Prove that
$\lim_{t \rightarrow \infty} \|e^{tA}\| = 0$ if and only if $\alpha(A) < 0 $, where $\alpha$ is the Spectral abscissa, defined as $\max{Re(\lambda_i)}$.
I tried to approach this problem by proving that $\|e^{tA}\| \leq \|e^{t\alpha(A)}\|$ but didn't succeed. The hints given is First show what happens when you exponentiate a triangular matrix. But I still don't know how to solve this problem. Any help would be appreciated.
Hint. You should use the Jordan decomposition $A=U^{-1}JU$, express the exponential as $\mathrm{e}^{tA}=U^{-1}\mathrm{e}^{tJ}U$, and realize the following two facts:
a. If all the real parts of the eigenvalues are negative, then indeed $\mathrm{e}^{tA}\to 0$.
b. If a real part of an eigenvalue is non-negative then $\mathrm{e}^{tA}\not\to 0$.
More specifically, $A \,=\, U^{-1}JU,$ where \begin{equation*} J\,=\,\left(\begin{array}{llll} J_0 & & & 0 \\ & J_1 & & \\ & & \ddots & \\ 0 & & & J_k \end{array}\right)\,=\,\mathrm{diag(}J_0,J_1,\ldots,J_k), \end{equation*} with $\,J_0=\mathrm{diag}\bigl(\mu_1,\ldots,\mu_{N_0}\bigr) \!\in\! \mathbb C^{N_0\times N_0}\,$ while for $\ell=1,\ldots,k$ \begin{equation*} J_\ell\,=\,\left(\begin{array}{lllllll} \lambda_\ell & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & \lambda_\ell & 1 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \lambda_\ell & 1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & \lambda_\ell \end{array}\right)\,\in\,\mathbb C^{N_\ell\times N_\ell}, \end{equation*} and $\,N_0+N_1+\cdots +N_k=N$. For the powers of $J$ we have \begin{equation*} J^m\,=\,\mathrm{diag}\left(J_0^m,J_1^m,\ldots,J_k^m\right), \end{equation*} and hence \begin{align*} \mathrm{e}^{tJ} &=\, \sum_{m=0}^\infty \frac{t^m}{m!} J^m \,=\, \text{diag} \bigg( \sum_{m=0}^{\infty} \frac{t^m}{m!}J_0^m, \sum_{m=0}^{\infty}\frac{t^m}{m!}J_1^m, \ldots, \sum_{m=0}^{\infty}\frac{t^m}{m!}J_k^m \bigg) \\ &=\, \mathrm{diag} \big( \mathrm{e}^{tJ_0},\mathrm{e}^{tJ_1},\ldots,\mathrm{e}^{tJ_k} \big). \end{align*} For $\,\ell=1,\ldots,k$: \begin{equation*} J_\ell \,=\, \lambda_\ell{\mathcal I}+S, \end{equation*} where \begin{equation} S\,=\,\left(\begin{array}{lllllll} 0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \end{array}\right), \end{equation} and \begin{equation*} S^2\,=\,\left(\begin{array}{lllllll} 0 & 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \end{array}\right),\ldots,\, S^{N_\ell-1}\,=\,\left(\begin{array}{lllllll} 0 & 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\ \end{array}\right)\!, \end{equation*} and $$ S^m \,=\, 0, \quad \text{for all}\,\, \,\,m\ge N_\ell. $$ Hence the Neumann series corresponding to the exponential of $S$ is finite: \begin{align*} \mathrm{e}^{tS} & \,=\, \sum_{k=0}^\infty \frac{t^k}{k!} S^k\,=\,\sum_{k=0}^{N_l-1} \frac{t^k}{k!}S^k \\ & \,=\, {\mathcal I}+tS+\frac{t^{2}}{2!}S^2 +\cdots+ \frac{t^{N_l-1}}{(N_l-1)!} S^{N_l-1}, \end{align*} and since $S$ and ${\mathcal I}$commute we have \begin{align*} \mathrm{e}^{tJ_\ell} & \,=\, \mathrm{e}^{t(\lambda_\ell{\mathcal I}+S)} \,=\, \mathrm{e}^{\lambda_\ell t{\mathcal I}} \mathrm{e}^{tS} \,=\,\mathrm{e}^{\lambda_\ell t}{\mathcal I}\mathrm{e}^{tS} \,=\, \mathrm{e}^{\lambda_\ell t}\mathrm{e}^{tS} \\ & \,=\, \mathrm{e}^{\lambda_\ell t} \biggl( {\mathcal I}+tS+\frac{t^2}{2!} S^2 +\cdots+ \frac{t^{N_l-1}}{(N_l-1)!}S^{N_l-1} \biggr) \\ & \,=\, \left.\left( \begin{array}{ccccc} \mathrm{e}^{\lambda_\ell t} & t\mathrm{e}^{\lambda_{\ell} t} & \frac{t^2}{2!}\mathrm{e}^{\lambda_\ell t} & \cdots & \frac{t^{N_l-1}}{(N_l-1)!}\mathrm{e}^{\lambda_\ell t} \\ & \mathrm{e}^{\lambda_\ell t} & t\mathrm{e}^{\lambda_\ell t} & \cdots & \frac{t^{N_l-2}}{(N_l-2)!}\mathrm{e}^{\lambda_\ell t} \\ & & & \ddots & \\ 0 & & & & \mathrm{e}^{\lambda_\ell t} \\ \end{array} \right)\!.\right. \end{align*}
From the above we conclude that if $\,\mathrm{e}^{tA}=\bigl(b_{ij}(t)\bigr)_{i,j=1}^N$, then \begin{equation*} b_{ij}(t)\,=\,\sum_{\ell=1}^m p_\ell^{ij}(t)\,\mathrm{e}^{\lambda_\ell\,t}, \end{equation*} where $\lambda_\ell\!\in\!\mathbb C\,$ eigenvalue of $A$ and $p_\ell^{ij}(t)$ a polynomial of degree at most the difference between the algebraic and geometric multiplicity of $\lambda_\ell$.