Can we characterize the set $\{S \in GL_n(\mathbb C): SAS^{-1} \in M_n(\mathbb R)\}$

Question

Let $A \in M_n(\mathbb R)$ be a fixed real matrix. Let \begin{align*} \mathcal E = \{S \in GL_n(\mathbb C): S A S^{-1} \in M_n(\mathbb R) \}. \end{align*}

My questions:

1. The set should contain $GL_n(\mathbb R)$. But is this inclusion proper? In other words, is it true: for any real matrix $A$, there exists an invertible matrix with complex entries, such that $SAS^{-1}$ has only real entries?
2. Is this set $\mathcal E$ connected? I guess if the inclusion in question $1$ is always proper, we end up with a set $\mathcal E$ that is larger than $GL_n(\mathbb R)$ but in general smaller than $GL_n(\mathbb C)$. I am wondering whether this set contains "enough elements" to connect the two components of $GL_n(\mathbb R)$.

Answer 1

Let $n\geq 2,E_A=\{S\in GL_n(\mathbb{C});SAS^{-1}\in M_n(\mathbb{R})\}$ and $\lambda\in\mathbb{R},\alpha\in\mathbb{R}^*$.

Then $(*)$ $E_{A+\lambda I}=E_{\alpha A}=E_A$ and $E_{\lambda I}=GL_n(\mathbb{C})$.

Let $S=M+iN,S^{-1}=P+iQ$ where $M,N,P,Q$ are real matrices.

Then the condition $S\in E_A$ is equivalent to $(*)$ $NAP+MAQ=0,NP+MQ=0,MP-NQ=I$.

Thus $E_A$ is a real algebraic set of dimension $d\leq 2n^2$; since it contains $GL_n(\mathbb{R})$, $d\geq n^2$.

$\textbf{Proposition 1}.$ $d=2n^2$ iff $A$ is a scalar matrix.

$\textbf{Proof}.$ Assume that $A$ is not a scalar matrix. There is $x\in\mathbb{C}^n$ s.t. $\{x,Ax\}$ is a free system; we consider a basis in the form $\{x,-iAx,\cdots\}$; then the associated $S_0AS_0^{-1}$ has $[0,i,0,\cdots]^T$ as first column, $(S_0AS_0^{-1})_{2,1}\notin \mathbb{R}$ and $S_0\notin E_A$. Let $\mathcal{B}$ be the set of bases of $\mathbb{C}^n$.

Let $Z=\{(e_1,\cdots,e_n)\in\mathcal{B};Ae_1=\sum_ja_je_j \text{ with }a_2\in\mathbb{R}\}$; note that $Z\not= \mathcal{B}$ (cf. above).

If $T=[t_{i,j}]=[e_1,\cdots,e_n]$, then $(e_1,\cdots,e_n)\in Z$ iff $Im((T^{-1}AT)_{2,1})=0$. Since $Im((T^{-1}AT)_{2,1})$ is a rational fraction in the $(Re(t_{i,j}),Im(t_{i,j}))$, $Z$ is a real algebric set and, moreover, it is a Zariski closed strict subset of $\mathcal{B}$. We deduce that $d<2n^2$. $\square$

To find $d$ in the general case seems difficult (at least for me...). We consider the case $n=2$ and we will see that there are matrices $A$ s.t. $n^2<dim(E_A)<2n^2$.

$\textbf{Proposition 2}.$ Assume that $n=2$. If $A$ is a scalar matrix, then $d=8$, otherwise $d=6$.

$\textbf{Proof}$. We may assume that $A$ is in the form of its Jordan real form. According to $(*)$, it suffices to consider the three cases

i) $A=\begin{pmatrix}0&0\\0&1\end{pmatrix}$, ii) $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, iii) $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.

For i). We obtain $S\in E_A$ iff $S=\begin{pmatrix}a&b\\\beta a&\gamma b\end{pmatrix}$ where $a,b\in \mathbb{C},\beta,\gamma\in\mathbb{R},ab\not= 0,\beta\not= \gamma$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.

For ii). We obtain $S\in E_A$ iff $S=\begin{pmatrix}a&b\\\beta a&\beta(b+\gamma a)\end{pmatrix}$ where $a,b\in \mathbb{C},\beta,\gamma\in\mathbb{R},a\beta\gamma\not= 0$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.

For iii). We obtain $S=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ where $p=\dfrac{ac+bd}{ad-bc}\in\mathbb{R},q=\dfrac{a^2+b^2}{ad-bc}\in\mathbb{R}^*,r=\dfrac{c^2+d^2}{ad-bc}\in\mathbb{R}$ and $ad-bc\not= 0$ (or $p\in\mathbb{R},q\in\mathbb{R},r\in\mathbb{R}^*$)

We show that $dim(E_A)=6$. Note that $q\not=0$ and $r=\dfrac{1+p^2}{q}$ is real if $p,q$ are. Assume that $a,b,p,q$ are independently given (6 degrees of freedom). Then $ad-bc$ and $ac+bd$ are known; since $a^2+b^2\not= 0$ we deduce a unique couple $c,d$. $\square$

$\textbf{Proposition 3.}$ For every $A$, $dim(E_A)\geq n^2+1$.

$\textbf{Proof}$. If $S\in E_A$, then, for every $\theta\in\mathbb{R}$, $e^{i\theta}S\in E_A$.

EDIT. I do not know why you are passionate about connected sets... Here, the answer is no in general.

$\textbf{Proposition 4.}$ There are matrices $A$ s.t. $E_A$ is not connected.

$\textbf{Proof.}$ Let $A\in .M_2(\mathbb{R})$ s.t. $tr(A)=1,\det(A)=4$ and $\phi:S\in GL_n(\mathbb{C})\rightarrow SAS^{-1}\in M_n(\mathbb{C})$. Clearly $\phi(E_A)=\phi(GL_n(\mathbb{R}))\approx \{(p,q,r,s)\in\mathbb{R}^4;p+s=1,ps-rq=4\}$. In particular, $\phi(E_A)$ has one or two connected components.

Let $Z=\{(p,q,r)\in\mathbb{R}^3;p(1-p)-rq=4\}$. $Z$ has exactly $2$ connected components (draw with Maple if you are in hurry). Suppose $\phi(E_A)$ is connected; since $(p,q,r,s)\in E_A\rightarrow (p,q,r)\in Z$ is a continuous function, this would cause $Z$ to be connected, a contradiction. Finally $\phi(E_A)$ is not connected and $E_A$ too.

$\textbf{Remark.}$ We can generalize the above proposition to matrices in $ M_n(\mathbb{R})$ that have conjugate eigenvalues.

Answer 2

Assume that $SA=BS$ for some real matrix $B$. Write $S=M+iN$ where $M$ and $N$ are real. Let us plug in, expand, and then separate the real and imaginary parts of the equation.

$$MA+iNA=BM+iBN; \quad \text{or} \quad MA=BM, NA=BN.$$

Thus, for fixed $B$, complex solutions to the equation correspond to pairs of real solutions.