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What are the sufficient and necessary conditions that an even sized and invertible Hermitian matrix to have its (real) eigenvalues $\lambda$ to come in pairs as in $(\lambda,-\lambda)$ with $\lambda$ having the same algebraic multiplicity as that of $-\lambda$. This is similar to this question.

Hans
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1 Answers1

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One necessary and sufficient condition is that the characteristic polynomial is an even function. Another is that (if your matrix is $A$) there exists a matrix $S$ with $S^2 = I$ and $S A + A S = 0$.

Robert Israel
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  • Could you explain the second condition? – Hans Aug 10 '23 at 18:44
  • Suppose $A$ is diagonal, with the positive eigenvalues listed first and then the corresponding negative eigenvalues, so we can write it as a block matrix $\pmatrix{D & 0\cr 0 & -D}$. Take $S = \pmatrix{0 & I\cr I & 0\cr}$. Then $S^2 = \pmatrix{I & 0\cr 0 & I\cr}$ and $SA + AS = \pmatrix{0 & -D\cr D & 0\cr} + \pmatrix{0 & D\cr -D & 0\cr} = \pmatrix{0 & 0\cr 0 & 0\cr}$. If $A$ is not diagonal, we diagonalize it so that $ A = U \pmatrix{D & 0\cr 0 & -D} U^{-1}$ and take $S = U \pmatrix{0 & I\cr I & 0\cr} U^{-1}$. – Robert Israel Aug 11 '23 at 01:18
  • Right. I was dumb not to have realized that immediately. But, essentially, there is no criterion simpler than diagonalizing $A$. Is that right? – Hans Aug 11 '23 at 04:52
  • Computing determinants is in principle simpler than diagonalizing, though there may be numerical issues. Take some random $\lambda$ and check if $\det(A-\lambda I) = \det(A + \lambda I)$. If the characteristic polynomial is an even function these will be equal; if not, with probability $1$ they are different. – Robert Israel Aug 11 '23 at 14:13
  • This is interesting. Can you expand on the numerical issues, specially whether these issues are more severe than diagonalization? Does Hermiticity impact the complexity comparison? – Hans Aug 11 '23 at 15:38