Matrix exponential when $(Av.v) \leq -(v.v)$

Question

I have a question from a previous homework that I never ended up figuring out:

If $(Av.v) \leq -(v.v)$ for all $v \in V$ (where $V$ is finite dimensional vector space with an inner product), show that $e^{tA}(v)$ converges to $0$ as $t \to \infty$ for all $v\in V.$

What I think will be useful is:

$(Av.v) \leq (v.v) \implies ((A+I)v.v) \leq 0.$ Also, since $A$ and $I$ commute, then $e^{t(A+I)} = e^{tA}\circ e^{tI}.$ So, showing $e^{t(A+I)}(v)$ converges to zero might be the way to go.

To do this, I can show $||e^{t(A+I)}(v)|| \to 0$ which means $(e^{t(A+I)}v.e^{t(A+I)}v) = 0.$

This is where I get stuck because I have tried manipulating the sum, but I don't know how to deal with the $(A+I)^k(v)$ term. Do I need to use the matrix exponential sum anyways? Or can I used properties of the matrix exponential/inner products/adjoints/etc to prove this?

The question seems to be wrong. In particular, if $A = \frac 12 I$, then we see that $(Av,v) \leq (v,v)$, but $e^{tA}(v)$ diverges for all $v \neq 0$. — Ben Grossmann, Oct 29 '20 at 17:33
I suspect that either they wanted to ask about $A^t(v)$ or they meant for the condition to be $(Av,v) \leq 0$. — Ben Grossmann, Oct 29 '20 at 17:34
I have a feeling that looking at $(I+\Delta t A)^n$ would be a better approach — user619894, Oct 29 '20 at 17:34
Note also that your deduction is incorrect: we have $$ (Av,v) \leq (v,v) \implies ((\color{red}{A - I})v,v) \leq 0. $$ — Ben Grossmann, Oct 29 '20 at 17:35
I see that I made a mistake when writing the question...I have corrected it now. Thank you for pointing these mistakes out! — trex1009, Oct 29 '20 at 17:39
can you solve this, perhaps a couple different ways, when the field is $\mathbb C$? — user8675309, Oct 29 '20 at 18:05

score 1 · Answer 1 · answered Oct 29 '20 at 18:56

1

One approach is to show that $A$ must have eigenvalues with negative real part and proceed. See this post for details.

Alternatively, we have $$ \frac d{dt} \left\|e^{At}v \right\|^2 = 2 \left( e^{At}v,\frac d{dt} e^{At}v \right) = 2 \left( [e^{At}v],A [e^{At}v] \right) \leq -2 (e^{At}v,e^{At} v) = -2 \left\|e^{At}v \right\|^2. $$ Thus, $\|e^{At}v\|$ is either $0$ or strictly decreasing for all $t$. The conclusion follows.

answered Oct 29 '20 at 18:56

Ben Grossmann

225,327

Thank you! This is a very nice approach. I appreciate your help – trex1009 Oct 29 '20 at 19:17

Matrix exponential when $(Av.v) \leq -(v.v)$

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