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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?

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    @QiaochuYuan: I think the approach of my answer is elegant, though maybe not cute. It dutifully fails in infinite dimension.2013-12-07

4 Answers 4

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A different approach, by the principle of irrelevance of algebraic inequalities:

a) Over an infinite field:

1) Diagonalizable matrices: If $A$ is diagonalizable to $D$ via $P$ then it is similar to its transpose, since $D$ being symmetric implies $PAP^{-1}=D=D^T=(PAP^{-1})^T = P^{-T}A^TP^T,$ so that $A^T=QAQ^{-1}, Q:=P^TP.$

2) Similarity is polynomial: Two matrices A,B are similar iff they have the same invariant factors $\alpha_i(A)=\alpha_i(B)$, which are polynomials in the entries of the matrices.

3) Extension to all matrices: Let $f_i$ be the polynomial $\alpha_i(A)-\alpha_i(A^T)$ on the entries of $A$.

Consider the set of matrices with pairwise different eigenvalues, which are diagonalizable. These are precisely those which do not annihilate the discriminant of their characteristic polynomial, which is a polynomial $g$ in the entries of the matrix.

Thus we have that $g(A)\neq0$ implies $f_i(A)=0$. By the irrelevance of algebraic inequalities, $f_i(A)=0$ for all matrices, i.e., $A^T$ is similar to $A$.

b) Over a finite field:

Matrices over a finite field $K$ can be seen as matrices over the infinite field $K(\lambda)$ ($\lambda$ trascendental over $K$), so by part a) they also satisfy the result.

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This question has a nice answer using the theory of modules over a PID. Clearly the Smith normal forms (over $K[X]$) of $XI_n-A$ and of $XI_n-A^T$ are the same (by symmetry). Therefore $A$ and $A^T$ 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.

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I had in mind an argument using the Jordan form, which reduces the question to single Jordan blocks, which can then be handled using Ted's method ---in the comments.

There is one subtle point: the matrix which conjugates a matrix $A\in M_n(k)$ to its transpose can be taken with coefficients in $k$, no matter what the field is. On the other hand, the Jordan canonical form exists only for algebraically closed fields (or, rather, fields which split the characteristic polynomial)

If $K$ is an algebraic closure of $k$, then we can use the above argument to find an invertible matrix $C\in M_n(K)$ such that $CA=A^tC$. Now, consider the equation $XA=A^tX$ in a matrix $X=(x_{ij})$ of unknowns; this is a linear equation, and over $K$ it has non-zero solutions. Since the equation has coefficients in $k$, it follows that there are also non-zero solutions with coefficients in $k$. This solutions show $A$ and $A^t$ are conjugated, except for a detail: can you see how to assure that one of this non-zero solutions has non-zero determinant?

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    I believe any argument based on "prove similarity over algebraic closure, then argue that the same linear equation has a non degenerate solution over $k$" will fail for finite $k $. I would be happy to learn otherwise. Of course, proofs using the rational canonical form are unaffected.2016-11-07
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Theorem 66 of [1] proves that a square matrix (over an arbitrary field) is conjugate to its transpose via a symmetric matrix.

[1] Kaplansky, Irving Linear algebra and geometry. A second course. Allyn and Bacon, Inc., Boston, Mass. 1969 xii+139 pp.