Let \(V\) be a finite-dimensional inner product space. In Exercise 6, we showed how to construct some linear operators on \(V\) which are both self-adjoint and unitary. Now prove that there are no others, i.e., that every self-adjoint unitary operator arises from some subspace \(W\) as we described in Exercise 6.
Solution
Let \(U\) be both self-adjoint and unitary. Then \[ U^*=U \] and \[ U^*U=I, \] so \[ U^2=I. \]
1. Construct a projection from \(U\).
Define \[ E=\frac12(I+U). \] Using \(U^2=I\), \[ E^2 =\frac14(I+2U+U^2) =\frac12(I+U) =E. \] Also, since \(U^*=U\), \[ E^*=\frac12(I+U^*)=E. \] Thus \(E\) is an idempotent self-adjoint operator.
2. A self-adjoint projection is the orthogonal projection onto its range.
Put \[ W=\operatorname{range}E. \] Because \(E^2=E\), the operator \(E\) acts as the identity on \(W\). If \(x\in\ker E\) and \(w=Ey\in W\), then \[ (x\mid w) =(x\mid Ey) =(E^*x\mid y) =(Ex\mid y) =0. \] Hence \[ \ker E\subseteq W^\perp. \] Since an idempotent gives the direct sum \[ V=\operatorname{range}E\oplus\ker E, \] the dimensions force \[ \ker E=W^\perp. \] Therefore \(E\) is exactly the orthogonal projection onto \(W\).
3. Recover \(U\) as the reflection determined by \(W\).
From the definition of \(E\), \[ U=2E-I. \] If \(w\in W\), then \(Ew=w\), so \[ Uw=(2E-I)w=w. \] If \(z\in W^\perp=\ker E\), then \[ Uz=(2E-I)z=-z. \] Thus \(U\) fixes \(W\) pointwise and changes the sign on \(W^\perp\). Equivalently, if \(v=w+z\) with \(w\in W\) and \(z\in W^\perp\), then \[ Uv=w-z. \] This is precisely the reflection across \(W\) constructed in Exercise 6. Hence every self-adjoint unitary operator arises from such a subspace.