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A compact normal operator is diagonalisable.

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

lis

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

1 Answers

One Answer

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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