Vectorial Subspace

1. Jan 22, 2010

Fanta

There's not really a problem statement here.
I just want to know :
If I have a vector starting on the origin (like a position vector), then it will always correspond to a vectorial subspace, right?

For example:

$$(b, 2a + b ) : a, b \in R$$

is a vectorial subspace

but is

$$(b, 2a + 1 ) : a, b \in R$$

a subspace too?
And if not, why is that?

2. Jan 22, 2010

Staff: Mentor

For a subset of a vector space to actually be a subspace of that vector space, the subset has to satisfy three conditions:
The set has to have the 0 vector.
If u and v are in the subset, then u + v is also in the subset.
If u is in the subset, and c is any scalar, then cu is in the subset.

The set of vectors (b, 2a + 1) is not a subspace of R2, because at least one of the three conditions is not met.

3. Jan 22, 2010

Dick

Uh, but (b,2a+1) spans all of R^2.

4. Jan 22, 2010

Staff: Mentor

Right, but maybe we aren't talking about the same thing. I'm thinking in terms of the set {(a, 2b + 1) | a, b are real}. This set isn't closed under addition, so isn't a subspace of R2.

5. Jan 22, 2010

Dick

(a1,2b1+1)+(a2,2b2+1)=(c,2d+1) where c=a1+a2 and d=b1+b2+1/2. The question looks like a different kind of question (i.e. is (a,2b,1) a subspace?). But it's not. 2b+1 is ANY real number, just like b and independent of a.

6. Jan 22, 2010

Staff: Mentor

Dick, I wasn't confusing it with (a, 2b, 1). I think I got thrown by the lack of dependence of a and b.

7. Jan 22, 2010

Dick

That's true. (a,2a+1) would also be a whole different story.

8. Jan 22, 2010

Staff: Mentor

That's exactly where I was coming from. My eyes must have glazed over...

9. Jan 22, 2010

Fanta

you got me kinda lost here.

so it is indeed a subspace since it spans all R^2?

10. Jan 22, 2010

Dick

R^2 is a subspace of R^2, isn't it? Check the conditions to be a subspace Mark44 was referring to.