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For a given even dimension square complex matrix $A$ ($2N\times 2N$ dimension),
what's the sufficient and necessary condition for the matrix $A$ such that:
if $\lambda_{1}$ is an eigenvalue, then $\lambda_{2}=-\lambda_{1}$ is also an eigenvalue?

I have found one sufficient condition: skew-symmetric matrix, but obviously there is a lot of matrices having eigenvalues in pairs are not skew-symmetric.

So I wonder is there a sufficient and necessary condition?

Or is there a name for this family of matrices (skew-symmetric matrix is a member)?

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  • $\begingroup$ One more sufficient condition is raised in another question, and it's proved by @egreg . The matrix $A$ in the form of $\{\{B,C\},\{-C^{*},-B^{*}\}\}$, where $B$ is hermitian, $C$ is symmetric, and $M^{*}$ means the complex conjugate of $M$ rather than the transpose conjugate of $M$. $\endgroup$ – phchen Apr 1 '16 at 12:53
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Take any Hermitian matrix of the form $\mathfrak{B}=\left[\begin{array}{c|c}\mathbf{0}_n & A\\\hline A^* & \mathbf{0}_m\end{array}\right],$ where $A$ is $n\times m$. Then if $\lambda$ is an eigenvalue of $\mathfrak{B}$ with multiplicity $k$, then $-\lambda$ is also an eigenvalue with the same multiplicity. This is another sufficient condition.

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