# Is every matrix conjugate to its transpose in a continuous way?

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.

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.

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.