A property about normal operator Let $H$ be a Hilbert space, $L$ be a normal operator (i.e. $LL^*=L^*L$).
Prove that there exist a unitary operator $U$ such that $L^*=UL$.
My approach is to first define $U$ on the $\operatorname{range}(L)$ and try to extend it to the whole space. But it seems not easy to extend. To respond to the comments:


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*I do not know the spectral theorem for normal operators. I only know the spectral theorem for compact operators and bounded self-adjoint operators.

*I have tried to use that $H=\overline{\operatorname{ran}}(L)\oplus\ker(L)$ and extended $U$ by the identity on $\ker(L)$, but failed.

 A: Elaborating on my comment:
On $\mathrm{ran}(L)$ define $U$ by $ULx:=L^\ast x$. From $L^\ast L=LL^\ast$ it follows easily that $\|Lx\|=\|L^\ast x\|$ for all $x\in H$, in particular, $Lx=Ly$ implies $L^\ast x=L^\ast y$ for all $x,y\in H$.
Thus, $U$ is well-defined on $\mathrm{ran}(L)$, has image $\mathrm{ran}(L^\ast)$, and $\|ULx\|=\|L^\ast x\|=\|Lx\|$ for all $x\in H$. Hence it extends to a unitary operator $\overline{\mathrm{ran}}(L)\to\overline{\mathrm{ran}}(L^\ast)$.
Further notice that $\|Lx\|=\|L^\ast x\|$ for $x\in H$ implies $\ker L=\ker L^\ast$ and consequently $\overline{\mathrm{ran}}(L^\ast)=(\ker L)^\perp=(\ker L^\ast)^\perp=\overline{\mathrm{ran}}(L)$. Thus, $U$ is a unitary operator $\overline{\mathrm{ran}}(L)\to\overline{\mathrm{ran}}(L)$.
Now $H=\overline{\mathrm{ran}}(L)\oplus \ker L^\ast$. Define $U$ to be the identity on $\ker L^\ast$. Then $U$ is clearly surjective on $H$, and $\|Ux\|=\|x\|$ for all $x\in H$ follows easily from the orthogonality of the decomposition of $H$. Hence $U$ is unitary (more generally, unitary operators $U_i\colon H_i\to H_i$ induce a unitary operator $U_1\oplus U_2\colon H_1\oplus H_2\to H_1\oplus H_2$).
The relation $L^\ast=UL$ was obviously built into the definition of $U$.
