Show that $e^A$ is normal whenever $A$ is normal for $A\in\mathcal{M}_{n\times n}(\mathbb{R})$ 
Show that $e^A$ is normal whenever $A$ is normal for $A\in\mathcal{M}_{n\times n}(\mathbb{R})$ 

Suppose that $A$ is normal so that $AA^*=A^*A$. We wish to show that $e^A$ is normal, i.e. we wish to show that 
$$\sum_{j=0}^\infty \dfrac{A^j}{j!}$$
is normal. I am having trouble even figuring out where to begin on this proof. Any help would be appreciated.
 A: One approach: If $A$ is diagonalized by the unitary $U$, then so is $\exp(A)$.
Or if you want to do it directly: 
$$\exp(A^*) = \sum_{n=0}^\infty (A^*)^n/n! = \exp(A)^*$$
and each term here commutes with each term in $\exp(A)$.
A: $e^A (e^A)^* = \sum_i \sum_j {1 \over i!} {1 \over j!} A^i (A^j)^* = \sum_i \sum_j {1 \over i!} {1 \over j!}  (A^j)^* A^i = (e^A)^* e^A $.
A marginally different approach:
More generally, if $[A,B] = 0$ then $e^A e^B = e^{A+B}$ and so $e^A$ and $e^B$
commute.
Since $(e^A)^* = e^{A^*}$, and $[A,A^*] = 0$ we get the desired result.
A: Let $m_A(x) = (x-\lambda_1)\cdots(x-\lambda_k)$ be the minimal polynomial of $A$.
For any analytic function $f$ we have
$$f(A) = f(\lambda_1)P_1 + \cdots + f(\lambda_k)P_k$$
for some fixed matrices $P_1, \ldots, P_k$. Plugging in $f(x) = \prod_{j \ne i}(x - \lambda_i)$ we get $$P_i = \prod_{j \ne i} \frac{A-\lambda_j I}{\lambda_i - \lambda_j}$$
so for $f(x) = \exp x$ we get $$\exp A = e^{\lambda_1}\left(\prod_{j \ne 1} \frac{A-\lambda_j I}{\lambda_i - \lambda_j}\right) + \cdots + e^{\lambda_k}\left(\prod_{j \ne k} \frac{A-\lambda_j I}{\lambda_i - \lambda_j}\right)$$
In particular, $\exp A$ is a polynomial in $A$, so it is normal because polynomials of normal matrices are normal.
