Csdrhrt

It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrics) that for each $ntimes n$ matrix $M$ with real entries, there is a matrix $S_Min GL(n,mathbb{R})$ 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 $psicolon M_{n,n}(mathbb{R})longrightarrow GL(n,mathbb{R})$ such that$$bigl(forall Min M_{ntimes n}(mathbb{R})bigr):psi(M)^{-1}.M.psi(M)=M^T?$$

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$$M=begin{bmatrix}x&y\z&tend{bmatrix},$$then you can take$$S_M=begin{bmatrix}az&bz\bz&bt-bx+ayend{bmatrix},$$with $a$ and $b$ chosen such that $det(S_M)neq0$ but, of course, this will only work if $zneq0$. What if $z=0$? Then you can take$$S_M=begin{bmatrix}-at+ax&ay\ay&byend{bmatrix}$$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 $yneq0$. And so on. This looks like the problem of finding a logarithm for each $zinmathbb{C}setminus{0}$: 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 - $2times 2$ 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 $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ 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 $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, 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
$$
M_t=begin{cases}
tpmatrix{1&1\ 0&2}&text{ when } tge0,\
tpmatrix{1&0\ -1&2}&text{ when } t<0.
end{cases}
$$
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.