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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 $$\Sigma = \begin{bmatrix} 0 & k \\ 1 & 0 \end{bmatrix}$$,

Where $k \in {\cal R}$ and $k \ne 1$. I found that the non-symmetric matrix $A=\Sigma M \Sigma$ always has real eigen values for all k except $k=1$ (for $k=1$, it becomes a symmetric matrix) Numerically.

Can anyone prove that non-symmetric matrix $A=\Sigma M \Sigma$ will always have real eigen values ??

Matrix $A$ is pseudo-symmetric under the constant metric, $$\eta = \begin{bmatrix} 1/k & 0 \\ 0 & k \end{bmatrix}$$ as $\eta A \eta^{-1} = A^{T}$.

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  • $\begingroup$ @James Oh! Sorry, one typo was there, Now you please read it again. $\endgroup$ – math Jan 15 '18 at 14:48
  • $\begingroup$ I shall remove my comment as it is no longer needed. $\endgroup$ – James S. Cook Jan 15 '18 at 14:59
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In general, if $X$ and $Y$ are two rectangular matrices such that both $XY$ and $YX$ make sense, then $XY$ and $YX$ share the same nonzero eigenvalues.

So, the nonzero eigenvalues of $\Sigma M\Sigma$ are those of $\Sigma^2M=kM$. Now the conclusion follows.

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