TITLE: Show that the inverse of a projection matrix, when exists, is again a projection matrix.
QUESTION [1 upvotes]: 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.

REPLY [1 votes]: (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 $\{I_n\}$...
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