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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?

Rodrigo de Azevedo
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George
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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 07 '11 at 07:51
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    Yes I know Jordan Normal Form. Perhaps Mariano meant some cute simple approach. – George Sep 07 '11 at 07:56
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    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 07 '11 at 08:16
  • How would you prove it using the Jordan normal form, then? – George Sep 07 '11 at 08:22
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    @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 07 '11 at 08:35
  • @Ted: Ah, ok, thanks, I see. – George Sep 07 '11 at 08:43
  • I don't know what Mariano has in mind, but this follows from the classification of finitely generated modules over a PID. – Pierre-Yves Gaillard Sep 07 '11 at 10:08
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    Someone should go clean up the "proof" at http://wiki.answers.com/Q/How_do_you_show_that_a_square_matrix_A_is_similar_to_its_transpose – Gerry Myerson Sep 07 '11 at 13:50
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    Apropos... – J. M. ain't a mathematician Sep 09 '11 at 14:19
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    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
  • @QiaochuYuan: I think the approach of my answer is elegant, though maybe not cute. It dutifully fails in infinite dimension. – Marc van Leeuwen Dec 07 '13 at 14:00

5 Answers5

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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.

Marc van Leeuwen
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  • I cannot complete the proof. What I understand is $XI-A$ and $XI-A^t$ both are equivalent to a diagonal matrix in $K[X]$ whose entries has descending divisibility criterion along the main diagonal and positive degree polynomials are the invariant factors of $A.$ From here how to conclude that $A^t$ also has the same set of invariant factors. – user371231 May 28 '18 at 19:33
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    @user371231 Both those matrices are equivalent to the same diagonal matrix, since they are transposes of one another, and any diagonal matrix is its own transpose. The diagonal entries are the invariant factors. – Marc van Leeuwen May 28 '18 at 21:38
  • We can change the rational canonical form part with a more elementary argument: Once we know that $XI-A$ and $XI-A^T$ are equivalent to the same diagonal matrix in $M_n(K[X])$, we know that there are $P,Q\in M_n(K[X])$ such that $P(xI-A)Q=xI-A^T$ (1). By a degree argument we can pick $P,Q\in M_n(K)$. Again, a degree argument on (1) shows that $PQ=I$, $PAQ=A^T$, so that $Q=P^{-1}$ and $A^T=PAP^{-1}$. – Jose Brox Aug 10 '18 at 10:40
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    @JoseBrox: I dont really see how that is more elementary than using the rational canonical form, which is what the Smith normal form algorithm is really computing; in fact the "by a degree argument" looks a bit vague to me. But if you have a precise argument along these lines, by all means feel free to post your own answer. – Marc van Leeuwen Aug 13 '18 at 22:01
  • @MarcvanLeeuwen I meant "elementary" in the sense of being closer to purely algebraic computations (not involving the companion matrices, which are a "less elementary" nice idea), but of course this is subjective. The argument I was thinking about is like the one in Albert's Modern Higher Algebra, Theorem 3.21 (Theorem 4.1 is also relevant). Btw, I have posted another answer, based on the irrelevance of algebraic inequalities, I'd be glad if you took a look at it! :-) – Jose Brox Aug 14 '18 at 06:18
  • Nice! this is the way to do it. Goodman's book which is freely available online does it using Jordan Canonical form. But your solution is more elegant. – mathemather Apr 23 '19 at 09:03
  • @JoseBrox Can you fill in the step where you say "degree argument"? – wlad Apr 12 '23 at 20:27
  • @wlad For the first part, see Albert's Modern Higher Algebra, Theorem 3.21 (page 73 of the 2nd edition). For the second part, from $P(xI-A)Q = xI-A^T$ (1) with $P,Q$ matrices over $K$ we immediately get, by evaluation at $x=0$, that $PAQ=A^T$ (2), hence by evaluation in (1) at $x=1$ and using (2), $PQ=I$. – Jose Brox Dec 29 '23 at 01:29
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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?

Mariano Suárez-Álvarez
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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 08 '11 at 03:21
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    @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 07 '13 at 21:06
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    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 08 '13 at 05:37
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    @MarcvanLeeuwen Sorry for the necromancy here, but I think I have a slightly more direct argument:

    Using rational canonical form as you suggested, it suffices to show that the companion matrix $C_p$ to $p(t)=t^n+a_{n-1}t^{n-1}+\ldots+a_0$ is similar to its transpose. I think we can prove this with just elementary linear algebra by noting:

    1. The minimal polynomial of $C_p$ is $p(t)$
    2. In general, the minimal polynomials of $A$ and $A^t$ are equal
    3. Any two matrices with cyclic bases (i.e. $v,Av,\ldots, A^{n-1}v$) and equal minimal polynomials are similar
    4. $C_p^t$ has a cyclic basis
     –  Jul 26 '15 at 22:26
  • Be careful: This argument doesn't work for finite fields $k$. – darij grinberg Nov 07 '16 at 08:06
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    @darijgrinberg, I count at least three arguments in this answer and the comments attached to it… You should probably be more explicit about what argument you are talking about. – Mariano Suárez-Álvarez Nov 07 '16 at 09:43
  • 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. – darij grinberg Nov 07 '16 at 10:29
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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.

Jose Brox
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  • I am not sure about the reduction to the infinite case. The invertible matrix could have coefficients that are not in the finite field. See also the comments below Mariano's (incomplete) answer. – Martin Brandenburg Dec 27 '23 at 16:13
  • @MartinBrandenburg You raised a good point, but I think there is no problem: Suppose $A^T=P(\lambda)A(P(\lambda))^{-1}$ with $P(\lambda)$ a matrix over $K(\lambda)$. Then $\det(P(\lambda))=c$ is a nonzero constant of $K$, and by evaluation to $K$, putting $P:=P(0)$ we have $A^T=PAP^{-1}$ over $K$ (with $\det(P)=c\neq 0$). – Jose Brox Dec 29 '23 at 00:52
  • You cannot evaluate every rational function at 0. – Martin Brandenburg Dec 29 '23 at 01:37
  • @MartinBrandenburg Oops, sorry, I was working over $K[\lambda]$ instead of $K(\lambda)$. Indeed I don't see a good way to get an (invertible) solution for $K$ from $K(\lambda)$, different approaches hit different finitary walls. For $K:=F_q$ with $q=p^k$ and any prime $r$ we may take the union $F:=\bigcup_i F_{q^{r^i}}$, which is an infinite field, thus has a solution, which must lie in some finite subfield, hence for each prime $r$ there is $n_r$ such that there is a solution over $F_{q^{r^{n_r}}}$. But I don't know if these allow to build a solution over $K$. – Jose Brox Dec 30 '23 at 01:52
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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.

Glasby
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Although this reply is not answering the original question that was asked, I think it's worth pointing out that in the original question it's very important to be working over a field: there are counterexamples if you work over the integers. For instance, the integral matrix $$ A = \begin{pmatrix}1&-5\\3&-1\end{pmatrix} $$ is not conjugate to $A^\top$ in ${\rm M}_2(\mathbf Z)$, although of course $A$ and $A^\top$ are conjugate in ${\rm M}_2(\mathbf Q)$. More generally, two integral matrices with the same irreducible characteristic polynomial over $\mathbf Z$ need not be conjugate as integral matrices. See here, in particular Example 3.7.

This phenomenon (failure of integral matrices to be integrally conjugate to their transposes) is related to the behavior of ideals in rings of algebraic integers.

KCd
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