# Eigenvectors of $A^TA$ and $AA^T$

Let $$A$$ be a real $$m\times n$$ matrix, $$m\leq n$$. Now, what can we say about the eigenvectors of $$A^TA$$ and $$AA^T$$?
First of all, the rank of $$A\leq \min(m,n)=m$$.
Now, $$R=A^TA$$ is $$n\times n$$ matrix and $$L=AA^T$$ is $$m\times m$$ matrix.
What can we say about their eigenvectors and eigenvalues? (Like, how many eigenvectors corresponding to $$0$$ eigenvalue)

I know that the eigenvector of L corresponding to a non-zero eigenvalue is an eigenvector of R with the same eigenvalue and vice-versa. So if one has exactly $$x$$ eigenvectors corresponding to non-zero eigenvalues, then the other one also has exactly $$x$$ eigenvectors corresponding to those non-zero eigenvalues. But what about the rest (if any)?
Note: Sorry for the confusion, I meant equivalence only. For every eigenvector (non-zero eigenval.) of $$L/R$$, there is a corresponding eigenvector for the other.
Please give the reasoning behind the claims too, thank you!

• No, the eigenvectors of $L$ for nonzero eigenvalues are not eigenvectors of $R$: they don't even have the same dimension if $m \ne n$. Nov 5 '20 at 18:19
• Echoing @RobertIsrael. They're directly related, but not equivalent. Nov 5 '20 at 18:20

What is true is that if $$x$$ is an eigenvector of $$AA^T$$ for nonzero eigenvalue $$\lambda$$, then $$A^T x$$ is an eigenvector of $$A^T A$$ for $$\lambda$$, and if $$y$$ is an eigenvector of $$A^T A$$ for nonzero eigenvalue $$\lambda$$, then $$A y$$ is an eigenvector of $$A A^T$$ for $$\lambda$$.
There are $$r$$ linearly independent eigenvectors $$u_n$$ of $$A A^T$$ for nonzero eigenvalues, where $$r$$ is the rank of $$A$$, and $$r$$ linearly independent eigenvectors $$v_n$$ of $$A^T A$$ for nonzero eigenvalues. These may be chosen so that $$u_n = A v_n$$ for each $$n$$, and the corresponding eigenvalues are the same. Another $$n-r$$ linearly independent eigenvectors of $$A^T A$$ are for eigenvalue $$0$$, and $$m-r$$ linearly independent eigenvectors of $$A A^T$$ are for eigenvalue $$0$$. These form bases of the null spaces of $$A$$ and $$A^T$$ respectively.
• Namely, why are rank of $A$ and number of eigenvectors of $L$, related. And how can we say that there are $n-r$ and $m-r$ linearly independent eigenvectors for those. Nov 5 '20 at 18:37