Is it true that for an $n\times n$ matrix $A$, $AA^T$ and $A^TA$ have the same eigenvalues?
I am currently studying singular values my in linear algebra course and when finding left and right singular vectors for some matrix $A$, by finding unit eigenvectors of $AA^T$ and $A^TA$ respectively.
In computing respective characteristic polynomials for $n\times n$ matrices, this held true.
Does this also mean that $AA^T$ and $A^TA$ are similar?
In general, for rectangular matrices $A$ and $B$, if both products $AB$ and $BA$ make sense, then $AB$ and $BA$ have the same set of nonzero eigenvalues (counting multiplicity). See this question for more details.
So, in your case, $AA^T$ has the same spectrum as $A^TA$, but they are not necessarily similar. For a counterexample, consider $$ AA^T=\pmatrix{1&i\\ 0&0}\pmatrix{1&0\\ i&0}=0\ne \pmatrix{1&0\\ i&0}\pmatrix{1&i\\ 0&0}=A^TA. $$ It is true, however, that if $A$ is a real square matrix, $AA^T$ is similar to $A^TA$, because two real symmetric matrices with identical spectra are necessarily similar. You may also prove this via singular value decomposition. If $A=USV^T$ is a SVD, then $AA^T=US^2U^T$ and $A^TA=VS^2V^T$. Therefore $A^TA=(VU)(AA^T)(VU)^{-1}$ is similar to $AA^T$.
By a similar argument, it is also true that $AA^\ast$ is similar to $A^\ast A$ when $A$ is a complex square matrix, but as the previous example shows, it is essential to replace transpose by conjugate transpose when $A$ is a general complex square matrix.
