Let $A$ be diagonal with strictly positive (real) entries, and let $B$ be skew hermitian. Can it be shown that the eigenvalues of $AB$ are pure imaginary?

I suspect this also holds in the more general case that $A$ is symmetric positive definite.

  • $\begingroup$ In the general case, would you call $0\cdot i$ a pure imaginary number? $\endgroup$ – Roland Mar 8 '16 at 7:47
  • $\begingroup$ @Roland: yes, I would. $\endgroup$ – David Ketcheson Mar 8 '16 at 7:56

Your intuition is right. When $A$ is positive definite and $B$ is skew-Hermitian, $AB$ is similar to $A^{1/2}BA^{1/2}$ (as $A^{1/2}BA^{1/2}=A^{-1/2}(AB)A^{1/2}$), which is skew-Hermitian. Hence it has a purely imaginary spectrum.


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