The problem is this:
Let A be a square $n \times n$ matrix, and define $$e^A=\sum_{k=0}^\infty \frac{1}{k!}A^k$$
Find a bound for $\lvert e^A \rvert$ in terms of $\lvert A \rvert$ and $n$.
I was able to show that this converges. I know that once $k \ge \lvert A \rvert$ , the values of $\left \lvert \frac{A^k}{k!}\right \rvert$ are going to start decreasing. So I'd imagine the strategy would be to take the partial sum up to $k \ge \lvert A \rvert$, and then then find some upper bound for the decreasing series that remains. I'm not sure where to go from here.
Additionally, I'm struggling to see the relevance of the dimension of the matrix. I mean, I could bound $\lvert A \rvert$ by $\sqrt{c^2n^2}$ if I take c = the max absolute value in A, but since I'm already expressing the result in the norm of A this doesn't seem like it would be a helpful thing to do.
Best Answer
I'm also not sure what $n$ has to do with anything. The simplest thing that comes to mind is
$$\left | e^A \right | \leq \sum_{k=0}^\infty \left | \frac{A^k}{k!} \right | \leq \sum_{k=0}^\infty \frac{|A|^k}{k!} = e^{|A|}.$$
I'm not sure how much better a bound could possibly get, because for a diagonal matrix whose largest entry is positive (for example $A=I$), this bound is tight.