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Suppose that $A$ is a positive definite matrix and that $B$ is a matrix whose eigenvalues are complex but have strictly positive real parts. Can it be shown that the eigenvalues of $AB$ also have positive real parts? Is there an easy way to prove this?

Tomas Jorovic
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  • If complex includes strictly real, then this might be relevant. – Riley Jan 17 '20 at 04:06
  • If $B$ is symmetric, then I can show that the above is true. But I am assuming that $B$ is not symmetric... – Tomas Jorovic Jan 17 '20 at 04:15
  • Ah right. I realised also the answer in that link would tell you $AB$ need not satisfy $x^T(AB)x > 0$ for $x \neq 0$ but AFAIK this is not a necessary condition for the real parts of the eigenvalues being positive. – Riley Jan 17 '20 at 06:22

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For example: $$A = \pmatrix{1 & -1\cr -1 & 2\cr},\ B = \pmatrix{1 & 5\cr -1 & 1\cr},\ AB = \pmatrix{2 & 4\cr -3 & -3\cr}$$ Then $A$ is positive definite, $B$ has eigenvalues $1 \pm i \sqrt{5}$ with positive real part, but $AB$ has eigenvalues $-1/2 \pm i \sqrt{23}/2$ with negative real part.

Robert Israel
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  • I see. Are then any additional conditions for which the above is true? – Tomas Jorovic Jan 17 '20 at 14:53
  • If $B$ is a normal matrix with all eigenvalues of positive real part and $A$ is positive definite, then all eigenvalues of $A^{1/2} B A^{1/2}$, and therefore of $AB$, have positive real part. – Robert Israel Jan 17 '20 at 16:18
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No. Consider $B=\pmatrix{3&-5\\ 1&-1}$, whose eigenvalues (namely, $1\pm i$) have positive real parts. Let $A=\pmatrix{t&0\\ 0&1}$ with $t>0$. Since $AB\to\pmatrix{0&0\\ 1&-1}$ when $t\to0$, $AB$ has an eigenvalue with negative real part when $t$ is small.

user1551
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