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.

$R(A)=range(A), A^H:=\textrm{conjugate transpose of }A.$


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 $(QQ^H)^H = Q.$

  • $\begingroup$ Is that $R(A)=R(QR)=R(Q(Q^HQ)R), R(A)\subseteq R(QQ^H)$ and $R(QQ^H)=R(A(A^HA)^{-1}A^H)\subseteq R(A),$ so $R(QQ^H)=R(A)$? Time limit for edit so I repost it, sorry. $\endgroup$ – Ning Wang Sep 3 '18 at 5:05
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    $\begingroup$ That is a way to prove it, yes. But note that your proof is only valid for the case that $A^H A$ is inversible, while the projection matrix may be written as $Q Q^H$ for any matrix $A$, no matter if $A^H A$ is inversible or not. So you may find a more general proof for that statement. $\endgroup$ – Javi Sep 3 '18 at 5:14
  • $\begingroup$ Thank you so much sir, you're right that's an error, I'll try to figure it out! $\endgroup$ – Ning Wang Sep 3 '18 at 5:17
  • $\begingroup$ you are welcome! I would not call it an "error", though, it's kind of a limited proof, but it works in your case $\endgroup$ – Javi Sep 3 '18 at 5:20

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