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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.

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    Unitary operators are normal, so I believe the spectral theorem applies to them (I'm talking about the general spectral theorem here, not the compact one). Hence any unitary operator can be considered as multiplication by some measurable $\Bbb C$-valued function which has norm $1$ almost everywhere on some measure space. But the function must also be positive, hence $1$. So it must be the identity. There is undoubtedly a more elementary way to see this though. – 2017-02-14
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    Positive operators are symmetric, so the general spectral theorem definitely applies. Also, this means $I=U^{*}U=U^{2},$ so all eigenvalues are $\pm 1,$ and invoking positivity again, for any eigenvector with corresponding eigenvalue $\lambda,$ $0\leq\langle Ux,x\rangle=\lambda\langle x,x\rangle,$ so $\lambda\geq0,$ and thus is 1. – 2017-02-14

3 Answers 3

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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.

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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 $U=I$.

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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.