Proof Assistance In C* Algebras By Murphy

Question

I'm going over the following theorem and I'm stuck on the last step. The theorem is the following:

Theorem: If $u$ is a compact normal operator on a Hilbert space $H$, then it is diagonalisable.

Proof: By Zorn's lemma there exists a maximal orthonormal set $E$ of eigenvectors of $u$. If $K$ is the closed linear span of $E$, then $H = K \bigoplus K^{\perp}$, and $u(K) \subset K$ and $u(K^{\perp}) \subset K^{\perp}$. The restriction $u_{\perp}: K^{\perp} \longrightarrow K^{\perp}$ is compact and normal, and $\sigma(u|_{\perp}) \subset \sigma(u)$. By maximality of $E$, and using the fact that the spectrum of a compact operator is it's eigenvalues, we have that $\sigma(u|_{\perp}) = \{0\}$.

Since $u|_{\perp}$ is normal, $||u|_{\perp}|| = r(u|_{\perp}) = 0$, so $u|_{\perp} = 0$. Hence $K^{\perp} = 0$, so $K=H$.

My question is, why does $u|_{\perp} = 0$ imply that $K^{\perp}= \{0\}$?

Thank you.

Answer 1

$u|_{K^{\perp}}=0$ implies that, for all $v\in K^{\perp}$, $$u|_{K^{\perp}}v=0=0v$$ Thus $v$ is an eigenvector for $u|_{K^{\perp}}$, and so an eigenvector for $u$.

But $K^{\perp}$ is perpendicular to every eigenvector of $u$, so $v$ is perpendicular to itself. Thus $v=0$.