Let $t,\sigma,R \in \mathbb{R^+}$. Let $$ \mathrm{A} = \left\lbrack\begin{array}{cc}0 & -\sigma\\ \sigma &-R \end{array} \right\rbrack$$ want to show that the induced $\mathcal{l}_2$ norm of $\exp(\mathrm{A} t)$ is strictly less than one.

This is what I have tried.

$$\exp(\mathrm{A} t) = \exp(-R t/2) \left\lbrack\begin{array}{cc} \cosh (\omega t) + \frac{R}{2 \omega} \sinh( \omega t)& -\frac{\sigma}{\omega} \sinh(\omega t)\\ \frac{\sigma}{\omega} \sinh(\omega t) & \cosh(\omega t) - \frac{R}{2\omega} \sinh(\omega t) \end{array} \right\rbrack, $$

Where $\omega = \sqrt{R^2-4 \sigma^2}/2$. So the idea would be to show that

$$ \left\lbrack\begin{array}{cc} \cosh (\omega t) + \frac{R}{2 \omega} \sinh( \omega t)& -\frac{\sigma}{\omega} \sinh(\omega t)\\ \frac{\sigma}{\omega} \sinh(\omega t) & \cosh(\omega t) - \frac{R}{2\omega} \sinh(\omega t) \end{array} \right\rbrack $$ has norm strictly less than $\exp(R t/2)$.

This is pretty much where I am stuck. I attempted to consider if $\omega$ is purely rule or purely imaginary separately, but they were both seemingly dead ends. I also have used the eigen decomposition of $\mathrm{A}$ but that only tells us that the matrix exponential is eventually (for large enough $t$) strictly less than 1.

Any ideas are appreciated.


I would use the connection between matrix exponential and ODEs. Namely, for any vector $v$, the function $x(t)=\exp(tA)v$ is the solution of the ODE $$x'(t) = Ax, \quad x(0)=v$$ So you'd like to prove that $\|x(t)\|<\|v\|$ for all $t>0$. To this end, it suffices to show $\|x(t)\|^2$ is strictly decreasing with $t$. Calculate its derivative: $$\frac{d}{dt}\|x(t)\|^2 = 2\langle x'(t),x(t)\rangle = 2\langle Ax,x\rangle =-2Rx_2^2$$ So, $\|x(t)\|^2$ is nonincreasing. If it fails to be strictly decreasing, then it's constant on some interval $I$, hence $x_2(t)=0$ on $I$. But then $0=x_2'(t)=Ax_2(t)=\sigma x_1(t)$, which implies $x(t)=0$ on $I$. Since $v=\exp(-tA)x(t)$, it follows that $v=0$ in this case.

A more geometric explanation: the vector field $V(x) = Ax$ points inward on every circle with center at the origin; hence, the trajectories of this field do not escape from any disk centered at the origin.


Mice gave a good (and funny) answer. The proof works in dimension $n$ when $A+A^T\leq 0$ and $\ker(A+A^T)\cap\ker((A+A^T)A)=\{0\}$.

Proof. Note that $2<Ax,x>=x^T(A+A^T)x\leq 0$. If locally $<Ax,x>=0$, then $x^T(A+A^T)x=0$ and $(A+A^T)x=0$; moreover, locally, $(A+A^T)x'=(A+A^T)Ax=0$. Therefore, locally, $x=0$ and $v=0$.


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