# Explain how any square matrix A can be written as

1. Nov 29, 2012

### macaholic

1. The problem statement, all variables and given/known data
a) Explain how any square matrix A can be written as

$A = QS$
where Q is orthogonal and S is symmetric positive semidefinite.

b) Is it possible to write

$A = S_1 Q_1$
Where Q1 is orthogonal and S1 is symmetric positive definite?

2. Relevant equations

$A = U \Sigma V^T$

3. The attempt at a solution

For a) I've gotten to the point where I've written:

$A = U V^T V \Sigma V^T$

Which is just a rearrangement of the single value decomposition. From this I believe there is some logic as to why U V^T is orthogonal, and VΣV^T is symmetric positive definite, but I can't seem to figure out the reasoning. Any pointers?

For b) I've surmised this is possible given that it is simply the "left polar decomposition" (http://en.wikipedia.org/wiki/Polar_decomposition) But again, I can't think about how to show this mathematically.

2. Nov 29, 2012

### Dick

Try and show the product of two orthogonal transformations is orthogonal. What's the definition of orthogonal? Ditto for the second part. What's the definition of positive semidefinite? Definitions are a great place to start.

Last edited: Nov 29, 2012
3. Nov 29, 2012

### macaholic

Okay well can I start by saying U and V^T are both orthogonal? I think they are, right? If so, does this work?:
$(UV^T)^T=VU^T$
$(UV^T)((UV^T)^T) = UV^T VU^T = UIU^T=UU^T=I$

I'm not sure what to do about a good definition for positive semidefinite :/. The only one I know is that the eigenvalues are all 0 or positive. How can I use that on the second series of matrices?

4. Nov 29, 2012

### Dick

Yes, that shows your first factor is orthogonal. For the second one the definition I'm thinking of is that a matrix Q is positive semidefinite if x^TQx>=0 for all vectors x.

5. Nov 29, 2012

### macaholic

So I have to show that $x^T(V\Sigma V^T x) \geq 0$?
hmm... I'm not sure how to show this.
I know $\Sigma$ has all diagonal entries... and I know that $V, V^T$ are orthogonal. I can't seem to think of how that helps me though :/

6. Nov 29, 2012

### Dick

The singular value decomposition tells you can pick $\Sigma$ to be positive semidefinite, since it has all nonnegative entries along the diagonal. Try to use the definition I gave you.

7. Nov 30, 2012

### macaholic

Oh, right. So it would become a matter of showing that the $V\Sigma V^T$ yields a matrix that's also positive semidefinite?

8. Nov 30, 2012

### Dick

That would be correct.

9. Nov 30, 2012

### macaholic

Well I don't know how to show this, but I have a feeling that the product of an orthogonal matrix and a positive semidefinite matrix is positive semi definite... But alas I can't think of a way to reason this out. I'm no good at these matrix manipulation proofs it seems.

10. Nov 30, 2012

### Dick

You want to show x^T(VΣV^T)x≥0. Regroup that. I think you can do it. (V^T)x is just 'any vector'.

11. Nov 30, 2012

### macaholic

OH wait I think I may have something...
I know that $V\Sigma V^T = \Sigma$ (I can't figure how to show this though...)
So that makes my inequality $x^T(\Sigma x) \geq 0$
Was that at least one of the right steps? Though I can't justify the first part...

12. Nov 30, 2012

### Dick

No, they aren't equal. But if $x^T(\Sigma x) \geq 0$ is true for ANY x, then it must also be true if you replace x with $V^T x$, right?

13. Nov 30, 2012

### Ray Vickson

The _definition_ of positive semidefinite is that x^T A x >= 0 for any x in R^n. One *consequence* of that is: a symmetric matrix is positive semidefinite if and only if its eigenvalues are >= 0. This is a theorem, not a definition.