How do you find (continuous) bounds on the matrix exponential Let $A$ be a $n \times n$ real or complex matrix. I am interested in bounds on the matrix exponential $e^{A t}$, for $t \geq 0$. In particular: is there a continuous function $C: M_{n\times n} \rightarrow \mathbb{R}_+$ such that
$$\|e^{At}\| \leq C(A) e^{\Lambda(A) t} (1 + t^{n-1})$$
As some background: for any individual $A$, I know how to find the big-$\mathcal{O}$ behavior of $e^{A t}$. Namely, re-write $A$ as $T B T^{-1}$ where $B$ is in Jordan canonical form, compute the matrix exponential of $B$, then translate back to the matrix exponential of $A$. This gives the upper bound:
$$\|e^{A t} \| = \mathcal{O}(e^{\Lambda(A) t} (1+t^{n-1}))$$
where $\Lambda(A)$ is the largest real part of any eigenvalue of $A$.
Now, the eigenvalues of $A$ depend continuously on $A$, so $\Lambda(A)$ is continuous. However the big-$\mathcal{O}$ constant in this expression depends on the matrix $T$ used above. However, $T$ will not vary continuously with $A$, and may even be arbitrarily large when $\|A\|$ is bounded. So, this doesn't give a continuous bound on the matrix exponential. Hence my question: is there a continuous function $C(A)$ such that:
$$\|e^{At}\| \leq C(A) e^{\Lambda(A) t} (1 + t^{n-1})$$
 A: Yes. You can use a Schur decomposition instead of Jordan normal form, which keeps the transformation bounded.
Claim:
For a fixed $n$, for lower triangular matrices
$$
L = \begin{pmatrix}
\lambda_1 &  &  &  \\
l_{21} & \lambda_2 &  & \\
\vdots& & \ddots & \\
l_{n1}&l_{n2} &\dots &\lambda_n
\end{pmatrix}
$$
There is a closed-form expression for the entries of $\exp(tL)$ restricted to $t \geq 0$ where each entry is a linear combination, with coefficients polynomials in the $l_{ij}$, of $k$-fold convolutions of functions of the form
$$
G_{\lambda_i}(t) =
\begin{cases}
\exp(\lambda_i t) & t \geq 0 \\
0 & \text{otherwise}
\end{cases}
$$
with $k \leq n$.
Proof:
This is fairly easy to see by induction. A column $v(t) = \exp(tL) e_j$ of $\exp(tL)$ represents a solution to a differential equation $d/dt\ v(t) = L v(t)$, in which the derivative of the $k$'th entry in $v(t)$ the vector depends only on that entry and previous ones, i.e. influence always flows from earlier entries in $v(t)$ to later entries. So we can build the solution up from beginning to end.
Explicitly, if we have the block
$$
L = \begin{pmatrix}
\tilde L & \\
w & \lambda
\end{pmatrix},
$$
the relevant equation is
$$
\frac{d}{dt}
\begin{pmatrix}
v(t) \\
y(t)
\end{pmatrix}
=
\begin{pmatrix}
\tilde L & \\
w & \lambda
\end{pmatrix}
\begin{pmatrix}
v(t) \\
y(t)
\end{pmatrix}
$$
i.e.
$$
\dot v(t) = \tilde L v(t), \\
\dot y(t) = \lambda y(t) + w\ v(t).
$$
So by induction $v(t)$ has the closed form described above, and the equation for $y(t)$ is an inhomogenous linear equation whose homogenous form has Green's function $G_{\lambda}$. So the solution to the inhomogenous equation is some convolution of $G_{\lambda}$ with the rest. $\blacksquare$
For example, the $(3, 1)$ coefficient of $\exp(tL)$ is
$$
\left(\exp(tL)\right)_{31}
=
l_{31} (G_{\lambda_3} * G_{\lambda_1})(t) + l_{32} l_{21} (G_{\lambda_3} * G_{\lambda_2} * G_{\lambda_1})(t).
$$
(All the weird $t e^{\lambda t}$ terms that pop up when the matrix has repeated eigenvalues and isn't diagonalizable show up from the resulting convolutions of $G_\lambda$ with itself.)
Edit: This seems like it should be a "textbook" result, so I've asked a separate reference request question.
Now observe that a $k$-fold convolution $G_{\lambda_1} * \dots * G_{\lambda_k}$ is bounded above by $C t^{k-1} \exp(\Lambda t)$, where $\Lambda$ is the largest real part of any $\lambda_i$:
$$
| G_{\lambda_1} * \dots * G_{\lambda_k} (t) |
=
\left|\int_{t_1 + \dots + t_k = t} G_{\lambda_1}(t_1) \cdots G_{\lambda_k}(t_k) \ dV\right|
\\
\leq
\int_{t_1 + \dots + t_k = t,\ t_i \geq 0} |\exp(\lambda_1 t_1)| \cdots |\exp(\lambda_k t_k)| \ dV
\\
\leq
\int_{t_1 + \dots + t_k = t,\  t_i \geq 0} \exp(\Lambda t) \ dV
\\
=
C t^{k-1} \exp(\Lambda t).
$$
Here $Ct^{k-1}$ is the volume of the $(k-1)$-simplex we're integrating over. Since $k \leq n$, all the convolutions involved in our expression are bounded above by $C (1 + t^{n-1}) \exp(\Lambda t)$
Claim 2:
For any $n$ and any $K$ there is a constant $C$ for which: Given $A \in M_{n \times n}(\mathbb C)$ with $\|A\| < K$, we have $\| \exp(tA) \| \leq C (1+t^{n-1}) \exp(t \Lambda(A))$, where $\Lambda(A)$ is the largest real part of an eigenvalue of $A$.
Proof:
Write $A$ as a Schur decomposition form $U L U^*$, with $L$ lower triangular and $U$ unitary. Note that $\| L \| = \| A \|$ and $\Lambda(L) = \Lambda(A)$ (call this $\Lambda$).
By the above, there is a fixed expression for the entries of $\exp(tL)$ as a linear combination of products of entries of of $L$ and $k$-wise convolutions of the $G_{\lambda_i}$. Since we have a fixed bound $K$ on $\| L \|$ we have a fixed bound on the entries $l_{ij}$ of $L$. Hence we have a fixed bound on all the products of those entries involved in the expression.
So every coefficient $\exp(tL)$ is bounded above by $C (1+t^{n-1}) \exp(\Lambda t)$. (With a different $C$ from above.)
So, $\| \exp(tL) \|$ is bounded above by that expression as well. (Again, with a different $C$).
Hence,
$$
\| \exp(tA) \| = \|U \exp(tL) U^*\| = \| \exp(tL) \|
\\
\leq  C (1+t^{n-1}) \exp(\Lambda t)
$$
$\blacksquare$
It follows that we can define a continuous function $C(A)$ as in the problem statement, e.g. interpolate between the values of $C$ corresponding to $K = 1, 2, 3, ...$.
Or I think tracking the bounds on coefficients in the argument more carefully can directly give you something like $C(A) = (1 + \|A\|^{n-1})C$.
