# Show that the orthogonal projection onto $Range T$ is equal to $T(T^\ast T)^{-1}T^\ast$ given that $T: V \to W$ is injective

Given $$V$$ and $$W$$ as finite-dimensional inner product spaces and an injective linear map $$T: V \to W$$, how can we show that $$P_{Range T} = T(T^\ast T)^{-1}T^* \in \mathcal{L}(W)$$ where $$P_{Range T}$$ is the orthogonal projection onto $$Range T$$ and $$T^\ast$$ is the adjoint of $$T$$? Supposedly this result can be used to derive a useful matrix formula for orthogonal projection, but I'm not sure how to figure out how we can show the equality above. Does $$(T^\ast T)^{-1}$$ even make sense? How do we show that? And what is the intuition of the above formulation?

Among my ideas are trying to show that the right hand side is in the range of $$T$$ (which it seems to be) and trying to utilize the fact that $$P_{Range T}^2 = P_{Range T}$$. But I'm not sure how to proceed from there.

First, note that if $$T^{*}Tv = 0$$ then

$$0 = \left< T^{*}Tv, v \right> = \left < Tv, Tv \right> = \| Tv \|^2 = 0$$

so $$Tv = 0$$. Since $$T$$ is injective, it implies that $$v = 0$$. Hence, $$T^{*}T \colon V \rightarrow V$$ is invertible. Now,

$$P^2 = T(T^{*}T)^{-1}T^{*} T(T^{*}T)^{-1} T^{*} = T(T^{*}T)^{-1}T^{*} = P$$

and

$$P^{*} = \left( T(T^{*}T)^{-1}T^{*} \right)^{*} = \left( T^{*} \right)^{*} \left( (T^{*}T)^{-1} \right)^{*} T^{*} = \\T \left( (T^{*}T)^{*} \right)^{-1} T^{*} = T \left( T^{*} \left (T^{*} \right)^{*} \right)^{-1} T^{*} = T(T^{*}T)^{-1}T^{*} = P$$

so $$P$$ is self-adjoint. Hence, $$P$$ is an orthogonal projection onto $$\operatorname{Range}(P)$$. Finally, since $$T$$ is injective, we have that $$T^{*}$$ is surjective and since $$(T^{*}T)$$ is an isomorphism we have

$$\operatorname{Range}(P) = \operatorname{Range}(T).$$

Let's see how you can arrive the formula intuitively. Since $$P$$ is an orthogonal projection on $$\operatorname{Range}(T)$$, it should vanish on $$\operatorname{Range}(T)^{\perp} = \ker(T^{*})$$ so it makes sense to look for $$P$$ of the form $$P = AT^{*}$$. You also want that the range of $$P$$ will be the range of $$T$$ so that it makes sense to look for $$P$$ of the form $$P = TAT^{*}$$ with $$A \colon V \rightarrow V$$ being an isomorphism. In order for $$P$$ to be a projection, we should have

$$P^2 = T(AT^{*}T)AT^{*} = TAT^{*} = P$$

which will hold if $$AT^{*}T = \operatorname{id}|_{V}$$ or $$A = (T^{*}T)^{-1}$$.

An alternative proof, assuming $$T^*T$$ is invertible:

If $$w\in\mathrm{range}(T)$$, then $$w=Tv$$ for some $$v$$, so $$Pw=T(T^*T)^{-1}T^*\,Tv=Tv=w$$ while, if $$w\perp\mathrm{range}(T)$$, then $$\langle T^*w, v\rangle =\langle w, Tv\rangle =0$$ for all $$v$$, hence $$T^*w=0$$, implying $$Pw=0$$.

Hence, $$P$$ must be the orthogonal projection to $$\mathrm{range}(T)$$.