63
$\begingroup$

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?

$\endgroup$
11
  • 4
    $\begingroup$ 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. $\endgroup$
    – Qiaochu Yuan
    Sep 7, 2011 at 7:51
  • 2
    $\begingroup$ Yes I know Jordan Normal Form. Perhaps Mariano meant some cute simple approach. $\endgroup$
    – George
    Sep 7, 2011 at 7:56
  • 13
    $\begingroup$ 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). $\endgroup$
    – Qiaochu Yuan
    Sep 7, 2011 at 8:16
  • 14
    $\begingroup$ @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: $e_1, \ldots, e_n \mapsto e_n, \ldots, e_1$. $\endgroup$
    – Ted
    Sep 7, 2011 at 8:35
  • 4
    $\begingroup$ 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, $detV = 1$. $\endgroup$
    – DVD
    May 10, 2013 at 23:33

4 Answers 4

Reset to default
37
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – user371231
    May 28, 2018 at 19:33
  • 1
    $\begingroup$ @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. $\endgroup$
    – Marc van Leeuwen
    May 28, 2018 at 21:38
  • $\begingroup$ 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 $A^T=PAP^{-1}$. $\endgroup$
    – Jose Brox
    Aug 10, 2018 at 10:40
  • 1
    $\begingroup$ @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. $\endgroup$
    – Marc van Leeuwen
    Aug 13, 2018 at 22:01
  • $\begingroup$ @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! :-) $\endgroup$
    – Jose Brox
    Aug 14, 2018 at 6:18
35
$\begingroup$

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?

$\endgroup$
7
  • 3
    $\begingroup$ 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. $\endgroup$
    – Ted
    Sep 8, 2011 at 3:21
  • 3
    $\begingroup$ @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. $\endgroup$
    – Marc van Leeuwen
    Dec 7, 2013 at 21:06
  • 2
    $\begingroup$ 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. $\endgroup$
    – Marc van Leeuwen
    Dec 8, 2013 at 5:37
  • 1
    $\begingroup$ @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 $\endgroup$
    – user59193
    Jul 26, 2015 at 22:26
  • 2
    $\begingroup$ @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. $\endgroup$
    – Mariano Suárez-Álvarez
    Nov 7, 2016 at 9:43
9
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .