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

1. ## 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

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

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

conversely if $y - v_1 =w_1 \in V^{\perp},$ for some $v_1 \in V,$ then $y=v_1+w_1=v_0+w_0.$ thus $v_1=v_0, \ w_1=w_0.$ hence $||y-v_1||=||y-v_0||,$ and as we showed: $||y-v_0||=\min_{v \in V} ||y-v||. \ \ \ \Box$

3. It makes so much more sense now. Thank you very much!