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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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Are you familiar with Jordan normal form? I don't know a simple straightforward way to do this. I believe it's false in infinite dimensions, so you should need to use some finite-dimensional fact. –  Qiaochu Yuan Sep 7 '11 at 7:51
Yes I know Jordan Normal Form. Perhaps Mariano meant some cute simple approach. –  George Sep 7 '11 at 7:56
Again, I don't think there is one. I think any sufficiently cute simple approach should work in infinite dimensions, and the result is false there. In $\ell^2(\mathbb{N})$ with orthonormal basis $e_1, e_2, ...$ the left shift $e_i \to e_{i+1}$ and right shift $e_i \to e_{i-1}$ are adjoint, but not conjugate (compare kernels). –  Qiaochu Yuan Sep 7 '11 at 8:16
@George: It comes down to showing that the matrix which has 1's on one level above the diagonal and 0's elsewhere, is conjugate to the matrix which has 1's on one level below the diagonal and 0's elsewhere. But the former is the map $e_i \mapsto e_{i-1}$, while the latter is the map $e_i \mapsto e_{i+1}$ (here $1 \le i \le n$; interpret $e_0$ and $e_{n+1}$ as 0). These two maps are conjugate because they are related by the change of basis which reverses the entire basis: $e_1, \ldots, e_n \mapsto e_n, \ldots, e_1$. –  Ted Sep 7 '11 at 8:35
A good similar exercise: prove that a square matrix $V = A^T A^{-1}$, where $A$ is a square invertible matrix, is similar/conjugate to its inverse $V^{-1}$. In particular, $detV = 1$. –  DVD May 10 '13 at 23:33

2 Answers 2

up vote 9 down vote accepted

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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For the small detail: If this were false, then the hypersurface $\det X = 0$ (inside $k^{n^2}$) would contain an entire subspace of dimension $d$ > 0. But $GL(n^2,k)$ acts transitively on the subspaces of dimension $d$ while keeping $\det X = 0$ invariant. So $\det X = 0$ contains all subspaces of dimension $d$. The union of all subspaces of dimension $d$ is all of $k^{n^2}$. This is a contradiction. –  Ted Sep 8 '11 at 3:21
@Ted: That simple argument does not work, because the action of $GL(n^2,k)$ can only be acting on matrices as if they were only (unstructured) vectors, and this does not leave the determinant (or its zero set) invariant. (In fact $GL(n^2,k)$ acts transitively on $k^{n^2}\setminus\{0\}$.) Conjugation action by $GL(n,k)$ does leave the determinant invariant, but it does not act transitively on all $d$-dimensional subspaces of matrices. Indeed there exist large subspaces of matrices inside the set $\{\,X\mid\det X=0\,\}$, for instance the space of all strictly upper triangular matrices. –  Marc van Leeuwen Dec 7 '13 at 21:06
However it is true that square matrices over $k$ that are similar over some extension field $K$ are already similar over$~k$. This is because similarity is equivalent to having the same rational canonical form, and this form is both rational (no decomposition of polynomials is used, so everything stays in the field$~k$) and canonical (only one form exists for a given matrix); having found the r.c.f. $M$ and then extending the field from $k$ to $K$, one finds that $M$ still satisfies the requirements, so it must be the r.c.f. over $K$ as well. –  Marc van Leeuwen Dec 8 '13 at 5:37

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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Thanks for the heads up in the other question. I thought it was weird my answer was un upupvoted, undownvoted and uncommented. –  Git Gud Dec 9 '13 at 7:01

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