I'm reading a book with following theorem:
Given $A\in F^{m\times n}, \textrm{rank}(A)=n, W=R(A), \mathbf b\in F^{m\times1},$ then $proj_W\mathbf{b}=A(A^HA)^{-1}A^H\mathbf b.$
Let $P=A(A^HA)^{-1}A^H,$ my question is that how to prove $R(P)=R(A),$ i.e. the range of $P$ is the same as that of $A$? My current idea is since $(A^HA)$ is invertible and $A$ is one-one so I just have to check $A^H$ on the lhs of $\mathbf b$ is onto, but I have no idea about it.
$R(A)=range(A), A^H:=\textrm{conjugate transpose of }A.$
Edit:
Now I got a thought but I'm not sure whether it's correct: Since $\textrm{rank}(A)=n, \textrm{rank}(A^H)=n,$ and $A^H\in F^{n\times m}$ so clearly $A^H$ is onto.
I would suggest you to prove the following:
For any matrix $A$, the matrix $P_{R(A)}$ which projects onto $R(A)$, i.e. the subspace spanned by the columns of $A$, may be written as the product $QQ^H$, where $Q$ is the matrix formed with an orthonormal basis of $R(A)$ (the vectors of the orthonormal basis are the columns of $Q$).
Also, that orthonormal basis is easily obtained using Gram-Schmidt process, which equations by the way may be written in matrix form, yielding $A = QR$ which is called the QR factorization of $A$, where $Q$ is the same as before and $R$ is an upper triangular inversible matrix. Try to put Gram-Schmidt equations into this matrix form to convice yourself.
Finally, you can replace $A$ in your expression of $P$ by its $QR$ factorization, and after a couple of steps applying the $H$ and $-1$ operator you may check that $P_{R(A)} = QQ^H$ (which of course was necessarily true, if you had previously proven my first paragraph).
Note that, in order to prove the first the first paragraph, you need to show that $R(QQ^H) = R(A)$ as well as the two requirements for a projection matrix: $(QQ^H)^2 = QQ^H$ and $(QQ^H)^H = Q.$