# Finding Singular values of general projection matrices...

1. Nov 4, 2016

### Mattbringssoda

1. The problem statement, all variables and given/known data
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∗ .

2. Relevant equations

3. The attempt at a solution

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)

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)

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

Thanks for any help!!

2. Nov 5, 2016

### I like Serena

Hi Mattbringssoda!

(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. Nov 5, 2016

### Mattbringssoda

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. Nov 5, 2016

### I like Serena

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. Nov 6, 2016

### Mattbringssoda

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 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. Nov 6, 2016

### I like Serena

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. Nov 7, 2016

### Mattbringssoda

I'm turning in my assignment now. You've been an incredible help. Thanks!