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First we will handle the diagonalizable case.
Assume A is diagonalizable, A=VDV-1.
Thus AT=(V-1)TDVT,
and D= VT AT(V-1)T.
Finally we have that A= VVT AT(V-1)TV-1, hence A is similar to AT
with matrix VVT.
If A is not diagonalizable, then we must consider its Jordan canonical form,
A=VJV-1, where J is block diagonal with Jordan blocks along the diagonal.
Recall that a Jordan block of size m with eigenvalue at L is a mxm matrix having L along the diagonal and ones along the superdiagonal.
A Jordan block is similar to its transpose via the permutation that has ones along the antidiagonal, and zeros elsewhere.
With this in mind we proceed as in the diagonalizable case,
AT=(V-1)TJTVT.
There exists a block diagonal permutation matrix P such that
JT=PJPT, thus J=PTVT AT(V-1)TP.
Finally we have that A= VPTVT AT(V-1)TPV-1, hence A is similar to AT
with matrix VPTVT.
Q.E.D.
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