Eigenvalues of Normal Operator have Orthogonal Eigenspaces

Theorem
Let $\HH$ be a Hilbert space.

Let $\mathbf T: \HH \to \HH$ be a normal operator.

Let $\lambda_1, \lambda_2$ be distinct eigenvalues of $\mathbf T$.

Then:
 * $\map \ker {\mathbf T - \lambda_1} \perp \map \ker {\mathbf T - \lambda_2}$

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

Proof
Requisite knowledge: $\mathbf T^*$ is the adjoint of $\mathbf T$ and is defined by the fact that for any $\mathbf u, \mathbf w \in \HH$, we have


 * $\innerprod {\mathbf {T u} } {\mathbf w} = \innerprod {\mathbf u} {\mathbf T^* \mathbf w}$

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

Claim: We know that for $\mathbf v \in \HH$:
 * $\mathbf {T v} = \lambda \mathbf v \iff \mathbf T^* \mathbf v = \overline \lambda \mathbf v$

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

and so:

Since $\mathbf T$ is normal, $\mathbf T - \lambda \mathbf I$ is also normal.

Thus:

Let $\mathbf v_1$ and $\mathbf v_2$ be non-zero eigenvectors of $\mathbf T$ with corresponding eigenvalues $\lambda_1$ and $\lambda_2$, respectively.

Then:

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