Let \(V\) be a finite-dimensional inner product space, and let \(E\) be an idempotent linear operator on \(V\), i.e., \(E^2 = E\). Prove that \(E\) is self-adjoint if and only if \(EE^* = E^*E\).
Solution
1. A self-adjoint idempotent commutes with its adjoint.
If \(E=E^*\), then \[ EE^*=E^2=E^*E. \] Thus the forward implication is immediate.
2. If \(EE^*=E^*E\), then \(\ker E=\ker E^*\).
Assume \(E\) commutes with \(E^*\). For any \(v\in V\), \[ \|E^*v\|^2 =(E^*v\mid E^*v) =(EE^*v\mid v), \] while \[ \|Ev\|^2 =(Ev\mid Ev) =(E^*Ev\mid v). \] Because \(EE^*=E^*E\), these norms are equal. Hence \[ Ev=0\iff E^*v=0, \] so \[ \ker E=\ker E^*. \]
3. The range of \(E\) is orthogonal to its kernel.
From the general adjoint identity proved in Exercise 4, \[ \operatorname{range}E=(\ker E^*)^\perp. \] Using step 2, \[ \operatorname{range}E=(\ker E)^\perp. \] Thus the range and kernel are orthogonal complements.
4. Therefore \(E\) is self-adjoint.
Since \(E^2=E\), every vector has the decomposition \[ v=Ev+(v-Ev), \] with \(Ev\in\operatorname{range}E\) and \[ E(v-Ev)=Ev-E^2v=0, \] so \(v-Ev\in\ker E\). Hence \(E\) is the projection onto \(\operatorname{range}E\) along \(\ker E\).
Step 3 shows this is in fact an orthogonal projection. Orthogonal projections are self-adjoint [@HK71]. Therefore \[ E=E^*. \] Combining with step 1 gives \[ \boxed{E=E^*\iff EE^*=E^*E.} \]