Eigenvalues of Normal Operator have Orthogonal Eigenspaces

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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:

\(\displaystyle \left\Vert{T v}\right\Vert^2\) \(=\) \(\displaystyle \left\langle{T v, T v}\right\rangle\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\langle{T^* T v, v}\right\rangle\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\langle{T T^* v, v}\right\rangle\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\langle{T^* v, T^*v}\right\rangle\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\Vert{T^* v}\right\Vert^2\) $\quad$ $\quad$


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


\(\displaystyle T v\) \(=\) \(\displaystyle \lambda v\) $\quad$ $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle \left\Vert{\left({T - \lambda I}\right) v}\right\Vert\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle \left\Vert{\left({T - \lambda I}\right)^* v}\right\Vert\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle \left\Vert{T^* v - \overline \lambda v}\right\Vert\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle T^* v\) \(=\) \(\displaystyle \overline \lambda v\) $\quad$ $\quad$


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:

\(\displaystyle \lambda_1 \left\langle{v_1, v_2}\right\rangle\) \(=\) \(\displaystyle \left\langle{\lambda_1 v_1, v_2}\right\rangle\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\langle{T v_1, v_2}\right\rangle\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\langle{v_1, T^* v_2}\right\rangle\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\langle{v_1, \overline{\lambda_2} v_2}\right\rangle\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lambda_2 \left\langle{v_1, v_2}\right\rangle\) $\quad$ $\quad$


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.

$\blacksquare$


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