Given are two matrices $A_0, A_1$, whose symmetric part is negative definite:
$A_{0}^{T} + A_{0} < 0$,
$A_{1}^{T} + A_{1} < 0$
How could one proof that: $A_{2}^{T} + A_{2} < 0$ for
$A_{2} = (A_{1}A_{0}^{-1})^{\alpha}A_{0}$ and $\alpha \in [0, 1]$?
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Sorry, what is $(A_1A_0^{-1})^\alpha$? If $A_1A_0^{-1}$ has positive spectrum, then I know how to define this. But in general, it will be massively multivalued. But perhaps there is a definition I don't know. – Theo Johnson-Freyd Dec 18 '13 at 17:36
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In my case $A_{1}A_{0}^{-1}$ has positive spectrum. Then it can be written: $(A_{1}A_{0}^{-1})^{\alpha}=\mathrm{e}^{\alpha \mathrm{log}(A_{1}A_{0}^{-1})}$. – Lyapunov123 Dec 19 '13 at 09:19
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So I assumed. But it would be best to specify this. – Theo Johnson-Freyd Dec 19 '13 at 16:48
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I mean, if $A_0= \epsilon + i$ and $A_1 = \epsilon - i$, for $\epsilon$ small and positive, then $A_1A_0^{-1} = -1 - 2\epsilon i + o(\epsilon)$. Here by $\epsilon$ I mean $\epsilon \bigl( \begin{smallmatrix} 1 & \ & 1\end{smallmatrix}\bigr)$ and $i = \bigl( \begin{smallmatrix} & 1 \ -1 & \end{smallmatrix}\bigr)$. – Theo Johnson-Freyd Dec 19 '13 at 17:37