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I'm trying to verify that $W$ (being the set of all vectors in $\mathbb R^3$ whose third component is $-1$) is not a subspace of the vector space.

You can have a vector $(0,0,-1)$ and through a scalar of $4$, you can obtain vector $(0,0,-4)$.

I understand to prove a subspace one only needs to prove that set is closed under both addition and multiplication.

However, I don't understand, why $(0,0,-1)$ and $(0,0,-4)$ are not in the same vector space.

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$(0,0,-1)$ and $(0,0,-4)$ are both elements of $\Bbb R^3$ but they're NOT both elements of the set $W$. – Bye_World Jan 10 at 23:29
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@Tommyixi Do you know how to tell whether a vector is in the set $W$? – Erick Wong Jan 10 at 23:32
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Oh, is it because the third component of the set (0,0,-4) is not -1? – Tommyixi Jan 10 at 23:32
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That's correct (except I wouldn't call $(0,0,-4)$ a set, I'd call it a vector). – Bye_World Jan 10 at 23:33
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@Bye_World: tuple would be even less controversial, seeing as it is here considered particularly as an element of a set that's not a vectorspace. – leftaroundabout Jan 11 at 1:37
up vote 6 down vote accepted

A subspace has to contain the $0$-vector. All vectors of $W$ have $-1$ as their third component.

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Cheers, thank you! – Tommyixi Jan 10 at 23:33
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This is true but I honestly don't see why this got so many upvotes. OP has a perfectly good counterexample to closure in his question. I think a better answer would point that out. – Bye_World Jan 10 at 23:35
    
@Bye_World OP's counterexample was already being discussed in the comments, which is why I saw no reason to bring it up. I am suggesting an approach that I find easier than OP's approach. I have no standpoint on whether or not my answer gets too many upvotes. – Mankind Jan 10 at 23:45
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I was mostly speaking to the $4$ people that immediately upvoted your answer. I think your answer is fine, but at the same time it sidesteps the real problem OP is having -- which is telling whether a vector is a member of a given set (and maybe the definition of closure, I'm not sure). – Bye_World Jan 10 at 23:48

I think that by your definition of $W$, $W$ is not a vector space, although it is s subset of a vector space. It is not a vector space, since the set must be closed under addition, and this is clearly not closed, since two vectors $(a_1, a_2, -1)$ and $(b_1, b_2, -1)$ add to $(a_1 + b_1, a_2 + b_2, -2)$ thus the sum of any two elements of $W$ is not in $W$ since $W$ contains only elements of the form $(x_1, x_2, -1)$. Thus it is not closed under addition, nor as you point out, under scalar multiplication.

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The differences of any two elements of $W$ is a sub-vector space $W_0$. $W$ itself is but a translated of this vector space with equation $z=0$ by the vector $(0,0,-1)$, and is called an affine subspace with direction $W_0$.

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Recall the axioms of a vector space. In particular, for any vector space $V$, there is an element $0 \in v$, such that $v+0=0+v=v$ for any $v \in V$. Obviously, your set $W = \{ (x_1,x_2,x_3) \in \Bbb R^3 \mid x_3 = -1\}$, doesn't contain such an element, thus can't be a vector space.

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