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Okay, so my text required me to actually prove both sides; The non-empty subset $W$ is a subspace of a $V$ if and only if $a\mathbf{x}+b\mathbf{y} \in W$ for all scalars $a,b$ and all vectors $\mathbf{x},\mathbf{y} \in W$. I figured out one direction already (that if $W$ is a subspace, then $a\mathbf{x}+b\mathbf{y}$ is an element of $W$ since $a\mathbf{x}$ and $b\mathbf{y}$ are in $W$ and thus so is their sum), but I'm stuck on the other direction.

I got that if $a\mathbf{x}+b\mathbf{y} \in W$, then $c(a\mathbf{x}+b\mathbf{y}) \in W$ as well since we can let $a' = ca$ and $b' = cb$ and we're good, so $W$ is closed under scalar multiplication. But for closure under addition, my text states that I can "cleverly choose specific values for $a$ and $b$" such that $W$ is closed under addition as well but I cannot find any values that would work. What I'm mostly confused about is how choosing specific values for $a$ and $b$ would prove anything, since $a, b$ can be any scalars and $\mathbf{x},\mathbf{y}$ can be any vectors, so setting conditions like $a = b$, $a = -b$, $a = 0$ or $b = 0$ don't seem to prove anything.

Also something I'm not sure about is if they're saying that $a\mathbf{x}+b\mathbf{y} \in W$, am I to assume that that is the only form? So if I'm testing for closure under addition, I have to do something like $(a\mathbf{x}+b\mathbf{y})+(c\mathbf{z}+d\mathbf{w})$?

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4 Answers

up vote 3 down vote accepted

You seem to be misunderstanding $\forall x,y \in W, \forall a,b, ax+by \in W$. This doesn't say that for a specific pair $(x,y) \in W^2$, every element of $W$ would be of the form $ax+by$ for some scalars $a$ and $b$. It does not say anything like $"\ldots \forall w \in W \ldots w =ax+by"$

For example, when you proved that $\forall w \in W, \forall c, cw \in W$, you implicitly supposed that $w$ was of this form, while you never stated who $x$ and $y$ were. If you thought that $x$ and $y$ were fixed members of $W$, then it's false.

However, you can say that since $w \in W$, and $0$ is a scalar, your hypothesis says that if you choose $x=w, y=w, a=c, b=0$, you get $cw+0w \in W$. Thus $cw \in W$. This is a valid proof of $\forall w \in W, \forall c, cw \in W$, and if you understand it, you should be able to prove the rest.

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It seems to me that your question arises out of a misunderstanding of the expression "$a\mathbf{x}+b\mathbf{y} \in W$ for all scalars $a,b$ and all vectors $\mathbf{x},\mathbf{y} \in W$". You seem to be taking that to mean that all elements of $W$ are of the form $a\mathbf{x}+b\mathbf{y}$ (which they are, but that's irrelevant). It just means that if you pick any scalars $a$, $b$ and any vectors $\mathbf{x},\mathbf{y} \in W$, then $a\mathbf{x}+b\mathbf{y} \in W$. So you're not "setting conditions" if you pick particular values of $a$ and $b$, you're just using what's given -- and what's given is far more general than what you're trying to prove, so it makes sense that you have to pick specific values to prove it.

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alright, so I was just overthinking it! thanks. –  Eugene Bulkin Feb 27 '11 at 15:54
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Consider $a=b=1$ and $a=-b=1$.

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As you may be a little bit confused this, let me remember you that

$W\subset$ is a subspace of $V$ if and only if:

  1. $W\not=\emptyset$
  2. $w+w'\in W$ for all $w,w'\in W$
  3. $aw'\in W$ for all $w'\in W, \ a\in\mathbb{R}$

Thus if $W$ is a subspace of $V$, it is, as you say, easy to see that

$$aw+bw'\in W \ \forall \ w,w'\in W, \ a,b\in \mathbb{R}$$

if you combine 2 and 3.

For the other side, suppose that

$$aw+bw'\in W \ \forall \ w,w'\in W, \ a,b\in \mathbb{R}$$

and you have to prove that then the three points above are true.

  1. $W$ is not empty by definition.
  2. Take $w, w'\in W$. Letting $a=1, \ b=1$ gives that $w+w'\in W$
  3. Take $w\in W$. Letting $a=1, \ b=0$ gives that $1w+0w'=w\in W$

So you see that they are simply particular cases of the relation.

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