Assume $ A(v) \cdot v \leq -|v|^2$ for all $v \in V$. Show that $ \lim_{t\to\infty} e^{tA} = 0$

Question

$V$ 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!

Answer 1

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 $v \in V$.

Answer 2

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 $t\to\infty$.

Answer 3

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 $k=1, \ldots, n$.