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

[macaholic]

Homework Statement

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.

Homework Equations

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

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?

 

[Dick]

Homework Statement

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.

Homework Equations

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

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?

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

 

[macaholic]

Didn't you also prove rank(A)=rank(A^T)?

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 $\text{rank}(A^T A)$

 

[Dick]

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 $\text{rank}(A^T A)$

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

 

[macaholic]

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

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?

 

[Dick]

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?

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

 

[macaholic]

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

'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.

 

[Dick]

'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.

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.

 

[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.