Eigenvalues of Normal Operator have Orthogonal Eigenspaces

Theorem
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:
 * $\ker \left({A - \lambda}\right) \perp \ker \left({A - \mu}\right)$

where:
 * $\ker$ denotes kernel
 * $\perp$ denotes orthogonality.

Proof
Let $\mathcal V$ be an inner product space.

Let $T: \mathcal V \to \mathcal V$ 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


 * $\left\langle{T u, w}\right\rangle = \left\langle{T^* w}\right\rangle$

It is important to note the existence and uniqueness of adjoint operators.

Claim: We know that for $v \in \mathcal V$:
 * $T v = \lambda v \iff T^* v = \overline \lambda v$

This is true because for all normal operators, by definition:
 * $T^* T = T T*$

and so:

Since for normal $T$, $\left({T - \lambda I}\right)$ is normal, we have:

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

Since $\lambda_1 \ne \lambda_2$, this is only possible if $\left\langle{v_1, v_2}\right\rangle = 0$, which means the eigenvectors of our normal operator are orthogonal.