convergence of matrix exponential as matrix converges entrywisely

Question

I wonder that whether $\lim_{n\rightarrow\infty}e^{A_n}=e^A$ or not, if the matrix $A_n$ converges to $A$ entrywise. if not, does the convergence holds under some manner of convergence for $A_n\rightarrow A$?

Answer 1

By continuity of norm, for $n$ large enough, we have $\left \|A_n \right \| \leq 1+\left \|A \right \|$. One can show by induction that $$\left \| A^k-A_n^k \right \| \leq \left \|A-A_n \right \| \sum_{i=0}^{k-1}\left \|A \right \|^i \left \|A_n \right \|^{k-1-i} \leq k \left \|A-A_n \right \| (1+\left \|A \right \|)^k. $$ By definition, one has $$\left \| e^A-e^{A_n} \right \| = \left \| \sum_{k=0}^\infty \frac{1}{k!}(A^k-A_n^k) \right \| \leq \sum_{k=0}^\infty \frac{k}{k!} \left \|A-A_n \right \| (1+\left \|A \right \|)^k \leq \left \|A-A_n \right \|(1+\left \|A \right \|)e^{1+\left \|A \right \|},$$ which proves that $e^{A_n} \rightarrow e^{A}$ if $A_{n} \rightarrow A$.