Proving Theorem 10.3 on Steven Roman's Advanced Linear Algebra

Question

I want to prove the (3) and (4) of theorem which says $\textrm{im}(\tau \tau ^{*})=\textrm{im}(\tau )$ and $(\rho _{\textrm{S,T}})^{*}=\rho _{\textrm{T}^{\perp },\textrm{S}^{\perp }}$, here $\tau ^{*}$ is the adjoint of $\tau$ and $\rho _{S,T}$ is the projection onto $S$ along $T$, and $S^{\perp }$ is the orthogonal complement of S, the book doesn't give a proof about this and I don't know how to do it, can anyone help me, thank you in advance

Theorem 10.3 Let $\tau\in\mathcal L(V,W)$, where $V$ and $W$ are finite-dimensional inner product spaces.$\newcommand{\im}{\operatorname{im}}$

1. $$\ker(\tau^*)=\im(\tau)^\bot \qquad\text{and}\qquad \im(\tau^*)=\ker(\tau)^\bot$$ and so \begin{gather*} \tau\text{ surjective }\Leftrightarrow\tau^*\text{ injective } \\ \tau\text{ injective }\Leftrightarrow\tau^*\text{ surjective } \end{gather*}
2. $$\ker(\tau^*\tau)=\ker(\tau) \qquad\text{and}\qquad \ker(\tau\tau^*)=\ker(\tau^*) $$
3. $$\im(\tau^*\tau)=\im(\tau^*) \qquad\text{and}\qquad \im(\tau\tau^*)=\im(\tau) $$
4. $$(\rho_{S,T})^*=\rho_{T^\bot,S^\bot}$$

Answer 1

$\newcommand{\im}{\operatorname{im}}$I figured it out, because $\im(\tau ^{*})=(\ker(\tau ))^{\perp }=(\ker(\tau ^{*}\tau ))^{\perp }=\im((\tau ^{*}\tau )^{*})=im(\tau ^{*}\tau )$ by using (1) and (2) in theorem 10.3, and $\ker((\rho _{S,T})^{*})=\im(\rho _{S,T})^{\perp }=S^{\perp}$, similarly $\im((\rho _{S,T})^{*})=T^{\perp }$, because the adjoint of the projection is also idempotent we show that it is also a projection.