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

• Feb 17th 2011, 12:32 PM
Jskid
confusion about telling wether or not a subset is a subspace
Is the following subset of $\displaystyle R^3$ a subspace $\displaystyle R^3$? The set of all vectors of the form (a,b,2)

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

Two things confuse me. 1) Why is it $\displaystyle b_1+b_1$ not $\displaystyle b_1+b_2$? 2)I don't see how the property "$\displaystyle \vec u + \vec v$ is in V" fails to hold because $\displaystyle (a_1+b_1+b_1,4)$ is in 3 space isn't it?
• Feb 17th 2011, 12:36 PM
Plato
Quote:

Originally Posted by Jskid
Is the following subset of $\displaystyle R^3$ a subspace $\displaystyle R^3$? The set of all vectors of the form (a,b,2)

Look no further. Is $\displaystyle <0,0,0>$ in the set?
• Feb 17th 2011, 01:05 PM
Jskid
Quote:

Originally Posted by Plato
Look no further. Is $\displaystyle <0,0,0>$ 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? $\displaystyle 0 \dot \vec u = \vec 0$
And that still don't see where $\displaystyle b_1+b_1$ came from?

EDIT:All though I do see how that answers $\displaystyle (a_1+b_1+b_1,\textbf{4})$ not being in W
• Feb 17th 2011, 01:23 PM
Ackbeet
$\displaystyle \vec u + \vec v = (a_1+a_2, b_1+b_2, c_1+c_2)=(a_1+b_1+b_1,4)$

should be

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

Just fyi. The first expression loses a dimension (!), loses a term, and changes an index.
• Feb 17th 2011, 01:36 PM
Jskid
Quote:

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

should be

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

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 $\displaystyle b_1+b_1$ NOT $\displaystyle b_1+b_2$ That's what I'm trying to ask about?
• Feb 17th 2011, 02:29 PM
Jskid
I get it now, I think my solution manual has a typo that got me very confused.
• Feb 17th 2011, 03:34 PM
Ackbeet
Yeah, it's a typo.