trace and operator norm of $\exp^A$ If $A$ is an n by n complex matrix,
1.How to compute $tr(\exp^A)$ .Can we use the Taylor expansion as following:
$\exp^A=\sum_{k=0}^{\infty}\frac{A^k}{k!},$then $tr(\exp^A)=\sum_{k=0}^{\infty}tr(\frac{A^k}{k!})$.
2.How to compute the operator norm of $\exp^A$?

The trace of an finite-dimensional operator is defined the sum of its diagonal elements, which in turn equals the sum of its eigenvalues. So let $\{\lambda_j \in \mathbb{C}\ | j\in\{1,\ldots,n\}\}$ the set of eigenvalues of $A$ then we have 
$$\mathrm{tr}{A}=\sum_{j=1}^n A_{jj} = \sum_{j=1}^n \lambda_{j}$$. Because the trace is invariant w.r.t. similarity transforms of $A$, we have $$\mathrm{tr}{(\exp{A})} = \sum_{j=1}^n \exp{\lambda_{j}} .$$
