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I am trying to formally understand why $\|e^{A \tau}\|>1$, $\forall A \in\mathbb R^{n \times n}$, $\tau>0$ holds for every norm.

Attempt: Proof A: Use matrix norm property: $\|MN\|\le \|M\|\|N\|$. Take $M:=e^{A\tau}, N:=e^{-A\tau}$. This gives $\|I\|\le\|e^{A\tau}\|\|e^{-A\tau}\|$. If $\|e^{A\tau}\|=\|e^{-A\tau}\|$ (which is not true however for any $A$), $\forall A \in\mathbb R^{n \times n}$ (Proof B: not totally sure how to do this, except to try Taylor series, may be?. I appreciate the suggestion here too, please). Then, we got $\|I\|\le\|e^{A\tau}\|^2 \Rightarrow \|e^{A\tau}\| \ge 1$. For $\tau>0$ and $A \ne 0$, we prove our claim.

I appreciate anyone clarifying with rigor both Proof A and Proof B.

Added: Can anyone show the detailed steps for at least the 2-norm?

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    Please note this conspicuous difference: $$ ||M|| ||N|| \quad\text{versus} \quad \|M\|\|N\| $$ The latter usage is standard and I edited accordingly. – 2017-02-02
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    @MichaelHardy Thank you for the change – 2017-02-02
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    "any norm" strikes me as exaggerated if it means _every_ norm. If you meant "Is there any norm for which this holds?" that's another matter. If the former, then changing "any" to "every" would disambiguate it. – 2017-02-02
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    If this holds for every matrix $A\in\mathbb R^{n\times n}$ and every $\tau>0$, then it holds if $A$ is replaced by $-A,$ so it holds for all $\tau<0$ as well. – 2017-02-02
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    @MichaelHardy I have changed "any" to "every" norm as per your suggestion since that is what I am looking for the proof. – 2017-02-02
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    There is a norm $A\mapsto \|A\|$ defined by $$ \|A\| = \sup_{x\in\mathbb R^{n\times1}} \frac {\|Ax\|}{\|x\|} $$ where the vector norm $x\mapsto \|x\|$ is the usual Euclidean norm. Is the proposed statement supposed to be about _that_ norm? If so, "any norm" isn't right. – 2017-02-02
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    @MichaelHardy The proposed statement is norm independent as I read in a paper since it doesn't specifically make a comment on the type of the norm that the result holds. Moreover, I tried simulating the condition in MATLAB for different norms and different values for $A$ and $\tau>0$. They all seem to validate the claim. However, I need a formal proof to the claim. – 2017-02-02
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    Can you quote exactly what the paper said? – 2017-02-02
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    Suppose for some particular matrix $A$ and some $\tau>0,$ the value of $\|e^{\tau A}\|$ is a particular number $c$. Then define a new norm as $\displaystyle B\mapsto \frac{\|B\|} {2c}.$ This is a norm but it maps $e^{\tau A}$ to $1/2$. – 2017-02-02
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    @MichaelHardy I am referring this paper: New predictive scheme for the control of LTI systems with input delay and unknown disturbances. On page 4, right column above Eq. (25). The authors do not clarify what norm is it. But in the context of the paper, I guess 2-norm is relevant and that is what I validated for in MATLAB. – 2017-02-02

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The conjecture is not true for a diagonal matrix $A=-3I$: $$ e^{t(-3I)} = e^{-3t}I \implies \|e^{tA}\|=e^{-3t} < 1 \mbox{ for all } t > 0. $$

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    Nice counterexample. Apparently, the conjecture holds for any (an overstatement since I validated it only finite # of 1000 matrices: random and constructed) non-diagonal $A$ and $\tau>0$. Can you let me know how to prove this? – 2017-02-02
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    @ems: It can not be true for all non-diagonal $A$. Try $A = \begin{bmatrix}-1 & 0.1\\0.1 & -1\end{bmatrix}$. – 2017-02-04