Bound on the norm of a matrix exponential in Jordan Form

Question

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

Answer 1

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 $t \ge 0$.