It seems that there is no way of computing
$$ \lVert e^{tA}\lVert $$
for an arbitrary real square matrix $A$ and any matrix norm $\lVert\cdot\lVert$, in terms of $t \in \mathbb{R}$ and $\lVert A \lVert$, i.e.
$$f:(t,\lVert A \lVert ) \mapsto \lVert e^{tA}\lVert$$
without explicitly performing the matrix exponential (exactly or approximately). Is that right?
Because so far I could only find upper-bound estimates. If this problem in its full generality is not solved, can anyone say something about its solvability?
For my particular case it would suffice to restrict to the case
$$A=A^\mathrm{T}, \quad \mathrm{tr}(A) = 0$$
meaning $\left(e^A\right)^\mathrm{T} = e^A$ and $\det(e^A) = 1$.
Thanks for any suggestions.
edit: since it is very unlikely that such an $f$ exists, I think the requirement should be eased to
$$f:(t, A ) \mapsto \lVert e^{tA} \lVert$$
just somehow circumventing exponentiation.
Best Answer
In your case (symmetric matrix) the matrix can be diagonalized and the exponential be computed:
$$A = U D U^T$$
with a orthogonal matrix $U$ and diagonal matrix $D$ containing the eigenvalues of A.
Since then
$$e^{At} = U e^{Dt} U^T$$
we get
$$||e^{At}|| = ||e^{Dt}||$$
Since the matrix norm $||M||_2$ is the maximum absolute eigenvalue of a symmetric matrix $M$ we have in fact
$$||e^{At}||_2 = \max_i( e^{\lambda_i t}) = e^{||D||t} = e^{||A||t}$$
Does this make sense?
I guess the property $||e^{At}|| = e^{||A||t}$ holds for any normal matrix, but for the non-normal case it is probably false.