Let $P$ be a projection mapping onto the hyperplane trough the origin which is normal to $v$. How do you show that $Px=x-\dfrac{vv^T}{v^Tv}x$
Any intuition?
[Math] equation of projection onto hyperplane
geometryreal-analysis
geometryreal-analysis
Let $P$ be a projection mapping onto the hyperplane trough the origin which is normal to $v$. How do you show that $Px=x-\dfrac{vv^T}{v^Tv}x$
Any intuition?
Best Answer
$Px$ is just $x$ minus the portion that is parallel to $v$.
$$Px=x-\langle x, \frac{v}{|v|}\rangle \frac{v}{|v|}=x-\frac{x^T v}{v^Tv}v.$$