I know the formal proof of the fact that a Projection Matrix is singular. From which it follows that the only invertible projection is the identity.
But I still don't understand intuitively why. I know that a singular matrix carries parallelepipeds into lines, but why a singular transformation must have $det = 0?$
On
A projection matrix $P$ is a projection, when it fulfills $PPx=Px$. This is the same as saying "If I project twice, it doesn't change a thing the second time". Assuming, that this matrix $P$ has an inverse, you arrive at $Px=x$ which only holds true for the identity $I$.
Another way to see it, is given by the geometric interpretation. In a projection, you always loose dimensions. Your shadow is a 2D-projection of your 3D body. When you loose a dimension, you loose information and the transformation can not be undone.
One « geometrical » way to prove that the determinant of a projection $p$ vanishes is to remember that for a basis $(e_1, \dots,e_n)$ you have for any linear map $u$ the relationship
$$Vol(u(e_1), \dots, u(e_n))= \det(u) Vol(e_1, \dots, e_n)$$
where $Vol$ designates the volume of the parallelepipede based on the vectors.
For a projection, the volume $Vol(p(e_1), \dots, p(e_n))$ vanishes as all the images of the basis $(e_1,\dots,e_n)$ are in a proper subspace. Hence $\det(p)=0$.