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Let $A \in {\mathbb R}^{n\times n}$ with $\rho(e^A) > 1$, where $\rho(\cdot)$ returns the spectral radius for a given matrix. $\forall t,\epsilon > 0$, does the following inequality always hold?

$$\|e^{A(t+\epsilon)}\|_2 \geq \|e^{At}\|_2$$

Thanks!

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  • $\begingroup$ What is your definition of $A^{\alpha}$ with $ \alpha >0$ ? $\endgroup$ – Fred Nov 22 '17 at 13:17
  • $\begingroup$ @Fred Thanks for your kind reminder. You are right: $A^{\alpha}$ is indeed confusing... I have revised the description by using a matrix exponential form. $\endgroup$ – Ryan Nov 22 '17 at 13:30
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I presume that by $\|\cdot\|_2$ you mean the two-norm (largest singular value) but I don't think the result depends upon the choice of norm.

The answer is no: Take e.g. the matrix: $$ A = \left(\begin{matrix} -1 & 4 & 0\\ 0 & -1 &0 \\ 0 & 0 & 0.1\end{matrix}\right)$$ You have: $\rho(e^A)=e^{0.1}>1$. The 2-norm evaluates to: $\|e^A\|_2=1.558..$ (large because of the off-diagonal element 4) but $\|e^{2A}\|_2=1.221...$ (smaller because of the -1's which will reduce the influence of the off-diagonal element).

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  • $\begingroup$ A great non-example! Thanks very much for your answer [+1]: even though $\|e^{A(1+\epsilon)}\|_2$ will exceed $\|e^{A}\|_2$ for large enough $\epsilon$ (e.g., $\epsilon = 5$), there exists a "temporary" decrease. This indicates $\|e^{At}\|_2$ is not an increasing function of $t$. Thanks a lot! $\endgroup$ – Ryan Nov 29 '17 at 1:13
  • $\begingroup$ You are welcome. In the beginning I thought that as a function of $t$ the norm might be convex, but even that it not the case in the above example. $\endgroup$ – H. H. Rugh Nov 29 '17 at 9:16
  • $\begingroup$ Yes, the graph seems very interesting. Thanks again! $\endgroup$ – Ryan Nov 29 '17 at 12:33

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