Is proving that an $n \times n$ diagonalizable matrix $A$ is similar to $A^T$ different that proving that $A\sim A^T$ when $A$ is not $n \times n$ and diagonalizable?
I think I can work from $A=SDS^{-1}$ but I am not sure where to go from there.
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Hint: For some $P$ and $Q$, one has $A=PDP^{-1}$ and $A^T=QDQ^{-1}$.
Solution: One has $D=P^{-1}AP=Q^{-1}A^TQ$. Thus $$ A = PQ^{-1}A^TQP^{-1} = (PQ^{-1})\ A^T(PQ^{-1})^{-1} $$
Mohammad Khosravi
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Ooh, so the $D$ in $A^T$ is the same as $D$ in $A$? How do we know that? – Emilio Mar 27 '15 at 18:39
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$A$ and $A^T$ have same set of eigen-values – Mohammad Khosravi Mar 27 '15 at 18:43
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@Mohammed-Khosravi Okay, that makes perfect sense. Thanks a lot! – Emilio Mar 27 '15 at 18:44