Let $V$ be a finite dimensional inner product space. Let $T$ be a Hermitian unitary operator. Prove there's a subspace $W$ such that for each $v\in V$ we have $Tv=w-w^\prime$ where $w\in W,w^\prime \in W^\perp$ and $v=w+w^\prime$.
My attempt: $v+T(v)=2w, v-Tv=2w^\prime$, so it looks like $I+T=2P_W$ is twice the projection on $W$ and $I-T=2P_{W^\perp}$ is twice the projection on $W^\perp$. If $T$ is diagonalizable, projections onto eigenspaces are precisely given by the linear factors of the minimal polynomial, so $W,W^\perp$ must be eigenspaces corresponding respectively to the eigenvalues $-1,1$.
Thing is, it looks to me like all this just needs $T$ to be diagonalizable, so I'm sure I'm completely wrong, but really don't see where. Why do we need unitary and Hermitian?
Update: I think unitary + Hermitian ensures $\pm 1$ are eigenvalues and are the only eigenvalues.