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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.

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  • $\begingroup$ For symmetric $A$ this has been answered by Andreas H. The general question is still open. $\endgroup$ Commented Mar 29, 2017 at 10:51
  • $\begingroup$ Which matrix norm are you using? $\endgroup$ Commented Mar 30, 2017 at 21:23
  • $\begingroup$ I use the Frobenius norm. $\endgroup$ Commented Mar 31, 2017 at 8:47

1 Answer 1

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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.

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  • $\begingroup$ I can't believe I could not see this myself. Thanks! $\endgroup$ Commented Mar 29, 2017 at 10:49
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    $\begingroup$ Actually this is wrong. Notice that $e^{\|D\|t} = \max_i e^{|\lambda_i|t}$ which is not the same as $e^{\|D\|t} = \max_i e^{\lambda_it}$. So this proof is only true when the largest positive eigenvalue has a greater absolute value than the greatest negative eigenvalue. Otherwise it's definitely not the case that $e^{\|A\|t} = \|e^{At}\|$. $\endgroup$
    – shalop
    Commented Jul 4, 2017 at 6:02

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