I'm looking to prove the following lemma:
Let $A$ be a matrix in $\mathbb{R}^{n\times n}$. Then for any $\lambda^* > \max_{\lambda} \; \mathrm{Re} \; (\lambda)$ such that $ \lambda \in\sigma (A)$, there exists $\beta > 0$ such that
\begin{eqnarray*}
||e^{At}|| \leq \beta e^{\lambda^* t}\qquad \forall t\geq 0 .
\end{eqnarray*}
This is what I have
Proof:
We know there exists a transformation $T$ such that $e^{At}=Te^{Jt}T^{-1}$, where
\begin{eqnarray*}
T^{-1}AT&=&\begin{bmatrix}
J_1 & \: & 0 \\
\: & \ddots & \: \\
0 & \: & J_k
\end{bmatrix}
,\; J_i=\begin{bmatrix}
\lambda_i & 1 & 0 & \cdots & 0 \\
0 & \lambda_i & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots & \ddots & 0 \\
\vdots & \: & \ddots & \ddots & 1 \\
0 & … & 0 & \lambda_i
\end{bmatrix}\text{ for $i=1,\dots,k$,} \\
e^{Jt}&=&\begin{bmatrix}
e^{J_1 t} & \: & 0 \\
\: & \ddots & \: \\
0 & \: & e^{J_k t}
\end{bmatrix} \text{and,}\;
e^{J_i t}=e^{\lambda_i t}\begin{bmatrix}
1 & t & \cdots & \frac{t^{q_i -1}}{(q_i -1)!} \\
0 & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots & t \\
0 & \cdots & 0 & 1
\end{bmatrix}
\end{eqnarray*}
Therefore
\begin{eqnarray*}
||e^{At}||&=&||Te^{Jt}T^{-1}|| \\
&\leq &||T||\;||T^{-1}||\;||e^{Jt}|| \\
&=& \beta_1 ||e^{Jt}||
\end{eqnarray*}
Let $\epsilon = \lambda^* – \mathrm{max} \; \lambda\in\sigma (A)$, then $\epsilon > 0$ and
\begin{eqnarray*}
||e^{Jt}||&\leq & \left| \left| e^{(\lambda^* – \epsilon)t} \begin{bmatrix}
1 & t & \cdots & \frac{t^{n -1}}{(n -1)!} \\
0 & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots & t \\
0 & \cdots & 0 & 1
\end{bmatrix}
\right| \right| \\
&\leq & e^{\lambda^* t} \left| \left| e^{ – \epsilon t} \begin{bmatrix}
1 & t & \cdots & \frac{t^{n -1}}{(n -1)!} \\
0 & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots & t \\
0 & \cdots & 0 & 1
\end{bmatrix}
\right| \right|
\end{eqnarray*}
Then $b^*_t$, the largest term at time $t$ as
\begin{eqnarray*}
b^*_t = \max_i \left( e^{-\epsilon t} \frac{t^{i}}{(i-1)!} \right)
\end{eqnarray*}
Since $\epsilon>0$, the $\exists$ $b^*$ such that
\begin{eqnarray*}
b^* = \max_t \; b^*_t
\end{eqnarray*}
Then
\begin{eqnarray*}
||e^{Jt}||&\leq & e^{\lambda^* t} \left| \left| \begin{bmatrix}
b^* & \cdots & b^* \\
& \ddots & \vdots \\
0 & & b^*
\end{bmatrix}
\right| \right| \\
&=& \beta_2 e^{\lambda^* t}
\end{eqnarray*}
Putting it all together we get that
\begin{eqnarray*}
||e^{At}||\leq \beta e^{\lambda^* t} \qquad \forall t\geq 0
\end{eqnarray*}
My problem is that I'm not sure if I'm allowed to do my jump from
\begin{eqnarray*}
||e^{Jt}||&\leq & \left| \left| e^{(\lambda^* – \epsilon)t} \begin{bmatrix}
1 & t & \cdots & \frac{t^{q_n -1}}{(q_n -1)!} \\
0 & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots & t \\
0 & \cdots & 0 & 1
\end{bmatrix}
\right| \right|
\end{eqnarray*}
Any advice? Alternatively if you could provide a reference to this lemma it would be greatly appreciated! 🙂
Thanks is advance
Best Answer
Your approach is reasonable, but can be simplified a bit.
You need $\lambda^* > \max_{\lambda \in \sigma(A)} \operatorname{re} \lambda $, remember that the spectrum have complex eigenvalues.
However, I think it would be easier to show that $\|e^{-\lambda^* t} e^{At} \|$ is bounded for $t \ge 0$.
In particular, this reduces to showing that each entry in $e^{-\lambda^* t} e^{At}$ (or rather, $e^{-\lambda^* t} e^{Jt}$) is bounded.
Since $\lim_{t \to \infty} e^{(\lambda_i-\lambda^*) t} \frac{t^{k}}{k!} = 0$ for all $k$, we see that every entry in $e^{-\lambda^* t} e^{At} $ goes to zero, hence it is bounded.
So, a slightly tighter result would be $\lim_{t \to \infty} e^{-\lambda^* t} e^{At} = 0$.
Addendum: Since $\lim_{t \to \infty} e^{-\lambda^* t} e^{At} = 0$ and $t \mapsto e^{-\lambda^* t} e^{At}$ is continuous, we see that $\beta = \sup_{t \ge 0} \| e^{-\lambda^* t} e^{At}\| < \infty$. hence we have $\| e^{-\lambda^* t} e^{At}\| \le \beta$ for all $t \ge 0$, from which it follows that $\| e^{At}\| \le \beta e^{\lambda^* t}$ for $t \ge 0$.