Assume $A$ is diagonalizable. Show that $A$ is similar to $A^T$
Any help would be very appreciated. I know for a matrix to be diagonalizable it has to be a square matrix so the # of rows and columns stay the same for $A^T$.
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1Duplicate of (http://math.stackexchange.com/q/94599) which considers the general case (no need that $A$ is diagonalizable). – Jean Marie Nov 07 '16 at 07:51
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$A$ has the same eigenvalues as $A^T$ because $det(A−λI)=det(AT−λI)$ And then from ( if it is true) math.stackexchange.com/questions/556257/… we have that if two matrices have the same set of eigenvalues they are similar under condition that all eigenvalues are distinct. – Widawensen Nov 07 '16 at 19:21
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Hint: $$A=PDP^{-1}$$ $$A^T=(P^{-1})^TDP^T$$ Express $D$ in terms of $A^T$ and substitute to the first equation.
Note that $\left(P^{-1}\right)^{T}=\left(P^T\right)^{-1}.$ It is also commonly denoted by $P^{-T}$.
Siong Thye Goh
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2Just a little thing: it might be better to write $P^{-T}$ as $(P^{-1})^T$. – Sungjin Kim Nov 07 '16 at 21:55
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Given that $A$ is diagonalizable, so there exists an invertible matrix $P$ such that $$D=P^{-1}AP$$
where $D$ is a diagonal matrix.
So we obtain $$A=PDP^{-1}$$
Then $$A^T=(PDP^{-1})^T=(P^{-1})^TD^TP^T=(P^{-1})^TDP^T=(P^{-1})^T(P^{-1}AP)P^T=Q^{-1}AQ$$
where $Q=PP^T$ is an invertible matrix.
So by definition, $A$ is similar to $A^T$.
Wang Kah Lun
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