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

[7sqr]

I have the feeling that it is, but I am 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]

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

 

[7sqr]

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]

Yes, that's the right way to do it.

 

[7sqr]

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]

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.

 

[7sqr]

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

 

[Fredrik]

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]

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.

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