In the matrix $A = \begin{pmatrix}1 & 2 \\ 3 & 4 \\ 5 & 6\end{pmatrix}$ then eigenvalues of $A^TA$ are eigenvalues of $AA^T$, and $AA^T$ has an eigenvalue of 0. I've also experimented with other matrices enough to think this is generally true for any matrix $A$ of the same dimensions. I can understand that the dimensions explain a nonempty nullspace and therefore eigenvalue of zero.
But for the other eigenvalues, I'd like to understand better why this is true, perhaps by giving a proof assuming that the proposition is true.
So if $\lambda $ is an eigenvalue with eigenvector $x$ for the matrix $AA^T$ then
$$ AA^Tx = \lambda x$$
The matrix $AA^T$ is a symmetric matrix and therefore $\lambda\ge 0$. Also $AA^T$ is therefore orthogonally diagonalizable, so we could write
$$ AA^T=PDP^{-1} $$
where $P$ is an orthonormal matrix. Therefore $P^{-1}=P^T$. So
$$PDP^Tx=\lambda x$$
so
$$D = \lambda P^TxP$$
I have no idea if this is going anywhere good, I'm just trying stuff out. I can't think of any other direction to head in from here.
