Show that there are constants $K$ and $\alpha$ such that $|(e^{At})_{ij}|\leq e^{-\alpha t}K$. I want to prove that if all eigenvalues of $\textbf{A}$ in the sytem $\dot{\textbf{x}}=\textbf{Ax}$ have negative real parts then there exist constants $K$ and $\alpha$ such that $$|(e^{\textbf{A}t})_{ij}|\leq e^{-\alpha t}K$$ for $1\leq i,j\leq n.$ This has been my proof so far:
\begin{align*}
e^{\textbf{A}t}&=\textbf{X(t)X(0)}^{-1}\text{ where $\textbf{X}$ is the fundamental matrix of solutions,}\\
&=\begin{pmatrix}
e^{\lambda_1t}x_{11}(t)&\cdots&e^{\lambda_nt}x_{1n}(t)\\
\vdots&\ddots&\vdots\\
e^{\lambda_1t}x_{n1}(t)&\cdots&e^{\lambda_nt}x_{nn}(t)
\end{pmatrix}\begin{pmatrix}
a_{11}&\cdots&a_{1n}\\
\vdots&\ddots&\vdots\\
a_{n1}&\cdots&a_{nn}
\end{pmatrix}.
\end{align*} Thus, every component of $e^{\textbf{A}t}$ is of the form
\begin{align*}
(e^{\textbf{A}t})_{ij}=a_{1j}e^{\lambda_1t}x_{i1}(t)+\cdots+a_{nj}e^{\lambda_nt}x_{in}(t).
\end{align*}
So, \begin{align*}|(e^{\textbf{A}t})_{ij}|&=|a_{1j}e^{\lambda_1t}x_{i1}(t)+\cdots+a_{nj}e^{\lambda_nt}x_{in}(t)|\\
&=|a_{1j}e^{(-p_1+iq_1)t}x_{i1}(t)+\cdots+a_{nj}e^{(-p_n+iq_n)t}x_{in}(t)|\\
&=|a_{1j}e^{-p_1t}e^{iq_1t}x_{i1}(t)+\cdots+a_{nj}e^{-p_nt}e^{iq_nt}x_{in}(t)|\\
&\leq e^{-\alpha t}|a_{1j}e^{iq_1t}x_{i1}(t)+\cdots+a_{nj}e^{iq_nt}x_{in}(t)|
\end{align*} where $-\alpha=\min\{-p_1,\ldots,-p_n\}$. From this point, I am stuck at finding $K$. I'd appreciate if someone could help me figure this out and improve my proof.
 A: Here a somewhat different approach. It gives a bound on the norm $\parallel
\exp [\mathbf{A}t]\parallel $ of $\exp [\mathbf{A}t]$ and hence $\parallel
\mathbf{x}(t)\parallel $ rather than on individual matrix-elements.
I gather from the context that $\mathbf{A}$ is an $n\times n$-matrix and $%
\mathbf{x}\in \mathbb{C}^{n}$. Then we can equip $\mathbb{C}^{n}$ with the
standard inner product  $<\mathbf{x,y}>=\sum_{j=1}^{n}\overline{x_{j}}y_{j}$
and norm squared $\parallel \mathbf{x}\parallel ^{2}=\sum_{j=1}^{n}\overline{
x_{j}}x_{j}$. Next we write
\begin{equation*}
\mathbf{A}=\sum_{j=1}^{n}\lambda _{j}|\mathbf{u}_{j}><\mathbf{v}_{j}|,
\end{equation*}
where $\{|\mathbf{u}_{j}><\mathbf{v}_{j}|\}$ is a bi-orthogonal basis $\ $($<
\mathbf{v}_{j}|\mathbf{u}_{h}>=\delta _{jh}$). Now we expand and estimate
\begin{eqnarray*}
\exp [\mathbf{A}t] &=&\sum_{j=1}^{n}\exp [\lambda _{j}t]|\mathbf{u}_{j}><
\mathbf{v}_{j}| \\
&\parallel &\exp [\mathbf{A}t]\parallel \leqslant \sum_{j=1}^{n}|\exp
[\lambda _{j}t]|\parallel |\mathbf{u}_{j}><\mathbf{v}_{j}|\parallel
\leqslant \sum_{j=1}^{n}\exp [{Re}\lambda _{j}t]\leqslant n\exp [-{
Re}|\lambda _{0}|t]
\end{eqnarray*}
where $\lambda _{0}$ is the eigenvalue with real part closest to $0$. Then
\begin{eqnarray*}
\mathbf{x}(t) &=&\exp [\mathbf{A}t]\cdot \mathbf{x}(0) \\
&\parallel &\mathbf{x}(t)\parallel =\parallel \exp [\mathbf{A}t]\cdot
\mathbf{x}(0)\parallel \leqslant n\exp [-{Re}|\lambda _{0}|t]\parallel
\mathbf{x}(0)\parallel
\end{eqnarray*}
Here $\parallel \mathbf{F}\parallel $ is the sup-norm of the operator $
\mathbf{F}$ (in this case an $n\times n$-matrix)
\begin{equation*}
\parallel \mathbf{F}\parallel =\sup_{\parallel \mathbf{x}\parallel
=1}\parallel \mathbf{F\cdot x}\parallel
\end{equation*}
A: If I apply the triangle inequality and use the fact that $\left|e^{iqt}\right|=1$, then the right side of the equation becomes $$e^{-\alpha t}\underline{(\left|a_{1j}x_{i1}(t)\right|+\cdots+\left|a_{nj}x_{in}(t)\right|)}$$. However, I am not certain if the underlined part is the constant $K$ that I am looking for.
