1 1 1 1 1 1 0 0

(c)True(d)False

1 1counter example : let A I eigenvalue of A 1,1

0 1

(f)True

pf : if all the singular values ⇒

= ⇒ = ⇒ =

=

of A are 0 , than rankA=0 A=0.

(g)False

0 1counter example : let A eigenvalue of A 0,0 number of nonzero eigenvalue 0

0 0

but rank(A) 1

(h)True

(i)True

−

⇒

1

( j)False

1 2 1 3counter example : let A= , B=

0 4 0 4

1 2 2 0 1 3 2 0= A and B are similar

0 4 0 3 0 4 0 3

singular values of A are 4.4954 and 0.8898

singular values of B are 5.0368

⇒ ≠and 0.7942

sigular values of A singular values of B

= σ σ σ σ = × × × =

= = σ + σ + σ + σ = + + + =

×

= = = =

−

T 2 2 2 2 2 2 2 241 2 3

T T 2 2 2 2 2 2 2 241 2 3

T

T T

# 4 (15pts)

(1)det(A A) 4 3 2 1 576

(2)trace(AA ) trace(A A) 4 3 2 1 30

(3)the size of AA is 5 5

rank(AA ) rank(A A) rank(A) # of nonzero singular value 4

by rank nullity theorem :

= =

=

≠

≠⊥

= − = − =

= = λ = λ = λ = σ =

⇒ = =

= λ = σ =

= σ = =

∵

T T

2 2T T T 2max max max maxx 1 x 1

x 1

T T2

x 0 min minT

22 2

W x 0 32x W

T

dim[N(AA )] 5 rank(AA ) 5 4 1

(4)max Ax max x A Ax (x x) x 16

max Ax 16 4

x A Ax(5)min 1

x xAx

(6)max min 2 4x

(7) A and A have the same nonzero sin

≠⊥

∴ = σ =T

W x 0 2x W

gular value

A xmin max 3

x

σ = = σ

× ⇒ × ⇒ =

σ ∴ = = = σ

∑ ∑

∑ ∑ ∑

∑∑ ∑ ∑ ∑∑

⋯

⋱ ⋯

⋮ ⋱ ⋮

⋮ ⋱ ⋱

⋯

∵

⋯

⋱ ⋯

⋮ ⋱ ⋮

⋮ ⋱ ⋱

⋯

1

Tr

T

21

T T 2r

#5 (15pts)(a)

0 0 00 0 0

let A U V , 00 00 0 0 0

the size of A is n n the size of is n n

0 0 00 0 0

00 00 0 0 0

= = =∑ ∑ ∑ ∑ ∑ ∑TT T T T T T T TA A (U V ) (U V ) (V U )(U V ) V V

− −

σ σ ⇒ = ⇒ = =σ σ

= = =

σ ⇒ = σ

∑ ∑

∑ ∑ ∑ ∑ ∑∑

⋯ ⋯

⋱ ⋯ ⋱ ⋯

⋮ ⋱ ⋮ ⋮ ⋱ ⋮

⋮ ⋱ ⋱ ⋮ ⋱ ⋱

⋯ ⋯

⋯

⋱ ⋯

⋮ ⋱ ⋮

⋮ ⋱ ⋱

⋯

2 21 1

T T T 1 T T 12 2r r

TT T T T T T T T

21

T 2r

0 0 0 0 0 00 0 0 0 0 0

A A V V V A A(V )0 00 0 0 00 0 0 0 0 0 0 0

AA (U V )(U V ) (U V )(V U ) U U

0 0 00 0 0

AA U 00 00 0 0 0

− −

− − − −

− − − − − − −

σ

⇒ = =σ

⇒ = = =

⇒ = = =

⇒

= = = ⇒ =

∑∑

∑∑ ∑ ∑

∑ ∑ ∑

⋯

⋱ ⋯

⋮ ⋱ ⋮

⋮ ⋱ ⋱

⋯

21

T T 1 T T 12r

1 T T 1 T T 1 T T 1

T 1 T T 1 T 1 T 1 T 1 T 1 1

T T

2 T T T 2 T

0 0 00 0 0

U U AA (U )00 00 0 0 0

U AA (U ) V A A(V )

AA UV A A(V ) U UV A A(UV ) UV A A(UV )

A A is similar to AA

(b)

B A A V V V V B V V

= = = ⇒ =∑

∑∑ ∑ ∑

T

2 T T T 2 T TC AA U U U U C U U

−

−

−

−

−

−

=

⇒ =

⇒ =

⇒ =

⇒ =

⇒ =

∑ ∑ ∑

∑ ∑ ∑

∑ ∑ ∑

∑ ∑ ∑

∑ ∑ ∑

⌢

⌢

⌢

⌢

⌢

⌢

T 1 T

T T T 1 T T

T T T 1 T T

T T 1 T T

T 1 T T T

T 1 T T

#6 (15pts)

x (A A) A b

x [(U V ) (U V )] (U V ) b

x [V U U V ] (V U )b

x [V V ] (V U )b

x [V( ) V ](V U )b

x V( ) U b

−

× ×

σ σ σ

σ σ

⇒ =

σ

⋯ ⋯ ⋯

⋯ ⋯ ⋯

⌢ ⋱ ⋮ ⋱ ⋮ ⋱ ⋮⋯

⋮ ⋱ ⋮ ⋱ ⋮ ⋱

⋮ ⋱ ⋱ ⋱ ⋮ ⋱ ⋱ ⋱ ⋮ ⋱ ⋱ ⋱

⋯ ⋯

⋱ ⋱ ⋱

⋯

1n1

n m

1 1 1

r r r

m n

0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0| |0 0 0 0 0 0 0 0 0x v v ( )

0 0 0 0 0 0 0 0 0 0 0 0| |0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ×

− − − − − −

⋮

1

2

m

n m

uu

b

u

−

××

=

− − − − ⇒ =

σ

− −

=

σ

σ

σ

σ

∑

⋯⋯

⋯⋯

⌢ ⋱ ⋮⋱ ⋮⋯

⋮⋮ ⋱⋮ ⋱

⋮ ⋱ ⋱ ⋱⋮ ⋱ ⋱ ⋱

⋯⋯

⋱⋱

21

1

1

2n1

m

n mn n

Tri

ii 1

r

i

1

2r

0 0 0 00 0 0 0u0 0 0 00 0 0 0| |u0 0 00 0 0x v v b

0 0 0 00 0 0 0| |u0 00 0

0 0 0 0 00 0 0 0 0u b

v