$U=I+C$ is a unitary operator on a Hilbert space $H$, where $C$ is compact operator. Then eigenvectors of operator $U$ creates a orthonormal basis for the Hilbert space $H$.

It is classical theorem that the above fact is correct for the normal compact operators.

My tries:

If $U$ is unitary then $U^*U=I$, then we conclude: $C+C^*+CC^*=0$. Therefore, $CC^*=C^*C$. It means that $C$ is compact and normal. Therefore there is a complete eigenvectors of operator $C$ for Hilbert space $H$. Eigenvectors of $U$ and $C$ are the same. Hence the proof completes.

Is there other approches?

  • 1
    $\begingroup$ You actually use both $U^{\ast}U=I$ and $UU^{\ast}=I$ but besides that I think that there is no simpler argument. $\endgroup$ – Mateusz Wasilewski Nov 24 '17 at 16:46

I don't think there's a radically different approach. But it can be simplified just a bit: mainly, in that you don't need that $U$ is a unitary, only that it is normal. And then just the fact that $I-A$ is normal if $A$ is.

In summary, with exactly the same effort, one can show that if $A$ is normal and $A=I+C$ with $C$ compact, then $A$ admits an orthonormal basis of eigenvectors.


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