Real eigenvalues of a non-symmetric matrix $A$ ?.

Question

Let $M$ be a $2 \times 2$ real symmetric matrix and it is well known/proved that it will always have two real eigenvalues.
Now let $$\Sigma(k) = \begin{bmatrix} 0 & k \\ 1 & 0 \end{bmatrix}$$,

Where $k \in {\cal R}$ and $k \ne 1$. I find that the non-symmetric matrix $A=\Sigma(k) M \Sigma(l)$ has real eigenvalues for $~kl>0$ (for $k=l=1$, it becomes a symmetric matrix) Numerically.

Can anyone prove that $A=\Sigma(k) M \Sigma(l)$ will have real eigenvalues for $~kl>0$ ??

Answer 1

Since $kl>0$, $\Sigma(l)$ is invertible and $A=\Sigma(k) M \Sigma(l)$ is similar to $$ B=\Sigma(l)\Sigma(k) M=\pmatrix{l&0\\ 0&k}M=\operatorname{sgn}(l)\pmatrix{|l|&0\\ 0&|k|}M, $$ which in turn is similar to the real symmetric matrix $$ C=\operatorname{sgn}(l)\pmatrix{\sqrt{|l|}&0\\ 0&\sqrt{|k|}}M\pmatrix{\sqrt{|l|}&0\\ 0&\sqrt{|k|}}. $$ Hence the result.