====== Proof of the Spectral Theorem for Normal Operators in Finite Dimensions ======

As a reminder, a //**normal operator**// is an operator $\hat{A}$ that satisfies $[\hat{A},\hat{A}^{\dagger}] = 0$, and the spectral theorem is as follows.

**Theorem**
Let $a_1,a_2,\cdots$ be the (distinct) eigenvalues of an operator $\hat{A}$.  Then, $\hat{A}$ is a normal operator if an only if all of the following hold:
  - Its eigenspaces are orthogonal: $\hat{P}_{a_j}\hat{P}_{a_k} = \delta_{jk}\hat{P}_{a_j}$.
  - Its eigenspaces are //**complete**//: $\sum_j \hat{P}_{a_j} = \hat{I}$.
  - The //**spectral decomposition**// holds: $\hat{A} = \sum_j a_j \hat{P}_{a_j}$.


**Proof**
The if direction is easy, so we will prove it first. by the spectral decomposition $\hat{A} = \sum_j a_j \hat{P}_{a_j}$ and  $\hat{A}^{\dagger} = \sum_j a_j^* \hat{P}_{a_j}$, where we have used the fact that projection operators are Hermitian, so $\hat{P}_{a_j}^{\dagger} = \hat{P_{a_j}}$.  Hence, we have
\begin{align*}
[\hat{A},\hat{A}^{\dagger}] & = \left [ \sum_j a_j \hat{P}_{a_j}, \sum_k a_k^* \hat{P}_{a_k}\right ] \\
& = \sum_{jk} a_j a_k^{*} \left [ P_{a_j}, P_{a_k} \right ].
\end{align*}
Now, we will show that $\left [ P_{a_j}, P_{a_k} \right ] = 0$ for all $j,k$.  From this it follows that all the terms in the sum are zero, and hence $[\hat{A},\hat{A}^{\dagger}] = 0$ so the operator is normal.

To see this, we will use the orthogonality property $\hat{P}_{a_j}\hat{P}_{a_k} = \delta_{jk}\hat{P}_{a_j}$.  Specifically,
\begin{align*}
[P_{a_j}, P_{a_k}] & = P_{a_j}P_{a_k} - P_{a_k}P_{a_j} \\
& = \delta_{jk}P_{a_j} - \delta_{kj}P_{a_k} \\
& = \delta_{jk}P_{a_j} - \delta_{jk}P_{a_j} \\
& = 0.
\end{align*}
In the third line, we used the fact that $\delta_{kj} = \delta_{jk}$ and, since $\delta_{jk}$ is only nonzero when $j=k$, $\delta_{jk}P_{a_k} = \delta_{jk}P_{a_j}$.