# Linearly independence question

1. Nov 3, 2013

### Karnage1993

Say I have a matrix $A$ that has linearly independent columns. Then clearly $A^T$ has lin. indep. rows. So what can we say about $A^TA$? Specifically, is there anything we can say about the rows/columns of $A^TA$? I'm thinking there has to be some sort of relation but I don't know what that is (if there is indeed any).

2. Nov 3, 2013

### R136a1

We have $$\textrm{rank}(A) = \textrm{rank}(A^T A)$$ So the number of linear independent columns/rows of $A$ is the same as the number of linear independent columns/rows of $A^T A$.

3. Nov 3, 2013

### Karnage1993

Isn't $\textrm{rank}(A) = \textrm{rank}(A^TA)$ only true if $A$ is symmetric? Also, I forgot to include that $A$ is not necessarily a square matrix. Let's have $A$ be an $n$ x $k$ matrix. Does your conclusion still follow with these new conditions?

4. Nov 3, 2013

### R136a1

Yes, it is true in general. Indeed, by rank-nullity is suffices to show that the nullity of $A$ equals the nullity of $A^T A$.

But take $Ax = 0$, then obviously $A^T A x = 0$.
Conversely, if $A^T A x = 0$, then $x^T A^T A x$. But then $|Ax| = 0$. Thus $Ax= 0$.

So the nullspace of $A$ equals the nullspace of $A^T A$.

5. Nov 3, 2013

### Karnage1993

I was under the impression that $\textrm{rank}(A) = \textrm{rank}(A^TA)$ is only true if $A$ is symmetric, but it appears you are right, and Wikipedia confirms it. It is indeed true in general for any $A$, so I guess I misread it somewhere. Thanks for the help!