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

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.
