Exercise 6.7.22

Observe that $\mathit{UT}$ is the orthogonal projection on $R(\mathit{UT})$ by Theorem 6.24 since it’s self-adjoint and $\mathit{UTUT}=\mathit{UT}$. We have that $R(\mathit{UT})=R(T^{\text{∗}}{U}^{\text{∗}})\subset R(T^{\text{∗}})$. Also, since

we have $R(T^{\text{∗}})\subset R(\mathit{UT})$ and hence $R(T^{\text{∗}})=R(\mathit{UT})$. This means $\mathit{UT}$ and $T^{\text{†}}T$ are both orthogonal projections on $R(T^{\text{∗}})=R(\mathit{UT})$. By the uniqueness of orthogonal projections, we have $\mathit{UT}=T^{\text{†}}T$.

Next, observe that $\mathit{TU}$ is the orthogonal projection on $R(\mathit{TU})$ by Theorem 6.24 since it’s self-adjoint and $\mathit{TUTU}=\mathit{TU}$. We have that $R(\mathit{TU})\subset R(T)$. Also, since

we have $R(T)\subset R(\mathit{TU})$. By the same reason, we have $\mathit{TU}=T{T}^{\text{†}}$ and they are the orthogonal projection on $R(T)=R(\mathit{TU})$.

Finally, since we have $\mathit{TU}-T{T}^{\text{†}}=T_0$, we may write is as

We want to claim that $R(U-T^{\text{†}})\cap N(T)=\left\{0\right\}$ to deduce that $U-T^{\text{†}}=T_0$. Observe that $R(T^{\text{†}})=N{(T)}^{\text{⊥}}=R(T^{\text{∗}})$. Also, we have $R(U)\subset R(T^{\text{∗}})$ otherwise we may pick $x\in W$ such that $U(x)\in R(U)\setminus R(T^{\text{∗}})$ and get the contradiction that

since $\mathit{UT}$ is the orthogonal projection on $R(T^{\text{∗}})$. Now we already have $R(U-T^{\text{†}})\subset R(T^{\text{∗}})=N(T)$. Hence the claim

holds. This means that $T(U-T^{\text{†}})(x)=0$ only if $(U-T^{\text{†}})(x)=0$. Since we have $T(U-T^{\text{†}})=T_0$, now we know that actually $U-T^{\text{†}}=T_0$ and hence $U=T^{\text{†}}$.