# Linear algebra subspace proof

1. Feb 10, 2013

### Mdhiggenz

1. The problem statement, all variables and given/known data

Prove that if S is a subspace of R1, then either S={0} or S=R1.

Trying to come up with a proof I dissected each statement, I know that in order for S to be

a subspace the zero vector must lie within the subset. So I know S={0} is true. I then

checked an arbitary vector x1 which lies on R1 to make sure it

was closed under scalar multiplication, and addition, and that checked out as well.

Not sure if I am on the right track.

Thanks

2. Relevant equations

3. The attempt at a solution

2. Feb 10, 2013

### jbunniii

You need to specify what field of scalars you are using. The statement is true if the field is $\mathbb{R}$, false if it is, for example, $\mathbb{Q}$.

Assuming you are using $\mathbb{R}$ for the field of scalars, use the fact that a subspace must be closed under scalar multiplication.

3. Feb 10, 2013

### Karnage1993

Your ideas are right, but it's not really a proof unless you do everything step-by-step.
For example, for the first possible set: $S = \{0\}$, show that all three properties of a subspace are satisfied.

4. Feb 10, 2013

### jbunniii

Why is that necessary? The problem statement tells you that $S$ is a subspace. All you need to do is show that if it's not $\{0\}$ then it must be all of $\mathbb{R}^1$.

So: if $S$ is not $\{0\}$, then $S$ contains a nonzero element, say $s \neq 0$. Since $S$ is a subspace, it is closed under scalar multiplication. Thus it must contain $\alpha s$ for every $\alpha \in F$ where $F$ is the scalar field (presumably $\mathbb{R}$). Therefore...?

5. Feb 11, 2013

s=r1?