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Consider the generalized eigenvalue equation $$A \mathbf{v}=\lambda S \mathbf{v}$$ where $S$ is a real square symmetric matrix and $A$ a real square anti-symmetric matrix.

I seek a necessary and sufficient condition for the eigenvalues $\lambda$ to be purely imaginary.

sufficient conditions include

  • $S$ is positive definite
  • $S$ is negative definite
  • $S$ and $A$ commute, and $S$ is invertible

That none of these are necessary is shown by this example $$S=\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$ $$A=\left( \begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & -2 \\ 0 & 2 & 0 \\ \end{array} \right)$$ $$\lambda=0,\pm i\sqrt{3}$$

On the other hand, symmetry of $S$ and anti-symmetry of $A$ does not suffice, even if $S$ is invertible and diagonalizable $$S=\left( \begin{array}{ccc} 6 & 4 & 10 \\ 4 & 12 & 4 \\ 10 & 4 & 12 \\ \end{array} \right)$$ $$A=\left( \begin{array}{ccc} 0 & 7 & 3 \\ -7 & 0 & -1 \\ -3 & 1 & 0 \\ \end{array} \right)$$ $$|A-\lambda S|=0\implies 304 \lambda^3-418 \lambda =0\implies \lambda=0,\pm \frac{\sqrt{\frac{11}{2}}}{2}$$ In this case, all eigenvectors have $<v,S v>=0$ and $<v,A v>=0$, but $S v$ and $A v$ themselves are not both $0$: the eigenvalue problem remains well-defined.

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  • $\begingroup$ Don't we just always have $(Av,v)\in i\mathbb R$ and $(Sv,v)\in \mathbb R$, so the only danger is that we directly hit the surface $(Sv,v)=0$ in which case we must also have $(Av,v)=0$ and we cannot obtain the result by passing to the limit from a generic position, so everything smells of degeneracy? In other words, do you have an example when it is not the case and the generalized eigenvalue problem is still meaningful? $\endgroup$ – fedja Jun 17 '18 at 20:23
  • $\begingroup$ @fedja Yes, I edited in such an example. If $S$ is indefinite, there are nonzero $v$ with nonzero $S v$ such that $<v,S v>=0$. $\endgroup$ – Wouter Jun 18 '18 at 6:07
  • $\begingroup$ @fedja Your comment does suggest a useful answer: there must be no simultaneous nonzero solutions of $<v,A v>=0$ and $<v, S v>=0$ $\endgroup$ – Wouter Jun 18 '18 at 6:17

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