# Showing the closest point in a set to a point outside the set is orthogonal to

• Oct 1st 2008, 12:05 PM
cutmeout
Showing the closest point in a set to a point outside the set is orthogonal to
Hi everyone,

I have a bit of an algebra/convexity type problem. I've been asked to show that if V is a linear subspace in R^n and y is not in V, then show that x* in V is the closest point in V to y if and only if y-x* is orthogonal to V. I'm a bit stuck on this. I know any subspace must be a hyperplane, so intuitively, this makes sense, but I'm having trouble bridging the gap to make it rigorous. Any ideas?

Thanks for any help!

Best,

Alex
• Oct 1st 2008, 08:18 PM
NonCommAlg
Quote:

Originally Posted by cutmeout
Hi everyone,

I have a bit of an algebra/convexity type problem. I've been asked to show that if V is a linear subspace in R^n and y is not in V, then show that x* in V is the closest point in V to y if and only if y-x* is orthogonal to V. I'm a bit stuck on this. I know any subspace must be a hyperplane, so intuitively, this makes sense, but I'm having trouble bridging the gap to make it rigorous. Any ideas?

Thanks for any help!

Best,

Alex

let $\displaystyle V^{\perp}=\{z \in \mathbb{R}^n: \ <z,v>=0, \ \forall v \in V \}$ be the orthogonal complement of $\displaystyle V$ in $\displaystyle \mathbb{R}^n.$ we know that $\displaystyle \mathbb{R}^n=V \oplus V^{\perp}.$ so: $\displaystyle y=v_0+w_0,$ for some unique $\displaystyle v_0 \in V, \ w_0 \in V^{\perp}.$ now let $\displaystyle v$ be any element
of $\displaystyle V.$ then: $\displaystyle ||y-v||^2=||v_0 - v +w_0||^2=<v_0-v+w_0, \ v_0-v+w_0>=||v_0-v||^2 + ||w_0||^2.$ (note that $\displaystyle <v_0-v, \ w_0>=<w,\ v_0-v>=0,$ because $\displaystyle v_0-v \in V$ and $\displaystyle w_0 \in V^{\perp}.$)

so we showed that for all $\displaystyle v \in V: \ ||y-v||^2=||v_0-v||^2+||w_0||^2.$ now $\displaystyle v_0,w_0$ are constants. thus $\displaystyle ||y-v||$ is minimum if and only if $\displaystyle ||v_0-v||$ is minimum, i.e. $\displaystyle v=v_0.$ but then $\displaystyle y-v=w_0 \in V^{\perp}.$

conversely if $\displaystyle y - v_1 =w_1 \in V^{\perp},$ for some $\displaystyle v_1 \in V,$ then $\displaystyle y=v_1+w_1=v_0+w_0.$ thus $\displaystyle v_1=v_0, \ w_1=w_0.$ hence $\displaystyle ||y-v_1||=||y-v_0||,$ and as we showed: $\displaystyle ||y-v_0||=\min_{v \in V} ||y-v||. \ \ \ \Box$
• Oct 5th 2008, 07:43 AM
cutmeout
It makes so much more sense now. Thank you very much!