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?
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$\begingroup$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 $$(UV^T)A^TA(VU^T)=AA^T$$
$\endgroup$ 2 $\begingroup$In general the statement is false. E.g. consider the complex matrix $A=\pmatrix{1&i\\ 0&0}$, for which $AA^T=0\ne A^TA$. However, the statement is true for real square matrices. A proof was given in the other answer here, but that proof can actually be made simpler if you are allowed to use polar decomposition: let $A=PU$, where $P$ is symmetric positive semidefinite and $U$ is real orthogonal (so that $U^T=U^{-1}$). Then $A^TA=U^TP^2U$ is similar to $AA^T=P^2$.
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