$A(A^T A)^{-1} A^T$ is the matrix of the orthogonal projection to im A Let $A \in \mathbb{R}^{n\times m}$, ker $A = {0}$.
Show that $A(A^T A)^{-1} A^T$ is the transformation matrix of the orthogonal projection to the image of $A$.
Any advice is greatly appreciated.
 A: Well, it's clearly a projection onto the image of $A$ since if $y=Ax$ is in the image of $A$ then $(A(A^T A)^{-1} A^T)y = A(A^T A)^{-1} A^T)Ax = A(A^T A)^{-1} (A^TA)x = Ax = y$ and since clearly it maps anything into the image of $A$.  To show it is orthogonal projection, you need to show it kills anything orthogonal to the image of $A$.  But $z$ is orthogonal to the image of $A$ if $z^TA=0$. And in that case also $A^Tz=0$ and so $(A(A^T A)^{-1} A^T)z=(A(A^T A)^{-1})( A^Tz)=0$.  
The transformation that is the identity on the image of $A$ and kills everything orthogonal to it is the orthogonal projection onto the image of $A$.  So $(A(A^T A)^{-1} A^T)$ is it.  QED
A: Let's denote $M = A (A^T A)^{-1} A^T$. 
Then $M^2 = A (A^T A)^{-1} A^T A (A^T A)^{-1} A^T = A (A^T A^{-1}) A^T = M$.
And, $M^T = M$. These two properties imply that $M$ is the orthogonal projection onto its image. ( An idempotent operator $P$ is an orthogonal projection iff it is self adjoint ) 
$im(M) \subset im(A)$ should be straightforward to see. (Since $M$ is a composition of linear transformations, the last of which is $A$.)
To show that $im(A) \subset im(M)$, it suffices to show that $(A^T A)^{-1} A^T$ is surjective onto $R^m$ (the domain of $A$). Since $(A^T A)^{-1}$ is an invertible map, it suffices to show that $A^T$ is surjective. In fact, $A^T$ is surjective whenever $A$ is injective ( The transpose of a linear injection is surjective. ) , so we are done. 
