Linear Alg: Projection is closest vector in subspace

[VeryEvilPettingZoo]

You've proven that every vector in the subspace is at least as far from x as its projection (denoted Proj(V, x)). You could complete this insight by using that your inequality is strict ("less than", instead of "less than or equal to") whenever the length of b is not 0, hence whenever b is not 0. But b = 0 only if v = Proj(V, x). Thus any vector in the subspace OTHER than Proj(V, x) is FARTHER from x than Proj(V, x). Thus the projection vector is THE closest vector in the subspace.

[tadm123]

@VeryEvilPettingZoo he missed a "the"? wtf *unsubscribed*

Sarcasm obviously.

[VeryEvilPettingZoo]

@tadm123

Establishing uniqueness is actually a big deal in math. He'd done all the setup work to establish it (it was just a brief comment away), but forgot to make that final observation. Yeah - no need to unsubscribe - he does good stuff, including this vid.