I get stuck in the follow problem:
Let $A \in R^{m \times n}$ and $\operatorname{rank}(A)=n$ (It's not written but I think $m \geq n$). Show that $\left\lVert{A(A^TA)^{-1}A^T}\right\rVert_2 = 1$ where $B \in R^{m \times n} \Rightarrow \left \lVert {B}\right \rVert_2 = \sup_{||x||=1}\left \lVert {Bx}\right \rVert_2$
My try:
I've decided look to $\left \lVert{A(A^TA)^{-1}A^T}\right\rVert_2^2$
And I get the following expression:
$\left \lVert{A(A^TA)^{-1}A^T}\right\rVert_2^2 = \sup_{||x||=1} ({x^TA(A^TA)^{-1}A^Tx})$
If we call $A(A^TA)^{-1}A^T=B$ then we have
$\left \lVert{A(A^TA)^{-1}A^T}\right\rVert_2^2 = \sup_{||x||=1} <x,Bx>$ ???
And here I get stuck, any suggestion?
Thank you!