Stability in linear systems
Theorem.
Let $A$ be an $n\times n$ matrix for which all eigenvalues satisfy $\real\lambda
\lt \delta$. Then there is a constant $C\lt \infty$ such that
\[
\forall t\ge0:\; \|e^{tA}\| \leq C e^{\delta t}
\]
Proof (in the case that $A$ has distinct eigenvalues)
Suppose that $A$ has $\ell$ real eigenvalues $\lambda_1, \ldots, \lambda_\ell$ and
$k$ pairs of complex eigenvalues $ \alpha_1 \pm i\omega_1$, $\dots$,
$\alpha_k\pm i\omega_k$.
Let $D$ and $V$ be the matrices defined in the the complex diagonalization theorem.
Then $A = V D V^{-1}$, and
the matrix exponential $e^{tA}$ is given by $e^{tA} = Ve^{tD}V^{-1}$.
We find that the norm, or maximal row sum, of the matrix $e^{tA}$ satisfies
\[
\|e^{tA}\| = \|V e^{tD} V^{-1}\|
\leq \|V\| \, \|V^{-1}\|\, \|e^{tD}\|.
\]
To compute $\|e^{tD}\|$ we first find $e^{tD}$:
\[
e^{tD} =
\begin{pmatrix}
e^{\lambda_1t} & & & & & & \\
& \ddots\\
& & e^{\lambda_\ell t} & & & & & \\
& & & e^{\alpha_1 t}\cos\omega_1t & e^{\alpha_1 t}\sin \omega_1 t & & \\
& & & -e^{\alpha_1 t}\sin\omega_1 t& e^{\alpha_1 t}\cos\omega_1 t & & \\
& & & & & \ddots & \\
& & & & & & e^{\alpha_k t}\cos\omega_kt & e^{\alpha_k t}\sin \omega_k t \\
& & & & & & -e^{\alpha_k t}\sin\omega_k t& e^{\alpha_k t}\cos\omega_k t
\end{pmatrix}
\]
Each row has either one or two non zero entries, which makes it easier to compute the maximal row sum. It is
\begin{align*}
\|e^{tD}\|
&= \max\Bigl\{ e^{\lambda_1t}, \dots, e^{\lambda_\ell t},
e^{\alpha_1 t}\bigl(|\cos\omega_1t| + |\sin\omega_1 t| \bigr), \dots,
e^{\alpha_k t}\bigl(|\cos\omega_kt| + |\sin\omega_k t| \bigr)
\Bigr\}\\
&\leq \sqrt{2}\max\Bigl\{ e^{\lambda_1t}, \dots, e^{\lambda_\ell t},
e^{\alpha_1 t}, \dots, e^{\alpha_k t}
\Bigr\}
\end{align*}
By assumption $\delta$ is larger than each of the $\lambda_i$ and
$\alpha_i$, so
\[
e^{\lambda_1 t} \leq e^{\delta t}, \dots,
e^{\lambda _\ell t}\leq e^{\delta t},
e^{\alpha_1t}\leq e^{\delta t}, \dots,
e^{\alpha_k t}\leq e^{\delta t},
\]
and therefore $\|e^{tD}\| \leq 2e^{\delta t}$ for all $t\geq 0$.
Hence we have
\[
\|e^{tA}\| \leq \sqrt{2} \|V^{-1}\| \, \|V\|\, e^{\delta t} = Ce^{\delta t}
\]
for all $t\geq 0$.
An example with a repeated eigenvalue
Consider the matrix
\[
A=
\begin{pmatrix}
0& 1 \\ 0 & 0
\end{pmatrix}.
\]
Both eigenvalues are $\lambda_1=\lambda_2 = 0$, so the maximal real part of the eigenvalues is $0$.
Since $A^2=0$ we find that $A^3=A^4=A^5=\cdots=0$, and thus
\[
e^{tA} = I + tA =
\begin{pmatrix}
1 & t \\ 0 & 1
\end{pmatrix}.
\]
Hence for all $t\geq 0$ we have
\[
\|e^{tA}\| = 1+t,
\]
The theorem predicts that for any $\delta\gt 0$ there will be a constant
$C_\delta \lt \infty$ for which
\begin{equation}
1+t = \|e^{tD}\| \leq C_{\delta}e^{\delta t}
\qquad\text{ for all }t\geq0.
\label{eq:exponential-growth-of-1plust}
\end{equation}