Seemingly strange exercise on unitary Hermitian operator II 
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 $v=w+w^\prime$.

$v+T(v)=2w, v-Tv=2w^\prime$, 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 $W^\perp$. $T$ is Hermitian, hence diagonalizable with real eigenvalues, and also unitary, so the eigenvalues are in $ \left\{ \pm 1 \right\} $. Hence there's a (possibly trivial) pair of eigenspaces $W,W^\prime$ for $1,-1$. In this case, however, $T+I=P_W,T-I=-P_{W^\prime}$ without any factors of two or minus signs. So how do I get the factors to fit the answer?
 A: Your reasoning is almost flawless, but then in the end you pull the equalities $T+I=P_W$, and $T-I=-P_{W'}$ out of thin air. 
You know that $T$ has eigenvalues $1,-1$ (possibly one of them with zero multiplicity), and that it is Hermitian. Then, by the Spectral Theorem, 
$$
T=P-Q
$$
for two orthogonal projections $P,Q$ with $P+Q=I$. That's exactly the decomposition the question is talking about: $P$ is the projection onto the subspace
$$
\{v:\ Tv=v\}
$$
and $Q$ is the projection onto the subspace 
$$
\{v:\ Tv=-v\}. 
$$
A: Because $T$ is unitary Hermitian, then $T^*T=I$ and $T=T^*$, which leads to $T^2=I$. Then you can write
$$
         I = \frac{1}{2}(I+T)+\frac{1}{2}(I-T).
$$
Every $v$ can be written as
$$
       v = w+w',\;\;\;\;\; w=\frac{1}{2}(I+T)v, w'=\frac{1}{2}(I-T)v
$$
And $(T-I)w=0$, $(T+I)w'=0$, which gives $Tw=w$ and $Tw'=-w'$. Automatically $w'\perp w$ for all $w\in(I+T)V$ and $w'\in (I-T)V$ because
$$
    \langle (I+T)u,(I-T)v\rangle=\langle (I-T)(I+T)u,v\rangle=\langle 0,v\rangle.
$$
