How do I prove that singular values of A are the square roots of eigenvalues of $AA^T$ $A \in \mathbb{R}^{m\times n}$
My task is to show that the singular values of A are the square roots of the eigenvalues of $AA^T$ if $m \geq n$ or of $A^TA$ if $ m \leq n$
I'm able to show that since $A = U\Sigma V^T$
$(AA^T)U = U\Sigma\Sigma^T$
and letting $\lambda_i$ denote the $i$th diagonal element of $\Sigma\Sigma^T$. 
Then this equation says that
$ (AA^T)u_i = \lambda_i u_i $ so  each squared singular value $\sigma_j^2$ is indeed an eigenvalue of $AA^T$ with corresponding eigenvector $u_j$
But i can't seem to see how $ m \leq n$ or $ m \geq n$ makes a difference
please help
 A: I think the question is wrong. I think you should consider $A^T A$ if $m\geq n$ and $AA^T$ if $m \leq n$. 
Note that $\Sigma$ is a diagonal  $m\times n$ matrix, so it's not necessarily square. The singular values are equal to the diagonal elements of $\Sigma$, so the number of singular values is $\min\{m,n\}$. 
Also, note that $A A^T$ is a $m\times m$ matrix, but $A^T A$ is a $n\times n$ matrix. So we have to make a distinction between the cases $m\geq n$ and $m\leq n$.
Case 1: If $m\geq n$, there are $n$ singular values, so we have to find the eigenvalues of the $n\times n$ matrix $A^T A$.
Case 2: If $m\leq n$, there are $m$ singular values, so we have to find the eigenvalues of the $m\times m$ matrix $A A^T$.
Example: Consider a $2\times 3$ matrix $A$ with $\Sigma$ given by
$$
  \Sigma = \begin{bmatrix}
 2 & 0 & 0 \\
 0 & 1 & 0 
  \end{bmatrix}.
$$
Then 
$$
\Sigma \Sigma^T = \begin{bmatrix}
   4 & 0 \\
   0 & 1
  \end{bmatrix}, \\
\Sigma^T \Sigma = \begin{bmatrix}
   4 & 0 &  0 \\
   0 & 1 & 0 \\
   0 & 0 & 0 
  \end{bmatrix}.
$$
Note that $\Sigma \Sigma^T$ indeed has two eigenvalues, as expected.
