# Eigenvalues of Normal Operator have Orthogonal Eigenspaces

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## 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

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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.

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**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:

\(\ds \norm {\mathbf {T v} }^2\) | \(=\) | \(\ds \innerprod {\mathbf{T v} } {\mathbf{T v} }\) | Definition of Inner Product Norm | |||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod {\mathbf T^* \mathbf {T v} } {\mathbf v}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod {\mathbf T \mathbf T^* \mathbf v} {\mathbf v}\) | Definition of Normal Operator | |||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod {\mathbf T^* \mathbf v} {\mathbf T^* \mathbf v}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm {\mathbf T^* \mathbf v}^2\) | Definition of Inner Product Norm |

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

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Thus:

\(\ds \mathbf {T v}\) | \(=\) | \(\ds \lambda \mathbf v\) | ||||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds \bszero\) | \(=\) | \(\ds \norm {\paren {\mathbf T - \lambda \mathbf I} \mathbf v}\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm {\paren {\mathbf T - \lambda \mathbf I}^* \mathbf v}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm {\mathbf T^* \mathbf v - \overline \lambda \mathbf v}\) | ||||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds \mathbf T^* \mathbf v\) | \(=\) | \(\ds \overline \lambda \mathbf v\) |

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:

\(\ds \lambda_1 \innerprod {\mathbf v_1} {\mathbf v_2}\) | \(=\) | \(\ds \innerprod {\lambda_1 \mathbf v_1} {\mathbf v_2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod {\mathbf T \mathbf v_1} {\mathbf v_2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod {\mathbf v_1} {\mathbf T^* \mathbf v_2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod {\mathbf v_1} {\overline{\lambda_2} \mathbf v_2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \lambda_2 \innerprod {\mathbf v_1} {\mathbf v_2}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \bszero\) | \(=\) | \(\ds \paren {\lambda_1 - \lambda_2} \innerprod {\mathbf v_1} {\mathbf v_2}\) |

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.

$\blacksquare$

## Sources

- 1990: John B. Conway:
*A Course in Functional Analysis*(2nd ed.) ... (previous) ... (next): $\text{II}.5.7$