Vector space projection and minimization...?

Question

In R^4 let U= span { (1,1,0,0), (1,1,1,2) }

Find u in U such that

|| u-(1,2,3,4)|| is as small as possible.

I think that this means use Gram-Schmidt on the basis vectors and computing the projection of something on U.

Answer 1

You could find the projection of (1, 2, 3, 4) on U, yes. If you use Gram-Schmidt on the basis vectors to get new orthonormal basis vectors v and w, the projection is just (x.v)v + (x.w)w [where x = (1, 2, 3, 4)] and this point is u.

Here's an alternative approach: let u in U be given by u = (1, 1, 0, 0) s + (1, 1, 1, 2) t = (s+t, s+t, t, 2t).
Then D = ||u - (1, 2, 3, 4)||^2 = (s+t-1)^2 + (s+t-2)^2 + (t-3)^2 + (2t-4)^2.
dD/ds (partial) = 2(s+t-1) + 2(s+t-2) = 4s + 4t - 6
dD/dt (partial) = 2(s+t-1) + 2(s+t-2) + 2(t-3) + 2(2t-4).2
= 4s + 14t - 28
So 4s + 4t - 6 = 0 = 4s + 14t - 28 => 22 = 10t
=> t = 11/5 and s = 3/2 - 11/5= -7/10.
So the point is (3/2, 3/2, 11/5, 22/5).

Compare the basis approach. I'll do it slightly differently, skipping the normalisation part and writing the projection as
(x.v)v / ||v||^2 + (x.w)w / ||w||^2.
So, v = (1, 1, 0, 0).
w = (1, 1, 1, 2) - (1, 1, 1, 2).(1, 1, 0, 0) (1, 1, 0, 0) / ||(1, 1, 0, 0)||^2
= (1, 1, 1, 2) - 2 (1, 1, 0, 0) / 2
= (0, 0, 1, 2).
Then u = (1, 2, 3, 4).(1, 1, 0, 0) v / 2 + (1, 2, 3, 4).(0, 0, 1, 2) w / 5
= 3/2 (1, 1, 0, 0) + 11/5 (0, 0, 1, 2)
= (3/2, 3/2, 11/5, 22/5).