Prove $\| A(A^TA)^{-1}A^T\|_2 = 1$ when rank of matrix $A$ is $n$ Given a matrix $A \in R^{m \times n}$ and whose rank is $n$. I need to show $\| A(A^TA)^{-1}A^T\|_2 = 1$. Can any hint me the direction in which I should solve this problem. Should I use any decomposition of matrix $A$ to show the result?
 A: I will explain it from a geometric perspective.
The matrix $P=A(A^TA)^{-1}A^T$ is an orthogonal projection matrix satisfying $P^2=P$ and $P^T=P$. For any $x\in\mathbb{R}^m$, $Px$ is the orthogonal projection of $x$ onto the column space of $A$.


*

*Consider an orthogonal basis $\{x_i\}_{i=1}^n$ of $\mathrm{Range}(A)$. Then $Px_i=x_i$. 

*Consider an orthogonal basis $\{y_i\}_{i=1}^{m-n}$ of the orthogonal complement of $\mathrm{Range}(A)$, then $Py_i=0$.


Therefore, $P$ has $n$ eigenvalues equal to 1 and $m-n$ eigenvalues equal to 0. Since $P$ is symmetric positive semi-definite, its singular values are equal to its eigenvalues. As a result, $\|P\|_2=\sigma_{\max}(P)=1$.
A: An argument without geometry goes like this:
As Shiyu said $P^2=P$ and hence, $\|P\| = \|P^2\|\leq \|P\|^2$ and therefor $1\leq \|P\|$. Moreover, $P^T=P$ and hence,  $\|Px\|^2 = \langle Px,Px\rangle = \langle Px, x\rangle \leq \|Px\|\|x\|$ which gives $\|P\| \leq 1$.
A: Let $H = A(A^TA)^{-1}A^T$.  To see that $x\mapsto Hx$ (for $x\in\mathbb{R}^m$) is the orthogonal projection onto the column space of $A$, it suffices to prove two things:


*

*If $x$ is in the column space of $A$, then $Hx=x$.

*If $x$ is orthogonal to all columns of $A$, then $Hx=0$.


To prove the second statement, notice that if $x$ is orthogonal to all columns of $A$, then $A^T x = 0$.  Therefore $A(A^TA)^{-1}A^Tx = 0$.
To prove the first statement, notice that $x$ is in the column space of $A$ iff $x = Aw$, for some $w$.  Therefore
$$
Hx = HAw = \Big(A(A^TA)^{-1}A^T\Big) Aw = A(A^TA)^{-1}\Big(A^T A\Big)w = Aw = x.
$$
Now let $x$ be any vector in $\mathbb{R}^m$.  Decompose $x$ into a component in the column space of $A$ and a component orthogonal to the column space of $A$.  The component in the column space of $A$ is $u=Hx$.  The component orthogonal to the column space of $A$ is $v=(I-H)x$.  What then is $\|Hx\|_2$?  It is $\|u\|_2 \le \|u+v\|_2 = \|x\|_2$.  Since $\|Hx\|_2 \le \|x\|_2$, we have $\|H\|_2 \le 1$.  But since $\|Hu\|_2= \|u\|_2$, we have $\|H\|_2\ge1$.
A: I'm probably missing something, but since $(AB)^{-1}=B^{-1}A^{-1}$, why not just
$$A(A^TA)^{-1}A^T = A(A^{-1}(A^T)^{-1})A^T = (AA^{-1})((A^T)^{-1}A^T) = II = I$$
?
