In an infinite dimensional Hilbert space the orthogonal complement of the (closure) of the direct sum of eigenspaces of a compact normal operator is finite dimensional. Why is this the case?


  • $\begingroup$ Because the span of the zero vector is zero-dimensional and $0<\infty$ ;) $\endgroup$
    – Neal
    Feb 10, 2014 at 12:59

1 Answer 1


More is true, in fact. If $H$ is a Hilbert space, and $T\colon H \to H$ a compact normal operator, then $H$ is the closure of the direct sum of the eigenspaces of $T$.

For $\lambda \in \sigma_P(T)$, let $E_\lambda$ be the eigenspace of $T$ corresponding to the eigenvalue $\lambda$, and let $S = \bigoplus\limits_{\lambda\in\sigma_P(T)} E_\lambda$. Then $S$ is clearly a $T$-invariant as well as $T^\ast$-invariant subspace of $H$, and hence so is $E = \overline{S}$. Since $T$ is normal, $N = E^\perp$ is $T$-invariant too, for $n\in N$ and $e\in E$, we have

$$\langle Tn,e\rangle = \langle n, T^\ast e\rangle = 0$$

since $T^\ast e\in E$.

Let $R\colon N \to N$ be defined by $Rn = Tn$, i.e. $R$ is the restriction of $T$ with its codomain also restricted to $N$. Then $N$ is also a Hilbert space, and $R$ is a compact normal operator. By construction, $R$ has no eigenvectors, since an eigenvector of $R$ is also an eigenvector of $T$.

But a compact normal operator on a nontrivial ($\neq \{0\}$) Hilbert space always has an eigenvector (Rudin, FA, Thm 12.31 (a)), so we must have $N = \{0\}$.

  • $\begingroup$ What happens if 0 is an eigenvalue? Then the eigenspace of 0 (the kernel of T) could be infinite dimensional. Then a basis of eigenvectors may not exist? $\endgroup$
    – user108605
    Feb 10, 2014 at 15:55
  • 1
    $\begingroup$ That is no problem. Even if $\ker T$ is infinite-dimensional, for the direct sum of eigenspaces, we don't need bases on the one hand, and on the other, the kernel is a closed subspace, so itself a Hilbert space, and it has Hilbert bases. $\endgroup$ Feb 10, 2014 at 17:29

You must log in to answer this question.