Suppose $T$ is diagonalizable in $\mathbb{C}$. Show $e^T = \sum_{\lambda \in sp(T)} e^\lambda P_\lambda$ is the matrix exponential series. 
Suppose $T$ is diagonalizable in $\mathbb{C}$ or in $\mathbb{R}$. Show 
  $$e^T = \sum_{\lambda \in sp(T)} e^\lambda P_\lambda$$
  is the same as the exponential series
  $$e^T = \sum_{k=0}^\infty \frac{1}{k!} T^k.$$

For an attempt, I probably want to somehow use $T^k = \sum_\lambda \lambda^k P_\lambda$ but I don't see the connection between this and the matrix exponential. Is there an obvious relation that I'm missing?
 A: If $T$ is a matrix, $e^T$ is defined as the matrix series $$e^T = \lim_{n\rightarrow \infty} \sum_{k=1}^n\frac{1}{k!}T^k$$
Since we can use $T^k = \sum_{\lambda \in \sigma(T)} \lambda^k P_\lambda$ (where this sum is a finite sum - for an $n\times n$ matrix, this has at most $n$ summands), then we have:
$$\sum_{k=1}^n \frac{1}{k!}T^k = \sum_{k=1}^n \frac{1}{k!} \left( \sum_{\lambda \in \sigma(T)} \lambda^k P_\lambda\right) =   \sum_{\lambda \in \sigma(T)} \left(\sum_{k=1}^n \frac{1}{k!} \lambda^k\right) P_\lambda,$$
taking the limit as $n \rightarrow \infty$, the bracketed sum tends to $e^\lambda$.
A: Hint: make the diagonalisation explicit with a rotation $U$ and its inverse, then do the calculation in a basis where the matrix is actually diagonal (so that projections are coordinate projections and matrix powers are just powers of diagonal elements).
A: Hint: Plug $T^k$ in the definition of $e^T$ and use the fact that
$$\sum_{k=0}^\infty \frac{1}{k!} \lambda^k P_\lambda=\left(\sum_{k=0}^\infty \frac{1}{k!} \lambda^k \right)P_\lambda=e^{\lambda} P_\lambda$$ 
