I saw in a proof somewhere that a square matrix $AA^T$ is similar to $A^T A$, so I thought about it and I don't know why (or whether) it's true.
I tried using the fact that every matrix is similar to its transpose and maybe transpose the entire expression $AA^T$ but what I get is $(AA^T)^T=A^{T^T} A^T=AA^T$ which is obvious because $AA^T$ is symmetric.
I tried to run some examples like
$$
A =
\begin{bmatrix}
1 & 4 \\
3 & 2
\end{bmatrix} \qquad \qquad \qquad
A^T =
\begin{bmatrix}
1 & 3 \\
4 & 2
\end{bmatrix}
$$
And I get that $AA^T$ and $A^TA$ have the same characteristic polynomial so obviously they have the same trace, eigenvalues and determinant.
But is it true for the general case?
Best Answer
If $A$ is a square real matrix and let $A=U D V^T$ be the SVD decomposition.
$$A^TA=VD^2V^T$$
$$AA^T=UD^2U^T$$
Notice that $$(UV^T)A^TA(VU^T)=AA^T$$