SPRING 2011
Correspondence # 8
March 3, 2011
Solution Set 3
21.
Let us check the four axioms:
(i)
(
B,A
) =
Tr
[
B
T
Q
¯
A
] =
Tr
[
¯
B
T
¯
QA
] =
Tr
[
A
T
¯
Q
T
¯
B
] = (
A,B
), where the third equality holds
because
Tr
[
C
] =
Tr
[
C
T
], and the last one holds because
Q
is Hermitian.
(ii) (
A
+
B,C
) =
Tr
[
A
T
Q
C
] +
Tr
[
B
T
Q
C
] = (
A,C
) + (
B,C
) – holds for all
A
,
B
and
C
.
(iii) (
λA,B
) =
λTr
[
A
T
Q
¯
B
] =
λ
(
A,B
) – holds for all
A
and
B
.
(iv) (
A,A
) =
Tr
[
A
T
Q
¯
A
]
→
this is positive for all
A
̸
= 0 because the matrix
Q
is Hermitian
and positive deﬁnite, i.e., all its eigenvalues are positive.
Hence, (
A,B
) =
Tr
[
A
T
QB
] is indeed an inner product.
22. (a)
It is
not
an inner product on
X
, because
(
x,x
) =
±
±
±
±
∫
4
1
s
2
x
(
s
)
ds
±
±
±
±
2
can be made zero without
x
being the zero function.
(b)
Yes
,
it is an inner product on
X
, as all four axioms of an inner product are satisﬁed:
(i) (
x,y
) = (
y,x
) for all
x,y
∈
X
.
(ii) (
x
+
y,z
) = (
x,z
) + (
x,z
) for all
x,y,z
∈
X
.
(iii) (
λx,y
) =
λ
(
x,y
) for all real numbers
λ
and all
x,y
∈
X
.
(iv) (
x,x
) =
∫
4
1
t
3
x
2
(
t
)
dt
≥
0 and is equal to zero iﬀ
x
(
t
) = 0 for all
t
∈
[1
,
4], which is the
zero element in
X
.
23. (a)
Here the Hilbert space
H
=
L
2
[
−
1
,
2] and
M
=
{
m
∈
H
:
m
(
t
) =
a
+
bt, a,b
∈
IR
}
. Note
that
M
is a 2-dimensional subspace of
H
, which is also closed (it is in fact isomorphic to IR
2
).
Hence the problem can be viewed as the optimization problem of minimizing
∥
x
−
m
∥
over
m
∈
M
, where
x
(
t
) =
t
3
is an element of
H
. The Projection Theorem directly applies here,
leading to the conclusion that there exists a unique
m
o
∈
M
that solves this minimization
problem, and that
x
−
m
o
⊥
M
.
(b)
We have to solve for
a
and
b
from the two relationships:
x
−
m
o
⊥
1
⇒
(
t
3
−
m
o
,
1) = 0
and
x
−
m
o
⊥
t
⇒
(
t
3
−
m
o
,t
) = 0
Using
m
o
(
t
) =
a
+
bt
, we have
5
4
=
a
+
1
2
b,
21
5
=
1
2
a
+
b
⇒
a
= 0
.
2
, b
= 2
.
1
(c)
The minimum value of
F
is
∥
x
−
m
o
∥
2
= (
x
−
m
o
,x
−
m
o
) = (
x
−
m
o
,x
) =
∥
x
∥
2
−
(
m
o
,x
) =
∥
x
∥
2
− ∥
m
o
∥
2
where