# Thread: confusion about telling wether or not a subset is a subspace

1. ## confusion about telling wether or not a subset is a subspace

Is the following subset of $R^3$ a subspace $R^3$? The set of all vectors of the form (a,b,2)

W is not a subspace. To show this let $\vec u=(a_1, b_1, c_1)$ and $\vec v = (a_2, b_2, c_2)$ be in W. Then $c_1=c_2=2$ Now $\vec u + \vec v = (a_1+a_2, b_1+b_2, c_1+c_2)=(a_1,b_1+b_1,4)$ which is not in W

Two things confuse me. 1) Why is it $b_1+b_1$ not $b_1+b_2$? 2)I don't see how the property " $\vec u + \vec v$ is in V" fails to hold because $(a_1+b_1+b_1,4)$ is in 3 space isn't it?

2. Originally Posted by Jskid
Is the following subset of $R^3$ a subspace $R^3$? The set of all vectors of the form (a,b,2)
Look no further. Is $<0,0,0>$ in the set?

3. Originally Posted by Plato
Look no further. Is $<0,0,0>$ in the set?
No it's not but I thought I didn't have to verify that rule if it's a subspace of a vector space? $0 \dot \vec u = \vec 0$
And that still don't see where $b_1+b_1$ came from?

EDIT:All though I do see how that answers $(a_1+b_1+b_1,\textbf{4})$ not being in W

4. $\vec u + \vec v = (a_1+a_2, b_1+b_2, c_1+c_2)=(a_1+b_1+b_1,4)$

should be

$\vec u + \vec v = (a_1+a_2, b_1+b_2, c_1+c_2)=(a_1+a_2, b_1+b_2, 4).$

Just fyi. The first expression loses a dimension (!), loses a term, and changes an index.

5. Originally Posted by Ackbeet
$\vec u + \vec v = (a_1+a_2, b_1+b_2, c_1+c_2)=(a_1+b_1+b_1,4)$

should be

$\vec u + \vec v = (a_1+a_2, b_1+b_2, c_1+c_2)=(a_1+a_2, b_1+b_2, 4).$

Just fyi. The first expression loses a dimension (!), loses a term, and changes an index.
Good catch, that should've been a comma not a plus.
The answer manual has $b_1+b_1$ NOT $b_1+b_2$ That's what I'm trying to ask about?

6. I get it now, I think my solution manual has a typo that got me very confused.

7. Yeah, it's a typo.