Unitary operator is identity

Question

I want to prove that the only positive unitary operator is the identity operator.

My attempt has been to been to consider a negative eigenvalue of this linear operator and yield a contradiction. But the operator may not even have an eigenvalue.

Answer 1

As ride the wavelet has pointed out, positive operators are self adjoint. Thus $U^2=U^*U=I$.

Now fix some $x$ in our Hilbert space. If $x\neq Ux$ then $x-Ux$ is an eigenvector of $U$ with eigenvalue $-1$. This contradicts positivity.

Answer 2

Since $U$ is normal, $U-I$ is normal, hence

norm of $U-I$ = spectral radius of $U-I$.

From $U^2=I$ we see that the spctrum of $U$ = $\{1\}$. Hence the spectral radius of $U-I$ is $=0$. This gives $U=I$.

Answer 3

The spectrum of a unitary lies in the unit circle. The spectrum of a positive operator lies on $[0,\infty)$. So the spectrum of a positive unitary is $\{1\}$, the single intersection point. The only normal operator with spectrum $\{1\}$ is the identity.