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)
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This link may contain what you are looking for: http://math.stackexchange.com/questions/421969/are-the-eigenvalues-of-a-top-a-equal-to-those-of-aa-top – Sarvesh Ravichandran Iyer Apr 04 '16 at 05:01
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2That link actually mentions the fact (proven on m.se several times) that if $A$ and $B$ are two $n \times n$-matrices, then the two matrices $AB$ and $BA$ have the same characteristic polynomial. This does not yet imply that $AB$ and $BA$ are similar. However, when $B = A^T$ and the matrices are over $\mathbb{R}$, we actually know more: The matrices $A^T A$ and $AA^T$ are symmetric and thus diagonalizable. Diagonalizable matrices sharing a characteristic polynomial must be similar. – darij grinberg Apr 04 '16 at 05:03
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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.
dineshdileep
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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.
copper.hat
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