Assume $ A(v) \cdot v \leq -|v|^2$ for all $v \in V$. Show that $ \lim_{t\to\infty} e^{tA} = 0$ 
$V$ is a finite inner product space, and $A$ is linear map $A:V\to V$. Assume
  $$ A(v) \cdot v \leq -|v|^2 \qquad \text{for all} \quad v \in V.$$
  Show that $ \lim_{t\to\infty} e^{tA} = 0$.

My thinking is: So $\operatorname{tr}(A) \leq -\dim(V) = -n $, and hence $\det(e^{tA})=e^{\operatorname{tr}(tA)} \leq e^{-tn} = 0$ as $t \to \infty$. So $ \exists v\in V$, $\lim_{t\to\infty} e^{tA}(v) = 0 $. And then I am stuck. Any suggestions?
Thanks! 
 A: 
Assumption. $V$ is a finite-dimensional inner product space and $A : V \to V$ is a linear operator such that the following condition holds:
$$\langle v, Av\rangle \leq -\|v\|^2 \qquad \text{for all }\quad v \in V. \tag{*}$$

Case 1. Suppose that $V$ is a $\mathbb{C}$-vector space. Since every eigenvalue $\lambda$ of $A$ has a unit eigenvector $v$, $\text{(*)}$ tells that $\lambda = \langle v, Av\rangle \leq -1$. Now by Jordan normal form, we know that
$$ A = P^{-1}(D + N)P, $$
where $P$ is an invertible matrix, $D$ is the diagonal matrix consisting of eigenvalues of $A$, and $N$ is a nilpotent matrix satisfying $DN = ND$. Then
$$ e^{tA} = P^{-1} e^{tD}e^{tN} P. $$
Write $\|\cdot\|$ for the operator norm. Then it is easy to check that $\| e^{tD} \| \leq e^{-t}$ and $\|e^{tN}\| \leq C t^{\dim V}$ for some constant $C > 0$. So it follows that
$$ \| e^{tA}\| \leq C' t^{\dim V} e^{-t} \to 0 $$
and the desired conclusion follows.
Case 2. Now suppose that $V$ is an $\mathbb{R}$-vector space. We prove the following lemma:

Lemma. Suppose that $V$ and $A$ be as in Assumption. Then $\operatorname{Re}(\lambda) \leq -1$ for every complex eigenvalue $\lambda$ of $A$.
Proof. Let $\lambda$ be an eigenvalue of $A$. Regard $A$ as a linear operator over a complex vector space. Then there exists a a non-zero complex eigenvector $v+iw$ of $A$, where $v, w$ themselves are real vectors. Then by writing $\sigma=\operatorname{Re}(\lambda)$ and $\xi=\operatorname{Im}(\lambda)$, the identity $(\sigma+i\xi)(v+iw)=A(v+iw)$ yields
  $$Av = \sigma v - \xi w, \qquad Aw = \xi v + \sigma w. $$
  From this,
  $$ \begin{gathered}
(1 + \sigma)\|v\|^2 - \xi\langle v, w\rangle = \|v\|^2 + \langle v, Av\rangle \leq 0\\
(1 + \sigma)\|w\|^2 + \xi\langle v, w\rangle = \|w\|^2 + \langle w, Aw\rangle \leq 0
\end{gathered}$$
  Summing two inequalities tells that $\sigma + 1 \leq 0$.

Given this lemma, the proof of complex case can be easily adapted to show that $e^{tA} \to 0$ as $t\to\infty$.
A: Solution without using spectral theory.
Choose $f:R\rightarrow R$ for given $v\in V $ with $f(t) = e^{tA}(v) \cdot e^{tA}(v) $
Then $f^{'}(t)=Ae^{tA}(v)\cdot e^{tA}(v)+e^{tA}(v)\cdot Ae^{tA}(v) = 2
(Ae^{tA}v \cdot e^{tA}(v)) \leq -2|e^{tA}(v)|^2=-2f(t)$
As $\forall t. f(t)>0 \implies \forall t. f^{'}(t)\leq-2f(t)\le0$, $f(t)$ is non-increasing and bounded below by $0$. 
Thus, $lim_{t\rightarrow \infty }f(t)=l$ for some limit $l$, and $lim_{t\rightarrow \infty } f^{'}(t)=0$. 
Suppose $l >0$, then $lim_{t\rightarrow \infty }\frac{1}{2}f^{'}(t)=Ae^{tA}(v)\cdot e^{tA}(v)  \leq - |e^{tA}(v)|^2=-l^2 < 0$.
Yet, $lim_{t\rightarrow \infty }f^{'}(t)=0$. Contradiction.
So, $lim_{t\rightarrow \infty }f(t)=e^{tA}(v) \cdot e^{tA}(v) =0$ and thus $lim_{t\rightarrow \infty }e^{tA}(v)=0$ for all $v \in V$.
A: I'm assuming $V$ is a complex vector space.
Notice that from $\langle Av,v\rangle \le -\|v\|^2, \forall v \in V$ follows that $A$ is hermitian. Indeed, we can write $A = B+iC$ for some hermitian matrices $B, C$. Then if $\lambda$ is an eigenvalue of $C$ with unit eigenvector $v$, we have
$$\overbrace{\langle Bv,v\rangle}^{\in\mathbb{R}}+i\lambda= \langle Bv,v\rangle + i\langle Cv,v\rangle = \langle Av,v\rangle \in \mathbb{R}$$
so $\lambda = 0$. It follows that $C = 0$ so $A = B$ which is hermitian.
Therefore $A$ is diagonalizable with eigenvalues $\lambda_1, \ldots, \lambda_n \le- 1$ (this follows easily from $\langle Av,v\rangle \le -\|v\|^2, \forall v \in V$). We have
$$A = P^{-1}\operatorname{diag}(\lambda_1, \ldots, \lambda_n)P$$
so
$$e^{tA} = e^{P^{-1}\operatorname{diag}(t\lambda_1, \ldots, t\lambda_n)P} = P^{-1}\operatorname{diag}(e^{t\lambda_1}, \ldots, e^{t\lambda_n})P \xrightarrow{t\to\infty} 0$$
since $t\lambda_k \to -\infty$ for all $k=1, \ldots, n$.
