# [Linear Algebra] rank(AT A) = rank(A AT)

1. Dec 11, 2012

### macaholic

1. The problem statement, all variables and given/known data
Does the equation $\text{rank}(A^T A = \text{rank}(A A^T)$ hold for all nxm matrices A? Hint: the previous exercise is useful.

2. Relevant equations
$\text{ker}(A) = \text{ker}(A^T A)$
$\text{dim}(\text{ker}(A) + \text{rank}(A) = m$

3. The attempt at a solution
The previous exercise it referring to asked to show that $\text{rank}(A) = \text{rank}(A^T A)$ holds for all nxm matrices A.
Which I did by stating:
$\text{ker}(A) = \text{ker}(A^T A)$
and then taking the dimension of both sides, using the rank-nullity theorem to get:
$n-\text{rank}(A) = n - \text{rank}(A^T A)$ which makes it clearly true.

I tried using this result to prove the stated problem like so:
$\text{rank}(A) = \text{rank}(A^T A)$
$\text{rank}(A^T A) = \text{rank}((A^T A)^T A^T A)$
$= \text{rank}(A^T A A^T A)$
But then I get promptly stuck because I'm not sure what to do with the right side of that. Any advice?

2. Dec 11, 2012

### Dick

Didn't you also prove rank(A)=rank(A^T)? I.e. column rank equal row rank?

Last edited: Dec 11, 2012
3. Dec 11, 2012

### macaholic

Funny you should mention that... That was the problem BEFORE the previous problem, which I still haven't figured out quite yet.

In any case, I suppose I can use that fact. But I'm not quite sure how it applies since $(A^T A)^T = A^T A$
So I guess I have these equalities:
$\text{rank}(A)=\text{rank}(A^T) = \text{rank}(A^T A)$
I'm not sure how to rearrange this to get $\text{rank}(A^T A)$

4. Dec 11, 2012

### Dick

But you also have rank(AA^T)=rank(A^T), don't you?

5. Dec 11, 2012

### macaholic

I do? I guess that what I'm missing, I can't currently see how that arises from the equalities I'm given.

It would make it trivial from there though, since then I would just say that $\text{rank}(A A^T) = \text{rank}(A) = \text{rank}(A^T A)$

I keep wanting to "sub in" $A A^T$ to one of the equalities as a replacement for A, I assume this is the wrong approach?

6. Dec 11, 2012

### Dick

Your theorem tells you rank(B^TB)=rank(B) for ANY matrix B. Put B=A^T.

7. Dec 11, 2012

### macaholic

'doh, thanks a bunch! I should have been able to try that on my own.

While I'm here, would you mind helping with the other proof I'm stuck on?

It wanted me to use: $(\text{im} A)^\perp = \text{ker}(A^T)$ to prove $\text{rank}(A)=\text{rank}(A^T)$

im is just the column space, and ker is the null space in my textbook.

I've gotten this far using rank-nullity but I got stonewalled:

$\text{im } A = \text{ker}(A^T)^\perp$
$\text{rank } A = \text{dim}(\text{ker}(A^T)^\perp ) = m - \text{dim } (\text{ker }A)$

I feel like this is probably the wrong way to approach this problem but I can't think of another.

8. Dec 11, 2012

### Dick

Sorry, it's kind of late here and for some reason this question is making me see double. I'll give you one thing you haven't used yet and that's if V is a subspace of a vector space of dimension n then $dim(V^\perp)+dim(V)=n$. Hope that helps. I'll have another look in the morning, and if you've gotten it I'll be really happy. I think you can.

9. Dec 12, 2012

### Dick

Yeah, it's easy enough. $m-dim(ker(A))=dim(ker(A)^\perp)$. Got it yet?

I just edited the above. I had a parenthesis is a bad place.

Last edited: Dec 12, 2012