Say you have a vector v in Cm and a matrix A in Cmxn. How do you obtain the projection of v onto range(A)?
Consider v once it has been decomposed into a component in range(A) and a component in ⊥range(A).
v = projA(v) + (v - projA(v))
Because proj_A(v) is in range(A), proj_A = Ax for some x in Cn.
v = Ax + (v - Ax)
v - Ax is orthogonal to range(A), so
(a_i)*(v - Ax) = 0 for all columns a_i. This is better shown in matrix form: A*(v - Ax) = 0
(A*A)x = A*v
When is A*A invertible? This is the case when the null space of A*A = {0}. If the n columns of A in Cm are linearly independent, then null(A) = {0} by the rank-nullity theorem: # columns of A = rank(A) + dim(null(A)). This means that the linear map A is injective (one-to-one), or an "embedding". The vectors Ax in Cm are all orthogonal to the left null space of A, so Ax -> A*Ax is injective. Thus, the entire map is injective, and is surjective too since the map goes from Cn to Cn. Thus, A*A is bijective and invertible.
x = (A*A)-1A*v, and projA(v) = Ax = A(A*A)-1A*v
What is the matrix, P = A(A*A)-1A*? P is the orthogonal projector onto range(A) since P2=P and P*=P.