Matrix is conjugate to its own transpose

Mariano mentioned somewhere that everyone should prove once in their life that every matrix is conjugate to its transpose.

I spent quite a bit of time on it now, and still could not prove it. At the risk of devaluing myself, might I ask someone else to show me a proof?

Answer

This question has a nice answer using the theory of modules over a PID. Clearly the Smith normal forms (over K[X]) of XInA and of XInAT are the same (by symmetry). Therefore A and AT have the same invariant factors, thus the same rational canonical form*, and hence they are similar over K.

*The Wikipedia article at the link badly needs rewriting.

Attribution
Source : Link , Question Author : George , Answer Author : Marc van Leeuwen

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