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EEE 424 - Digital Signal Processing
Fall 2023-
Homework 2
Assigned Date: 19 October 2023
Due Date: 27 October 2023
Remarks
- Show your reasoning and your work in your answers.
- Please scan your answers and upload them in a .pdf format.
Questions
Problem 1 [15 pts]
Consider the vector space equipped with the standard norm and inner product.
Prove that;
a) multiplication by an orthogonal matrix preserves lengths, i.e.,
for any.
b) multiplication by an orthogonal matrix preserves angles, i.e.,
for all ,.
c) Now consider the vector space equipped with the standard norm and inner
product. What would be the condition on the operator such that the two
conditions in (a) and (b) are satisfied?
Problem 2 [15 pts]
Prove that if and , then_._
Problem 3 [15 pts]
For some bounded linear operator with the adjoint , prove that the
following is true:
N
U
| | Ux | | = | | x | |
x ∈ ℝ
N
< Ux , Uy > = < x , y >
x ∈ ℝ
N
y ∈ ℝ
N
N
U
x ∈ l
p
(ℤ) p < q x ∈ l
q
(ℤ)
A : H
0
→ H
1
a) The operators and are self-adjoint.
b) If is invertible, the adjoint of the inverse and the inverse of the adjoint are
equal, i.e.,.
c) Let be a bounded linear operator. Then,.
Problem 4 [19 pts]
Consider the following set of vectors:
a) Write the matrix representation for the set, i.e., the synthesis operator associated
with the set.
b) Find the dual basis for the set. Sketch the original set and it's dual. Can you
comment on the relation between the original vectors and their duals?
c) Is the given set a basis? If it is, specify whether it is an orthonormal basis.
d) For , find the projection coefficients.
e) Using the same x, verify the expansion formula.
f) Verify the matrix version of the expansion formula.
g) Check whether.
h) Recall the example in the recitation notes, with the set of vectors
. Compare your answer for part (g) in the
homework and the recitation. Are they different? If they are, what property of the
set of vectors might be causing this difference? (Note that even though the signals
for are different for the two questions, it is still possible to do generalizations
for this part.)
You can find the recitation questions on Moodle.
A A * A * A
A
( A
− 1
)* = ( A *)
− 1
B : H
1
→ H
2
( BA )* = A * B *
S
0
= { ϕ 0,
, ϕ 0,
1
2
3
2
− 3
2
1
2
x =
[
]
α i , k
= < x ,
ϕ i , k
x =
k
α i , k
ϕ i , k
T
= I
| | (^) x | |
2
k
| (^) α i , k
2
S
1
= { ϕ 1,
, ϕ 1,
1
2
3
2
[
1 ]
x
First, take :
If ,. Then,
Hence, and. This implies that.
Assume this holds for.
Now take :
If ,. Then,
Since and. Hence, , and. This
implies that
Problem 3
a) For any in ,
Remember the adjoint relation for some operator with
adjoint. Then, is the adjoint of , i.e., is self-adjoint.
Similarly, for any in ,
Hence, is self-adjoint.
b) For any in ,
Then,. This shows that is the left inverse of. Applying
similarly,
Then,. This shows that is the right inverse of. Hence,
is the inverse of
c) For any in and in ,
Since the adjoint is unique, is the adjoint of.
p = 1
x ∈ l
1
(ℤ) | | x | |
1
| | x | |
2
2
i ∈ℤ
| x i
2
i ∈ℤ
| x i
2
= | | x | |
2
1
| | x | |
2
< ∞ x ∈ l
2
(ℤ) l
1
(ℤ) ⊂ l
2
(ℤ)
p = q
p = q + 1
x ∈ l
q
(ℤ) | | x | |
q
| | x | |
q + 1
q + 1
i ∈ℤ
| x i
q + 1
i ∈ℤ
| x i
q
| x i
i ∈ℤ
| x i
q
i ∈ℤ
| x i
| = | | x | |
q
q
| | x | |
1
x ∈ l
q
(ℤ) x ∈ l
1
(ℤ) | | x | |
q + 1
q + 1
< ∞ x ∈ l
q + 1
(ℤ)
l
q
(ℤ) ⊂ l
q + 1
(ℤ)
x , y H 1
< A A * x , y > = < A ( A * x ), y > = < A * x , A * y > = < x , A A * y >
< Bx , y > = < x , B * y > B
B * A A * A A * A A *
x , y H 0
< A * A x , y > = < A x , Ay > = < x , A * Ay >
A * A
x , y H 1
< x , y > = < A A
− 1
x , y > = < A
− 1
x , A * y > = < x , ( A
− 1
)* A * y >
( A
− 1
)* A * = I ( A
− 1
)* A *
< x , y > = < A
− 1
A x , y > = < A x , ( A
− 1
)* y > = < x , A *( A
− 1
)* y >
A *( A
− 1
)* = I ( A
− 1
)* A *
( A
− 1
)* A *
x H 0
y H 2
< BA x , y > = < A x , B * y > = < x , A * B * y >
A * B * BA
Problem 4
a).
b) To find the dual basis, need to solve. Solving the equation gives, and
taking the Hermitian transpose of gives,
c) The set is a basis for its span, as the two vectors are linearly independent. It is an
orthonormal basis, since the inner product.
d) For ,
e).
1
2
3
2
3
2
1
2
H
= I
H
1
2
− 3
2
3
2
1
2
< ϕ 1,
, ϕ 1,
x =
[
1 ]
α 1,
= < x ,
ϕ 1,
[
1 ]
1
2
3
2
α 1,
= < x ,
ϕ 1,
[
]
3
2
1
2
α =
3 + 3
2
1 − 3 3
2
x =
k
α i , k
ϕ i , k
1
2
3
2
− 3
2
1
2
[
]
= x
However, we assumed that is the solution to the problem ,
which requires that for any. Now, we have
such that , which raises a contradiction.
Then, for any. Since is orthogonal to any vector in
, it is orthogonal to :
b) Consider another vector in such that both and
, i.e., there are two vectors that minimize the given
distance, and. Consider. Since both and , any
linear combination of and are still in. Then, we can write,
Since and ,. Then,
Since norm is a positive definite function, , and
. Then,
Consequently,
Hence, is unique.
c) To prove :
If , since ,.
To prove :
| | y − ( y ̂ + cv ) | |
2
= | | y − y ̂| |
2
− 2 | c |
2
2
= | | y − y ̂ | |
2
− | c |
2
| | y − ( y ̂ + cv ) | |
2
= | | y − y ̂| |
2
− | c |
2
< | | y − y ̂ | |
2
y ̂ ∈ S 0
arg min
y ̂
| | y − y ̂ | |
| | y − y ̂ | | < | | y − l | | l ∈ S 0
y ̂+ cv ∈ S 0
| | y − ( y ̂ + cv ) | |
2
< | | y − y ̂| |
2
< y − y ̂, v > = 0 v ∈ S 0
y − y ̂
S
0
S
0
y − y ̂⊥ S 0
S
0
y = arg min
y ̂
| | y − y ̂ | |
v = arg min
y ̂
| | y − y ̂ | |
y ̂ ≠ v | | y − v | |
2
v ∈ S 0
y ∈ S 0
v y ̂ S 0
| | y − v | |
2
= | | y − y ̂ + y ̂− v | |
2
= < y − y ̂ + y ̂ − v , y − y ̂+ y ̂ − v >
= < y − y ̂, y − y ̂ > + < y − y ̂, y ̂ − v > + < y ̂− v , y − y ̂> + < y ̂ − v , y ̂ − v >
= | | y − y ̂ | |
2
- < y − y ̂, y ̂ − v > + < y ̂ − v , y − y ̂ > + | | y ̂− v | |
2
= | | y − y ̂ | |
2
- < y − y ̂, y ̂ − v > + < y − y ̂, y ̂ − v >* + | | y ̂ − v | |
2
= | | y − y ̂ | |
2
- 2 ℝ{ < y − y ̂, y ̂ − v > } + | | y ̂− v | |
2
y − y ̂ ⊥ S 0
y − v ∈ S 0
< y − y ̂, y ̂ − v > = 0
| | y − v | |
2
= | | y − y ̂ | |
2
2
| | y ̂ − v | |
2
| | y ̂ − v | |
2
= 0 ⟺ y ̂ = v
| | y − v | |
2
= | | y − y ̂ | |
2
2
≥ | | y − y ̂ | |
2
| | y − v | |
2
= | | y − y ̂ | |
2
⟺ v = y ̂
y ̂
y ̂ = y ⟹ y ∈ S 0
y = y ̂ y ̂ ∈ S 0
y ∈ S 0
y ̂ = y ⟸ y ∈ S 0
If , the problem to find with is solved by ,
since norm is a positive-definite function. As uniqueness was proved in (b), we can
say that implies that.
Hence,.
d) , and since implies that is orthogonal to any vector in ,
.
e) Similarly, since , constitutes a basis for ,. Since
, ,.
y ∈ S 0
arg min
y ̂
| | y − y ̂ | | y ̂ ∈ S 0
y = y ̂
y ∈ S 0
y = y
y ̂ = y ⟺ y ∈ S 0
y ̂ ∈ S 0
y − y ̂ ⊥ S 0
y − y ̂ S 0
y − y ̂ ⊥ y ̂
u i
i = 1,2,..., K S 0
u i
∈ S
0
y − y ̂ ⊥ S 0
y − y ̂ ⊥ u i
i = 1,2,..., K
