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

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:

\(\ds \norm {\mathbf {T v} }^2\) \(=\) \(\ds \innerprod {\mathbf{T v} } {\mathbf{T v} }\) 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.


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.

$\blacksquare$


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