19
$\begingroup$

It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrices) that, for each $n\times n$ matrix $M$ with real entries, there is a matrix $S_M\in GL(n,\mathbb{R})$ such that ${S_M}^{-1}MS_M=M^T$. My question is: can you choose $S_M$ in such a way that it depends continuously on $M$? In other words:

Is there a continuous map $\psi\colon M_{n,n}(\mathbb{R})\longrightarrow GL(n,\mathbb{R})$ such that$$\bigl(\forall M\in M_{n\times n}(\mathbb{R})\bigr):\psi(M)^{-1}.M.\psi(M)=M^T?$$

My guess is that the answer is negative even for $n=2$.

Note that, for each individual matrix $M$, there are plenty of choices for $S_M$. For instance, if $n=2$ and$$M=\begin{bmatrix}x&y\\z&t\end{bmatrix},$$then you can take$$S_M=\begin{bmatrix}az&bz\\bz&bt-bx+ay\end{bmatrix},$$with $a$ and $b$ chosen such that $\det(S_M)\neq0$ but, of course, this will only work if $z\neq0$. What if $z=0$? Then you can take$$S_M=\begin{bmatrix}-at+ax&ay\\ay&by\end{bmatrix}$$and, again, $a$ and $b$ should be chosen such that $\det(S_M)\neq0$; the problem now is that, of course, this will only work if $y\neq0$. And so on. This looks like the problem of finding a logarithm for each $z\in\mathbb{C}\setminus\{0\}$: there are plenty of choices for each individual $z$, but there is no continuous way of picking one.

$\endgroup$
11
$\begingroup$

My first thought is to look at a simple example - $2\times 2$ rotation matrices. Oops - rotation by $\theta$ and rotation by $-\theta$ are conjugate by any (real) reflection. No help there.

Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.

Now we're ready. Consider a matrix $A$ with some eigenvalue $\lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $\lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $\exp(tA)$ for any nonzero $t$. As $t\to 0$, we then have $S(I)w = \lim_{t\to 0}S(\exp(tA))w=\lim_{t\to 0}a(\exp(tA))v=cv$ for some $c$, possibly zero.

Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=\langle v,w\rangle v$ and $A^Tw = \langle v,w\rangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.

Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.

OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.

$\endgroup$
6
$\begingroup$

Consider the continuous function $$ M_t=\begin{cases} t\pmatrix{1&1\\ 0&2}&\text{ when } t\ge0,\\ t\pmatrix{1&0\\ -1&2}&\text{ when } t<0. \end{cases} $$ It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form $$ S_t=\begin{cases} \pmatrix{a&b\\ b&b}&\text{ when } t>0,\\ \pmatrix{b&b\\ b&a}&\text{ when } t<0. \end{cases} $$ It follows that if $S_t$ is chosen continuously, it must be in the form of $\pmatrix{b&b\\ b&b}$ at $t=0$ and hence it cannot remain non-singular.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.