Nonzero Eigenvalue of Compact Operator has Finite Dimensional Eigenspace

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Theorem

Let $H$ be a Hilbert space.

Let $T \in \map {B_0} H$ be a compact operator.

Let $\lambda \in \map {\sigma_p} T, \lambda \ne 0$ be a nonzero eigenvalue of $T$.


Then the eigenspace for $\lambda$ has finite dimension.


Proof


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$\def \sequence#1{\left({#1}\right)}$

Note that in the following, the notation for a sequence of terms in the form $e_n$ has been changed for this particular page to be $\sequence {e_n}$ rather than the $\mathsf{Pr} \infty \mathsf{fWiki}$ standard $\left\langle{e_n}\right\rangle$.

This is in order to avoid potential confusion with the notation $\innerprod {\lambda e_{n_k} } {\lambda e_{n_p} }$ for inner product.


Assume that there is an infinite orthonormal sequence $\sequence {e_j}$ in my eigenspace for the eigenvalue $\lambda$.

That is:

$\sequence {e_j: j \in J} \subset \map \ker {T - \lambda}$

Since $\norm {e_j} = 1$, then $\sequence {T e_n}$ will be in the closure image of $T$ of the closed unit ball.

Since the closure of this image is compact by assumption, then there is a subsequence $\sequence {T e_{n_k} }$ that converges, hence $\sequence {T e_{n_k} }$ is Cauchy.

Notice that

$\norm {T e_{n_k} - Te_{n_p} }^2 = \norm {\lambda e_{n_k} - \lambda e_{n_p} }^2 = 2 \size \lambda^2 > 0$

for $k \ne p$, we are using the fact that $\lambda \ne 0$ and $\innerprod {\lambda e_{n_k} } {\lambda e_{n_p} } = 0$ (since $\sequence {e_j}$ is an orthonormal sequence).

Since $\lambda$ is constant, then it contradicts the fact that $\sequence {e_{n_k} }$ is Cauchy.

Thus $\map \ker {T - \lambda}$ is finite-dimensional.

$\blacksquare$


Sources

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