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)


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*}
