Prove $A^TA$ is similar to $AA^T$ Let $A$ be a nxn matrix in $R^{nxn}$.\
Prove  $A^TA$ is similar to $AA^T$.
Could anyone give me a hint on how to prove this? Any help will be appreciated.
(I have tried using eigenvectors but I don't know if those matrices are diagonalizable)
 A: In case you are allowed to use singular value decomposition (SVD), Let $A=U\Sigma V^T$ be the SVD. Then $AA^T = U\Sigma^2 U^T$ and $A^TA = V\Sigma^2 V^T$. Since $U$ and $V$ are orthogonal, the similarity follows. 
A: Suppose $\lambda \neq 0$ is an eigenvalue of $AB$ with eigenvector $v$, then
$B(ABv)= (BA) Bv= \lambda Bv$ and so $\lambda$ is an eigenvalue of $BA$ (note that $Bv \neq 0$, otherwise $\lambda =0$).
Hence $AB$ and $BA$ have the same non zero eigenvalues.
Since $AA^T$ and $A^TA$ are real symmetric, they can be diagonalised with orthogonal matrices. It follows from the previous statement (since the geometric & algebraic multiplicities coincide) that $AA^T$ and $A^TA$ have
the same eigenvalues.
Hence we can find $U,V$ orthogonal such that
$AA^T U = U \Lambda$ and $A^TA V = V \Lambda$, where $\Lambda$ is a diagonal
matrix of the eigenvalues. Hence $A A^T = U V^T A^T A V U^T = (V U^T)^T A^T A (V U^T)$ and so $AA^T$ and $A^TA$ are similar.
A: Hint induction can be used or use the property $|A|=|A^{T}|$.  
