Show that the inverse of a projection matrix, when exists, is again a projection matrix.

Question

A proof in linear algebra. Show that the inverse of a projection matrix, when exists, is again a projection matrix.

My thinking:

The definition of a projection matrix is $PP=P$. So, $$ PP=\frac{1}{(v\cdot v)^2}vv^{-1}vv^{-1}=\frac{1}{(v\cdot v)^2}v(v^{-1}v)v^{-1} $$ this is what I have tried so far, not sure how to proceed or if what I have so far is correct.

Answer 1

Since $P=P^2$, if $(P)^{-1}$ exists, then $(P^2)^{-1}$ exists, so $(P^{-1})=(P^2)^{-1}=(P^{-1})^2$. In fact, it can be shown that $P=I$ is the only invertible projection matrix, so the result is obvious from this fact.

Answer 2

(In connection with the last sentence of @Dave)

If $P$ is invertible, it is a bijective, thus $Im(P)=\mathbb{R}^n$.

What does it mean? That the space on which one projects is $\mathbb{R}^n$ in its entirety. It is thus the identity $I_n$, which is its own inverse, thus is a projection matrix...

In other words (it is why this exercise is very special), the set of matrices that are concerned is reduced to $\{I_n\}$...