TITLE: Non-zero eigenvalues of $AA^T$ and $A^TA$
QUESTION [35 upvotes]: As a part of an exercise I have to prove the following:
Let $A$ be an $(n \times m)$ matrix. Let $A^T$ be the transposed matrix of $A$. Then $AA^T$ is an $(n \times n)$ matrix and $A^TA$ is an $(m \times m)$ matrix. $AA^T$ then has a total of $n$ eigenvalues and $A^TA$ has a total of $m$ eigenvalues.
What I need to prove is the following:
$AA^T$ has an eigenvalue $\mu \not = 0$ $\Longleftrightarrow$ $A^TA$ has an eigenvalue $\mu \not = 0$
In other words, they have the same non-zero eigenvalues, and if one has more eigenvalues than the other, then these are all equal to $0$.
How can I prove this?
Thanks and regards.

REPLY [4 votes]: SVD is definitely an overkill, but maybe it would be helpful to you (as it is for me) to draw the matrices that the decomposition gives us:

Recall that:

The columns of $V$ (right-singular vectors) are eigenvectors of
$A^TA$.
The columns of $U$ (left-singular vectors) are eigenvectors
of $AA^T$.
$V^T=V^{-1}$ and $U^T=U^{-1}$.

With some simple operations you can get:

And also:

Write down both of these equations, but for the $i^\text{th}$ column.
From there, the solution is not far.