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Possible Duplicate:
Matrix is conjugate to its own transpose

How can I prove that a matrix is similar to its transpose?

My approach is: if $A$ is the matrix then $f$ is the associated application from $K^n\rightarrow K^n$. Define $g:K^n\rightarrow (K^n)^*$ by $g(e_i)=e_i^*$, and define $f^T$ to be the transpose application of $f$. I proved that $f^T=gfg^{-1}$. What I don't understand is, what is the matrix associated to $g$, so I can write $A^T=PAP^{-1}$.

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    There's [this](http://projecteuclid.org/euclid.pjm/1103039127)...2011-12-28
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    This question has already been asked and answered here http://math.stackexchange.com/questions/62497/matrix-is-conjugate-to-its-own-transpose .2011-12-28
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    could you help me to complete my argument?2011-12-28
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    @J.M.: The article you linked to shows that $A$ and $A^T$ are always conjugate by a nonsingular symmetric matrix, and uses the "well known" fact that $A$ and $A^T$ are always conjugate. An interesting precision, but it does not really answer this question.2013-12-09

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