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It is known that every square matrix $A$ is conjugate to its transpose. When I heard this first time, the way I tried to prove this is the following way:

(1) let $V$ be an $n$-dimensional vector space over a field $K$. Fix a basis $\mathcal{B}_V$ of $V$. Then there exists a linear transformation $T\colon V\rightarrow V$, whose matrix w.r.t $\mathcal{B}_V$ is $A$.

(2) Then what about $A^t$? It is known that if $$\begin{array}{rl} V^*=& \mbox{dual space of } V,\\ \mathcal{B}_V^*=& \mbox{dual basis of } \mathcal{B}_V\\ T^*\colon V^*\rightarrow V^*=& \mbox{adjoint of the transformation } $T$, \end{array}$$ then $A^t$ is the matrix of the adjoint transformation $T^*$.

(3) Further we know that, since $V$ is finite dimensional, $V\cong V^*$.

BUT, I couldn't proceed further, towards the proof of $A$ and $A^t$ are conjugate. I don't know whether this way can be proceeded towards proof. Can you help me, to decide whether this is a right way?

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    Here's a hint: Suppose you start with $K=\Bbb C$. What tool do you have at your disposal now? – Ted Shifrin Sep 29 '15 at 15:58
  • This result annoys me (and probably others) with its simple formulation and lack of a comparably nice proof. Short of module theory over PIDs (which, to some extent, is a coordinate-free rewriting of what @TedShifrin is suggesting), no real "explanations" for this result have been seen. – darij grinberg Sep 30 '15 at 02:04

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