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Suppose $A$ and $B$ are conjugate invertible real $n \times n$-matrices. Does there always exist a path from $A$ to $B$ inside their conjugacy class?


I thought I had an easy proof for odd $n$ which goes as follows, but it was incorrect as pointed out in this answer. To show where my misunderstanding arised, here is the wrong argument.

Suppose there exists a real matrix $P$ such that $B = PAP^{-1}$. By replacing $P$ with $-P$ if necessary, we can assume that $\det P > 0$ (this is what goes wrong in even dimensions, see this question). Then we have that $P = e^Q$ for some real matrix $Q$ (since the image of the exponential map is the path-component of the identity in $\text{GL}_n(\mathbb R)$). But now the path $$t \mapsto e^{tQ}Ae^{-tQ}$$ is a path connecting $A$ to $PAP^{-1} = B$.

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This edit comes a bit late, but the way I see it is a bit different than the other answers so I'll write it anyway.

Again, I use the example from the question that you link to: $A$ is the rotation by $\frac \pi 2$ in the euclidean plane, and $B$ is the rotation by $-\frac \pi 2$.

Now any conjugate of $A$ by a matrix with positive determinant will correspond to a linear map $\varphi$ such that for any non-zero vector $v$, the vectors $v$ and $\varphi (v)$ in this order make a positive basis of the plane.

Conversely, a conjugate of $A$ by a matrix with negative determinant (which is the case of $B$) will correspond to a linear map $\varphi$ such that for any non-zero vector $v$, the vectors $v$ and $\varphi (v)$ in this order make a negative basis of the plane.

A path from $A$ to $B$ has to cross the set of matrices that have real eigenvalues, and such a matrix cannot be conjugate to $A$.

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  • $\begingroup$ I like this geometric approach, this is the solution I will not forget. $\endgroup$ – Levi Nov 4 '19 at 0:50
  • $\begingroup$ @Levi Thanks :) $\endgroup$ – Arnaud Mortier Nov 4 '19 at 0:51
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We can use the counterexample from your other question to answer this one. Let $A=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ must have trace $a+d$ equal to $0$ and determinant $ad-bc=-a^2-bc$ equal to $1$. This implies that such a matrix cannot have $b=0$, since $-a^2\leq 0$. Therefore the set of conjugates of $A$ is disjoint union of open sets defined by $b>0$ and $b<0$. Neither of those sets is empty, since one contains $A$ and the other contains $-A$. Thus the conjugacy class of $A$ is disconnected.

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  • $\begingroup$ This is much nicer than my argument! (But slightly less detailed, whence I'm leaving mine up.) $\endgroup$ – darij grinberg Nov 4 '19 at 0:30
  • $\begingroup$ This is an awesome way of seeing this. This particular example also shows that I have been trying to prove the wrong thing when answering the other question. So I am particularly grateful! $\endgroup$ – Levi Nov 4 '19 at 0:33
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    $\begingroup$ Sorry for unaccepting your answer, but I cannot resist going with coordinate-freeness :) $\endgroup$ – Levi Nov 4 '19 at 0:50
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Other people have given counterexamples, so I would like to demonstrate that counterexamples are somewhat rare. Here is an attempt at a conceptual explanation for both why this is not true in general and also when it is true. First, we note that if we have a path $P(t)$ in $GL_n(\mathbb R)$ with $P(0)=P_0$ and $P(1)=P_1$, then $P(t)AP(t)^{-1}$ gives us a path inside the conjugacy class of $A$. Since $GL_n(\mathbb R)$ has two path components (given by sign of determinant), this shows that the conjugacy class of $A$ is the union of two path connected subsets (conjugating by things with positive or negative determinant), and it will be path connected if these two subsets intersect.

If $PAP^{-1}=QAQ^{-1}$ where $\det(P)>0$ and $\det(Q)<0$, we have $A=(P^{-1}Q)A(P^{-1}Q)^{-1}$, so $A$ commutes with a matrix with negative determinant. The converse is also true, so we have the following result.

Lemma: The conjugacy class of $A$ is path connected if and only if $A$ commutes with some matrix with negative determinant.

This gives several conditions that would ensure conjugacy classes are path connected.

  • If $n$ is odd, since then $\det(-I_n)=-1$
  • If $\det(A)<0$
  • If $\mathbb R^n$ as a direct sum of two $A$-invariant subspaces where one summand is odd dimensional, (e.g, if the Jordan normal form of $A$ has a block of odd size with a real eigenvalue).
  • If $\mathbb R^{n}$ is the direct sum of two $A$-invariant subspaces, and the restriction of $A$ to either subspace has negative determinant.

One can check that Jordan blocks only commute with matrices that are upper triangular and have only a single eigenvalue, and if $n$ is even and that eigenvalue is real, such a matrix could not have negative determinant. So these give counterexamples.

I suspect one could give a nice characterization of all counterexamples in terms of JNF. However, I have not worked out the details.

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No.

Here is a counterexample: Let $n=2$, $A=\left( \begin{array} [c]{cc} 0 & -1\\ 1 & 0 \end{array} \right) $ and $B=-A=\left( \begin{array} [c]{cc} 0 & 1\\ -1 & 0 \end{array} \right) $. Then, $A$ and $B$ are conjugate, since $B=C^{-1}AC$ for the invertible matrix $C=\left( \begin{array} [c]{cc} 1 & 0\\ 0 & -1 \end{array} \right) $. Thus, there exists a conjugacy class $\mathcal{C}$ in $\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right) $ that contains both $A$ and $B$. However, there exists no path from $A$ to $B$ in $\mathcal{C}$. Why not?

Probably there is a nice conceptual reason [EDIT: yes, and @Wojowu explains it in his answer], but you can just as well brute-force it: I claim that every matrix in $\mathcal{C}$ has a nonzero $\left( 1,2\right) $-th entry. To see this, just notice that any arbitrary element of $\mathcal{C}$ has the form \begin{align} \left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) ^{-1}A\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) =\left( \begin{array} [c]{cc} -\dfrac{ab+cd}{ad-bc} & -\dfrac{b^{2}+d^{2}}{ad-bc}\\ \dfrac{a^{2}+c^{2}}{ad-bc} & \dfrac{ab+cd}{ad-bc} \end{array} \right) \end{align} for some $\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) \in\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right) $, and thus its $\left( 1,2\right) $-th entry $-\dfrac{b^{2}+d^{2}}{ad-bc}$ is nonzero (because $b^{2}+d^{2}$ can only be $0$ if both $b$ and $d$ are $0$, but then $\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) $ cannot belong to $\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right) $).

Thus, every matrix in $\mathcal{C}$ has a nonzero $\left( 1,2\right) $-th entry. But if there was a path from $A$ to $B$ in $\mathcal{C}$, then some point on this path would be a matrix in $\mathcal{C}$ with zero $\left( 1,2\right) $-th entry (since the $\left( 1,2\right) $-th entries of $A$ and $B$ have opposite signs). Thus, there cannot be a path from $A$ to $B$ in $\mathcal{C}$.

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  • $\begingroup$ There is also a geometric way to see it if you're interested. $\endgroup$ – Arnaud Mortier Nov 4 '19 at 0:48
  • $\begingroup$ @ArnaudMortier I'm interested. $\endgroup$ – Abhimanyu Pallavi Sudhir Nov 29 '19 at 13:22
  • $\begingroup$ @AbhimanyuPallaviSudhir Thanks! You will find it in my answer which is currently the accepted one I think. $\endgroup$ – Arnaud Mortier Nov 29 '19 at 13:47
  • $\begingroup$ Ah I missed that, thanks. $\endgroup$ – Abhimanyu Pallavi Sudhir Nov 29 '19 at 15:04
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Notice that the assertion "since the image of the exponential map is the path-component of the identity in $GLn(R)$" is false.

Firstly, the path-component of the identity in $GLn(R)$ is the set of matrices with $>0$ determinant. Thus (inside this component) there is a path linking your $P$ and $I$ (in fact, there is nothing to prove).

Secondly, $diag(-1,-2)$ has $>0$ determinant and is not in the image of the real matrices by the exponential map.

EDIT.

Let $SC(A)$ be the conjugacy class of $A\in M_n(\mathbb{R})$.

*In dimension $2$, $SC(A)$ is non-connected in 2 cases

a. The eigenvalues of $A$ are non-zero conjugate complex; then $SC(A)$ is homeomorphic to a hyperboloid of two sheets.

b. $A$ is non-zero and non-diagonalizable; then $SC(A)$ is homeomorphic to a conical surface with the apex cut off.

*In dimension $4$, we consider the matrices $A_1=diag(U,U)$ where $U=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $A_2=diag(V,V)$ where $V=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. Then, using the Aaron's test, we can prove that $SC(A_1),SC(A_2)$ are non-connected (it's easy for $A_1$ and more difficult for $A_2$).

*Assume that we randomly choose $A\in M_n(\mathbb{R})$ -the $(a_{i,j})$ follow iid normal laws- We deduce that follows

$\textbf{Proposition}.$ When $n\rightarrow +\infty$, the probability that $SC(A)$ is connected tends to $1$.

$\textbf{Proof}$. A random matrix $A$ has distinct complex eigenvalues with probabiity $1$. Then, up to a real change of basis, $A=diag(a_1I_2+b_1V,\cdots,a_pI_2+b_pV,\lambda_1,\cdots,\lambda_q)$, where $2p+q=n$, the $(b_i)$ are non-zero and the $(\lambda_j)$ are real distinct.

Thus $SC(A)$ is connected iff $q\not=0$ (Aaron's test). When $n$ tends to $+\infty$, the mean of the number of real zeroes of a polynomial of degree $n$ is in $\Omega(\sqrt{n})$; we can deduce that the probability that $A$ has, at least, one real eigenvalue tends to $1$ when $n$ tends to $+\infty$. $\square$

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  • $\begingroup$ Sorry for not reacting to this sooner. I have now finally managed to convince myself that what you say is true. Thanks for pointing it out! $\endgroup$ – Levi Nov 29 '19 at 13:09
  • $\begingroup$ This statement also struck me when I first read it but then I focused on answering the question and forgot about it. Thanks for this. $\endgroup$ – Arnaud Mortier Nov 29 '19 at 13:47
  • $\begingroup$ @Levi . More precisely, the real matrices that can be written in the form $e^Q$ are the squares of real matrices. $\endgroup$ – loup blanc Nov 29 '19 at 14:02
  • $\begingroup$ @Arnaud Mortier . Thanks Arnaud. I wrote this because I was surprised that no one would react. $\endgroup$ – loup blanc Nov 29 '19 at 14:05

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