Exercise 6.6.4

Since $T$ is an orthogonal projection, we have $N(T)=R{(T)}^{\text{⊥}}$ and $R(T)=N{(T)}^{\text{⊥}}$. Now we want to say that $N(I-T)=R(T)=W$ and $R(I-T)=N(T)=W^{\text{⊥}}$ and so $I-T$ is the orthogonal projection on $W^{\text{⊥}}$. If $x\in N(I-T)$, we have $x=T(x)\in R(T)$. If $T(x)\in R(T)$, we have

So we have the first equality. Next, if $(I-T)(x)\in R(I-T)$ we have

If $x\in N(T)$ we have $T(x)=0$ and so $x=(I-T)(x)\in R(I-T)$. So the second equality also holds.