# $A^tA$ and $A^t$ have same rank

Let $$A$$ be an $$n \times n$$ real matrix. prove $$A^tA$$ and $$A^t$$ have same rank.

I know it has an answer here but I don't wish to look into a solution. can somebody give me a hint?

I have no idea about the geometric interpretation of transpose since it involves dual spaces. So in fact, I don't have any idea where to start

• One thing to think about is how this problem looks over $\mathbb{C}$, if it doesn't still hold, you know your proof must involve some specific properties about $\mathbb{R}$ that $\mathbb{C}$ doesn't have. May 3, 2019 at 14:46

## 1 Answer

First, $$A$$ and $$A^t$$ have the same rank, and it's a bit easier to work with $$A$$, so that's what I would do.

Second, if two matrices (linear transformations) have the same domain and same kernel, then they must have the same rank.

It's easy to show that $$\ker(A)\subseteq \ker(A^tA)$$.

As for the other direction, take an $$x\in \ker(A^tA)$$, and consider the length of $$Ax$$.

• This will prove that $A^tA$ and $A^t$ have the same range, not just the same rank. Is it true? May 12, 2019 at 11:17
• @VinayDeshpande Yes, $A^tA$ and $A^t$ do have the same range. May 12, 2019 at 11:26