Relation between eigenvalues of $A^{\top}BB^{\top}A$ and $B^{\top}AA^{\top}B$ I have two real value matrices: $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{m \times p}$.
If I know which are the eigenvalues of $A^{\top}BB^{\top}A$, what can I say about the eigenvalues of $B^{\top}AA^{\top}B$?
I suspect that they coincide and that the extra eigenvalues are zeroes since this is what happens with the eigenvalues of $A^{\top}A$ and $AA^{\top}$.
Is this correct?
I will do some tests with Octave now.
Update:
the Octave results for a random $10 \times 10$ and a random $10 \times 20$ matrix seem to confirm my claim. Try it yourself:
m = 10, n = 10, p = 20;
A=rand(m, n);
B=rand(m, p);
eig(A'*B*B'*A)
eig(B'*A*A'*B)

 A: Yes you are correct, 
if you set $C=A^TB$ and $C^{T}=B^TA$.
Suppose that $CC^{T}$ is n x n matrix, and $C^{T}C$ is an m x m matrix, and the $n \geq m$.
If $\lambda$ is an eigenvalue for $CC^{T}$, where $\lambda$ is non-zero, then we have:
$CC^{T}x=\lambda x$, multiplying by $C^{T}$, we get : $C^{T}C(C^{T}x)=\lambda(C^{T}x)$.
Therefore, if $\lambda$ is a non-zero eigenvalue for $CC^{T}$, then $C^{T}C$ has the same eigenvalue too. 
Observe, that this holds for any matrices A,B ie, they need not be transposes of each other. 
A: Fact. The nonzero eigenvalues of $X^\top X$ and $XX^\top$ are the same.
We may apply this fact to your two matrices by taking $X=A^\top B$. Indeed, this gives
\begin{align*}
X^\top X &= (A^\top B)^\top(A^\top B)=B^\top AA^\top B &
XX^\top &= (A^\top B)(A^\top B)^\top = A^\top BB^\top A
\end{align*}
A: Take $C = A^\top B$. Then, $C^\top = B^{\top}A$. Therefore 
$A^{\top}B  B^{\top}A = C^\top C$ and 
$B^{\top}A  A^{\top}B= C^\top C$, so you have already answered the question.
