# A compact normal operator is diagonalisable.

Consider the following proof in Murphy’s book: "$$C^*$$-algebras and operator theory":

I try to understand why $$u(K^perp) subseteq K^perp$$. Equivalently, one can prove $$u^*(K) subseteq K$$.

I tried to calculate $$langle u(x), y rangle$$ with $$x in K^perp, y in K$$ and I hoped to calculate that this is equal to $$0$$ but I must be missing something.

Mathematics Asked by user745578 on November 12, 2021

At first observe that if $$u$$ is normal and $$u(x) = lambda x$$, then $$u^*(x) = overline{lambda} x$$ (it is a simple exercise).

Now we can prove the following proposition: if $$u$$ is normal, $$E$$ is arbitrary set of eigenvectors of $$u$$, and $$K = overline{span(E)}$$, then $$K^perp$$ is $$u$$-invariant. Indeed, $$x in K^perp$$ iff $$langle x, a rangle = 0$$ for all $$a in E$$. Thus, for all $$a in E$$, $$langle u(x), a rangle = langle x, u^*(a) rangle = 0$$ (by the foregoing observation). Therefore, $$u(x) in K^perp$$.

$$mathbf{Edit}$$: proof of the observation.

It is obvious that operator $$v = u - lambda I$$ is normal, since $$v^* = u^* - overline{lambda}I$$. Let $$v(x) = 0$$. Then $$0 = langle v(x), v(x) rangle = langle v^*(v(x)), x rangle = langle v(v^*(x)), x rangle = langle v^*(x), v^*(x) rangle = 0$$ Thus, $$v^*(x) = 0$$, i.e. $$u^*(x) = overline{lambda}(x)$$.

Answered by Matsmir on November 12, 2021

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