# Determine if all vectors of form (a,0,0) are subspace of R3

I have the feeling that it is, but Im not really sure how to start the proof. I know I have to prove both closure axioms; u,v ∈ W, u+v ∈ W and k∈ℝ and u∈W then ku ∈ W.
Do I just pick a vector arbitrarily say a vector v = (x,y,z) and go from there?

mathwonk
Homework Helper
pick vectors that are of the form (a,0,0), and otherwise arbitrary.

proving u+v
let u, v ∈ W
u=(a1,0,0) v= (a2,0,0)
u+v = (a1+a2, 0,0) ∈ W
∴u+v ∈ W

am i anywhere close to doing that right?

Fredrik
Staff Emeritus
Gold Member
Yes, that's the right way to do it.

Thank you guys, I really appreciate the help.

Svein
The usual way is to determine the subspace ⊥ (a, 0, 0) - which is the subspace spanned by all vectors (x, y, z) such that (a, 0, 0)⋅(x, y, z) = 0. Since the scalar product is ax, this means that x = 0 and thus the normal subspace is spanned by (0, 1, 0) and (0, 0, 1). Therefore the original subspace has dimension 1 and is spanned by (1, 0, 0).

Fredrik
Staff Emeritus
Gold Member
The usual way is to determine the subspace ⊥ (a, 0, 0) - which is the subspace spanned by all vectors (x, y, z) such that (a, 0, 0)⋅(x, y, z) = 0. Since the scalar product is ax, this means that x = 0 and thus the normal subspace is spanned by (0, 1, 0) and (0, 0, 1). Therefore the original subspace has dimension 1 and is spanned by (1, 0, 0).
This is true, but all you have to do to see it is to write ##(a,0,0)=a(1,0,0,)##.

Svein
This is true, but all you have to do to see it is to write (a,0,0)=a(1,0,0,)(a,0,0)=a(1,0,0,).

Yes, in this case it is easy. But if the specification had been more complicated, having a standard recipe is not a bad idea.

Svein thanks for that, it clears up the concept a little more.

Fredrik
Staff Emeritus
Gold Member
The orthogonal complement of a set ##S## is defined as the set ##S^\perp## of all vectors that are orthogonal to all the vectors in ##S##.

What Svein described is how to find the orthogonal complement of the orthogonal complement of the set W. This is always a subspace, even if W isn't. If you know this, you can find out if ##W## is a subspace by checking if ##W^{\perp\perp}=W##. This is rarely (never?) the easiest way to do it.

Svein
Well, in the $l_{2}$ space (with the standard scalar product), this algorithm is used in proofs of completeness...