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Let $A\in\mathbb{R}^{n\times n}$ be a matrix that has at least one positive eigenvalue with positive real part. Let $B\in\mathbb{R}^{n\times n}$ be a matrix where all its eigenvalues have positive real part.

Is it true that $B^{-1}A$ has at least one eigenvalue with positive real part?

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    $\begingroup$ Take $A$ to be any real matrix whose eigenvalues all have angles in $(\pi/4,\pi/2) \cup (-\pi/4, -\pi/2)$ and take $B=A^{-1}$. Then both $A$ and $B$ have eigenvalues whose real parts are all positive. However, $B^{-1} A = A^2$ will have eigenvalues whose angles will be in $(\pi/2,\pi) \cup (-\pi/2, -\pi)$, and consequently will all have negative real parts. $\endgroup$
    – alex
    Commented Apr 19, 2016 at 0:57
  • $\begingroup$ For a concrete example, take $A = \left( \begin{array}{cc} 1 & 9 \\ -1 & 1 \end{array} \right)$, $B=A^{-1}$. $\endgroup$
    – alex
    Commented Apr 19, 2016 at 1:05
  • $\begingroup$ What do you mean by "positive eigenvalue with positive real part"? Do you want it to be positive, in which case the "with positive real part" is redundant? $\endgroup$
    – Robert Israel
    Commented Apr 19, 2016 at 1:26
  • $\begingroup$ As others have pointed out, unless $A$ and $B^{-1}$ share the same eigenvectors (which I mistakenly assumed in my now deleted answer), that conclusion does not hold in general. $\endgroup$
    – Abhishek Halder
    Commented Apr 19, 2016 at 2:01

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