Math 104
Fall 2009
Homework 4: solution set
We use the notations below to ease readability. Matrices are bold capital, vectors are bold lowercase and
scalars or entries are not bold. For instance,
A
is a matrix and
a
ij
its (
i, j
)th entry. Likewise
x
is a vector
and
x
j
its
j
th component. The linear span generated by a group of vecotors
v
1
,
v
2
, . . . ,
v
n
is denoted by
span(
v
1
,
v
2
, ...,
v
n
).
Problem 1
It is true that
k
E
k
F
=
k
u
k
F
k
v
k
F
. Set
u
=
u
1
.
.
.
u
m
and
v
=
v
1
.
.
.
v
n
. Then
E
=
uv
*
=
u
1
¯
v
1
· · ·
u
1
¯
v
n
.
.
.
.
.
.
.
.
.
u
m
¯
v
1
· · ·
u
m
¯
v
n
.
By the definition of Frobenius norm, we have
k
E
k
2
F
=
X
1
≤
i
≤
m,
1
≤
j
≤
n

u
i

2

v
j

2
=
X
1
≤
i
≤
m

u
i

2
X
1
≤
j
≤
n

v
j

2
=
k
u
k
2
F
k
v
k
2
F
.
It follows that
k
E
k
F
=
k
u
k
F
k
v
k
F
.
Problem 2
Recall that
k
A
k
2
= sup
x
∈
C
n
k
x
k
2
=1
k
Ax
k
2
. Since
U
is unitary, we have
k
Uy
k
2
=
k
y
k
2
for any
y
∈
C
m
. Thus
k
UA
k
2
=
sup
x
∈
C
n
:
k
x
k
2
=1
k
UAx
k
2
=
sup
x
∈
C
n
:
k
x
k
2
=1
k
Ax
k
2
=
k
A
k
2
.
As for the Frobenius norm, We will use the convenient form
k
A
k
F
=
p
tr(
A
*
A
). Since
U
*
U
=
I
, we have
k
UA
k
F
=
p
tr((
UA
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 Fall '09
 Linear Algebra, Vectors, Matrices, Scalar, ax, Singular value decomposition, Orthogonal matrix, A∗, Frobenius norm

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