Eigenvalues of Normal Operator have Orthogonal Eigenspaces

Let $H$ be a Hilbert space.

Let $A \in B \left({H}\right)$ be a normal operator.

Let $\lambda, \mu$ be distinct eigenvalues of $A$.

Then $\operatorname{ker} \left({A - \lambda}\right) \perp \operatorname{ker} \left({A - \mu}\right)$.

Here $\operatorname{ker}$ denotes kernel, and $\perp$ signifies orthogonality.

Proof
Let $\mathcal{V}$ be an inner product space and $T : \mathcal{V} \rightarrow \mathcal{V}$ be a normal linear operator.

Requisite knowledge: $T^*$ is the adjoint of $T$ and is defined by the fact that for any $u, w \in \mathcal{V}$, we have

\begin{equation*}  = < ,T^*w> \end{equation*}

It is important to note the existence and uniqueness of adjoint operators. If you don't know about them, learn about them and come back!

Claim: We know that for $v \in \mathcal{V}$, \[ Tv = \lambda v \Leftrightarrow T^*v = \overline{\lambda} v. \] This is true because for all normal operators, by definition $T^* T = T T*$ and so

\[ \]
 * Tv||^2 =  =  = T T^* v, v> =  = ||T^* v||^2

and since for normal $T$, $(T - \lambda I)$ is normal (Check it! It's true!), we have

\begin{align*} Tv = \lambda v &\Leftrightarrow ||(T - \lambda I) v|| = 0 \\ & \Leftrightarrow ||(T - \lambda I)^* v|| = 0 \\ & \Leftrightarrow ||T^*v - \overline{\lambda}v|| = 0\\ & \Leftrightarrow T^*v = \overline{\lambda} v. \end{align*}

Now, if $Tv_1 = \lambda_1 v_1$ and $Tv_2 = \lambda_2 v_2$, where $\lambda_1 \ne \lambda_2$ and $v_1, v_2$ are eigenvectors (i.e. $v_1, v_2 \ne \vec{0}$), we have

\begin{align*} \lambda_1  & = <\lambda_1 v_1, v_2>\\ &= \\ &= \\ & = \\ &= \lambda_2 .\\ \end{align*}

Since $\lambda_1 \ne \lambda_2$, this is only possible if $ = 0$, which means the eigenvectors of our normal operator are orthogonal! Whoopdeedoo!