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I want to find a bound of the norm of matrix exponential $||e^{At}||\leq Ke^{\alpha t}$. However, the matrix $A$ has the following form

$$ A = \left[ \begin{matrix} 0 & A_1 & A_2 \\ A_3 & A_4 & 0 \\ A_5 & 0 & A_6 \end{matrix}\right] $$

Where all $A_i$ are $3\times 3$ matrices. Is there a way to find $K$ and $\alpha$ using characteristics, like eigenvalues, of all $A_i$ matrices?

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$K$ has no interest. The condition about $\alpha$ is $\alpha >M=\max_{\lambda\in spectrum(A)}(Re(\lambda))$.

When $n=1$, according to Gerschgorin, $M\leq \max(|a_1|+|a_2|,Re(a_4)+|a_3|,Re(a_6)+|a_5|)$.

If $U\in M_n$, then let $r(U)= \max_{\lambda\in spectrum(U)}(Re(\lambda))$ and $||U||=Norm(U,2)$.

Conjecture. $M\leq \max(||A_1||+||A_2||,r(A_4)+||A_3||,r(A_6)+||A_5||)$.

This conjecture works for randomly chosen matrices. I have no counter-examples.

EDIT. If "Conjecture" is true, then take $\alpha=\max(||A_1||+||A_2||,r(A_4)+||A_3||,r(A_6)+||A_5||)+\epsilon$ with $\epsilon >0$ as small as you want. Some value of $K$ depends on the values of $||e^{tA}||$ for small values of $t$. In general, $K$ is useless.

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  • $\begingroup$ Thanks for your answer but I have a question. If $\alpha > M$ and $M \leq \max(...)$, how do I build an upper bound for $||e^{At}||$ or, specifically, an upper bound for $\alpha$? $\endgroup$ – GilbertoD Mar 20 '18 at 17:42

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