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In my physical problem I have a matrix:

$\begin{bmatrix} 0 & -c_0 & -i b_1 & -c_1 \\ a_0 & 0 & a_1 & ib_1 \\ ib_1 & -c_1 & 0 & -c_0 \\ a_1 & -i b_1 & a_0 & 0 \\ \end{bmatrix}$

All symbols are reals, $i^2=-1$. The matrix has four eigenvalues satisfying $\lambda_1=-\lambda_2$ and $\lambda_3=-\lambda_4$, this can be shown symbolically.

It is seems that eigenvectors, corresponding to eigenvalues with the same absolute values, have the following properties so: $V_1 = [v_1,v_2,v_3,v_4]$ and $V_2 = [-v_3,-v_4,-v_1,-v_2]$ (and the same for $V_3$ and $V_4$) in orthonormal basis.

I tried to proof it with Mathematica without any luck. I made lots of numerical calculations with different values of parameters and my assumption was always valid.

Is it possible to prof it?

Moreover, it seems that this property is valid for matrices with higher dimensions.

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Note that the negative signs in $V_2$ are not important. Now consider

$$ J =\pmatrix{0&0&-1&0\\0&0&0&1\\-1&0&0&0\\0&1&0&0}$$ If you call the original matrix $A$ then you get $$ A V_1 = \lambda_1 V_1 \Leftrightarrow -J^{-1} A J \left(J^{-1}V_1\right) = -\lambda_1 J^{-1} V_1 $$

Your result follows from the observation that $$ A = -J^{-1} A J $$ Hence $-\lambda_1$ is an eigenvalue with eigenvector $J^{-1} V_1$. This answers your question.

Note: In my original answer I had $J$ in the place of $J^{-1}$. For this matrix the two are the same, so the answer was technically correct. I just fixed in the interest of correctness.

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    $\begingroup$ You mean $A=-J^{-1}AJ$ $\endgroup$ – galadog Jan 19 '14 at 5:20

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