Finding Singular values of general projection matrices....

In summary: Thanks!The matrix P is hermitese, meaning P=P*.As a consequence we have U=V.Note that ##P^*=(qq^*)^* = q^{**}q^* = qq^*=P##, so we also have that ##(UEV^*)^* = VE^*U^* = UEV^*##.Oh, I see. I forgot about that...So, now, I think I figured out part a, thanks to you, and it seems to work on paper. And I THINK I have figured out part b, setting it up along the lines of:And then solving
  • #1
Mattbringssoda
16
1

Homework Statement


Let q ∈ C^m have 2-norm of q =1.

Then P = qq∗ is a projection matrix.

(a) The matrix P has a singular value decomposition with U = [q|Q⊥] for some appropriate matrix Q⊥.

What are the singular values of P?

(b) Find an SVD of the projection matrix I − P = I − qq∗ . In particular, what are the singular values? Hint: Write I = UU∗ where U is as above and use the SVD of qq∗ .

Homework Equations

The Attempt at a Solution


[/B]
I'm afraid I'm at a loss for what I should aim for as far as an answer. Here's what I've been working on...

a)

mtnwuv.png


with the above matrix coming from the equation of U in the instructions.

So, I can answer that the singular values are the diagonals of Σ, which I now have an equation for...however it feels like I'm supposed to take this a step further...would anyone have any insight??

b)

mtnwuv.png


And again I can't help but feel this is too general or that I'm missing something.

Thanks for any help!
 
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  • #2
Mattbringssoda said:

Homework Statement


Let q ∈ C^m have 2-norm of q =1.

Then P = qq∗ is a projection matrix.

(a) The matrix P has a singular value decomposition with U = [q|Q⊥] for some appropriate matrix Q⊥.

What are the singular values of P?

(b) Find an SVD of the projection matrix I − P = I − qq∗ . In particular, what are the singular values? Hint: Write I = UU∗ where U is as above and use the SVD of qq∗ .

Homework Equations

The Attempt at a Solution


[/B]
I'm afraid I'm at a loss for what I should aim for as far as an answer. Here's what I've been working on...

a)

mtnwuv.png


with the above matrix coming from the equation of U in the instructions.

So, I can answer that the singular values are the diagonals of Σ, which I now have an equation for...however it feels like I'm supposed to take this a step further...would anyone have any insight??

b)

mtnwuv.png


And again I can't help but feel this is too general or that I'm missing something.

Thanks for any help!

Hi Mattbringssoda! :oldsmile:

(a)
Note that ##(qq^*)q=q(q^*q)=q\|q\|^2=q##.

Let ##q_\perp## be a vector perpendicular to ##q##, so ##\langle q_\perp, q \rangle = 0##.
Then ##(qq^*)q_\perp = q(q^*q_\perp) = q \langle q_\perp, q \rangle = q \cdot 0 = 0##.
So the singular values of ##P## are the column vectors of ##Q_\perp##.

Moreover, the singular value decomposition is:
$$P=UE_{11}U^*$$
where ##E_{11}## is the standard unity matrix with only zeroes and a single 1 at the top left.

(b)
We can write:
$$I-P=UU^* - UE_{11}U^*$$
Where can we go with this?
 
  • #3
I like Serena said:
Hi Mattbringssoda! :oldsmile:

(a)
Note that ##(qq^*)q=q(q^*q)=q\|q\|^2=q##.

Let ##q_\perp## be a vector perpendicular to ##q##, so ##\langle q_\perp, q \rangle = 0##.
Then ##(qq^*)q_\perp = q(q^*q_\perp) = q \langle q_\perp, q \rangle = q \cdot 0 = 0##.
So the singular values of ##P## are the column vectors of ##Q_\perp##.

Moreover, the singular value decomposition is:
$$P=UE_{11}U^*$$
where ##E_{11}## is the standard unity matrix with only zeroes and a single 1 at the top left.

(b)
We can write:
$$I-P=UU^* - UE_{11}U^*$$
Where can we go with this?

Thanks!

I think I'm starting to see a glimmer...

But, when you say P=UE11U∗, I'm not sure how the typical UEV* became UEU*,

in other words, why does V* = U*??

Really - thanks again!
 
  • #4
Mattbringssoda said:
Thanks!

I think I'm starting to see a glimmer...

But, when you say P=UE11U∗, I'm not sure how the typical UEV* became UEU*,

in other words, why does V* = U*??

Really - thanks again!

The matrix P is hermitese, meaning P=P*.
As a consequence we have U=V.
Note that ##P^*=(qq^*)^* = q^{**}q^* = qq^*=P##, so we also have that ##(UEV^*)^* = VE^*U^* = UEV^*##.
 
  • #5
Oh, I see. I forgot about that...

So, now, I think I figured out part a, thanks to you, and it seems to work on paper. And I THINK I have figured out part b, setting it up along the lines of:
7bb39a.png


And then solving for the Σ_orthog and using the orthogonal U and U* from the left side to work towards a final answer...hopefully I set that up correctly.

The next portion of the question is to get the SVD of I-2P. We just worked on I-P above, and it's orthogonality to P helped me get to the answer, but I'm not sure how to set up I-2P.

I know it's a reflection across the null, so that probably adds some "negative" values somewhere, but I'm not sure the proper way to show that in this symbolic representation problem that we're doing here...

Am I missing something obvious again, or over thinking it?
 
  • #6
Mattbringssoda said:
Oh, I see. I forgot about that...

So, now, I think I figured out part a, thanks to you, and it seems to work on paper. And I THINK I have figured out part b, setting it up along the lines of:
7bb39a.png


And then solving for the Σ_orthog and using the orthogonal U and U* from the left side to work towards a final answer...hopefully I set that up correctly.

The next portion of the question is to get the SVD of I-2P. We just worked on I-P above, and it's orthogonality to P helped me get to the answer, but I'm not sure how to set up I-2P.

I know it's a reflection across the null, so that probably adds some "negative" values somewhere, but I'm not sure the proper way to show that in this symbolic representation problem that we're doing here...

Am I missing something obvious again, or over thinking it?

Yep. Overlooking something.
We can use the property of distributivity to simplify - it applies to matrices as well.
That is, take the matrices out of the parentheses, so to speak.

##I-P = UU^* - UE_{11} U^* = U(U^* - E_{11} U^*) = U(I - E_{11})U^* = U\Sigma' U^*##
 
  • #7
I'm turning in my assignment now. You've been an incredible help. Thanks!
 
  • Like
Likes I like Serena

1. What is a general projection matrix?

A general projection matrix is a square matrix that maps a vector onto a subspace. It is often used in linear algebra to project a vector onto a lower-dimensional space.

2. What are singular values in a projection matrix?

Singular values in a projection matrix are the eigenvalues of the matrix. They represent the scaling factors by which the projection matrix transforms the input vector.

3. How do you find the singular values of a general projection matrix?

The singular values of a general projection matrix can be found by first computing the eigenvalues of the matrix. Then, the square root of the eigenvalues will give the singular values.

4. What is the significance of singular values in a general projection matrix?

The singular values in a general projection matrix provide important information about the transformation of a vector onto a subspace. They can also be used to determine the rank and invertibility of the projection matrix.

5. Can a general projection matrix have zero singular values?

Yes, a general projection matrix can have zero singular values. This means that the projection matrix is not invertible and has a rank of less than the dimension of the input vector space.

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