# Determining a set is a vector space

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• Feb 25th 2011, 03:44 PM
evant8950
Determining a set is a vector space
Hey everyone,

I'm a little confused. I am confused on how to determine if a set is a vector space. I've look at some examples and I am still a bit confused. How do you go about determining if a set is vector space?

Here is a problem I am confused about:
The set of all pairs of real numbers of the form (x,0) with the standard operations on R^2

Any help is appreciated.
Thanks
• Feb 25th 2011, 03:48 PM
Ackbeet
The set, together with the scalar multiplication and vector addition operations, has to satisfy the axioms of a vector space. Now, in your case, you know already (I assume?) that R^2 is a vector space with "the standard operations". When, therefore, you're trying to determine if a set is a vector subspace, your job is a little easier. All you have to do is determine three things:

1. The set is nonempty.
2. The set is closed under scalar multiplication.
3. The set is closed under vector addition.

If those three conditions are satisfied, you've got yourself a subspace. Do those three conditions hold for your set?
• Feb 25th 2011, 03:53 PM
evant8950
What is meant when it is "closed" by addition and multiplication?

Thanks
• Feb 25th 2011, 04:05 PM
Ackbeet
Ok, let's get into the notation. Let $\displaystyle V$ be a vector space over the scalar field $\displaystyle F.$ (You could think of $\displaystyle \mathbb{R}^{2}$ as the vector space, over the field of real numbers $\displaystyle \mathbb{R}$.) The word "closed", in this context, means the following:

Scalar multiplication: for any vector $\displaystyle \mathbf{v}\in V$ and scalar $\displaystyle f\in F,$ it follows that $\displaystyle f\mathbf{v}\in V.$ That is, multiplying a vector in $\displaystyle V$ by a scalar in $\displaystyle F$ keeps you inside the vector space $\displaystyle V.$

Vector addition: for any two vectors $\displaystyle \mathbf{u},\mathbf{v}\in V,$ it follows that $\displaystyle \mathbf{u}+\mathbf{v}\in V.$ Again, the addition of two vectors in $\displaystyle V$ is still inside $\displaystyle V.$

That's what closure means. Does that make sense?

Here's an example. Let

$\displaystyle V=\{\mathbf{v}\in\mathbb{R}^{2}|v_{1}+v_{2}=0\}.$

Take the reals as the field. We check if it's closed under the two operations.

Let $\displaystyle r\in\mathbb{R},$ and let $\displaystyle \mathbf{v}\in V$ be given by $\displaystyle \mathbf{v}=\langle v_{1},v_{2}\rangle.$ By assumption, we have that $\displaystyle v_{1}+v_{2}=0.$ Then $\displaystyle r\mathbf{v}=\langle r v_{1},r v_{2}\rangle.$ Does the property still hold? We check to see if $\displaystyle rv_{1}+rv_{2}=0,$ or $\displaystyle r(v_{1}+v_{2})=0,$ which we can see is true, since $\displaystyle v_{1}+v_{2}=0.$ Thus, we can see that we're closed under scalar multiplication, since the scalar and vector I chose was completely arbitrary.

The vector addition case will be very similar. Finally, you note that the zero vector is, indeed, in our space, and hence the set is nonempty.

Make sense?
• Feb 25th 2011, 04:12 PM
evant8950
I think I understand what you are explaining. If you add two vectors or multiply by a scalar the new vector is still in the space?
• Feb 25th 2011, 04:14 PM
Ackbeet
Right. If that condition holds for all vectors and scalars, you've got yourself a vector subspace.

Now, I should point out that if you're trying to see if a given set is a vector space, and it's not a subset of a known vector space, then you've got to verify every single one of the axioms to which I linked in post # 2. Only then, a full vector space, will you have.
• Feb 25th 2011, 04:18 PM
evant8950
So with the problem I posted:
The set of all pairs of real numbers of the form (x,0) with the standard operations on R^2

So (x,0) considered "u"
• Feb 25th 2011, 04:26 PM
evant8950
I understand your example, but how do I see if (x,0) is a vector space of R^2 if there is no condition like in your example such as v1+v2=0.

Thanks
• Feb 25th 2011, 04:33 PM
Ackbeet
If there's no "condition", your job is actually easier.

1. Is the zero vector in your set? (Really just need nonempty, but the zero vector is often the easiest to check).

2. Closed under scalar multiplication?

• Feb 25th 2011, 04:47 PM
evant8950
So would this set no be a vector space because of the axiom (c+k)u = cu + ku

(cx + kx, 0) is not equal to (cx + kx)?
• Feb 25th 2011, 04:48 PM
Ackbeet
No, no, no. Just check what I mentioned in post # 9. Is the zero vector in your set?
• Feb 25th 2011, 04:51 PM
evant8950
So 0 + (x,0) = (x,0)? I think I am making this harder than it really is.

Thanks
• Feb 25th 2011, 04:54 PM
Ackbeet
Is the vector $\displaystyle \mathbf{0}=\langle 0,0\rangle$ in your set? That is, does $\displaystyle \langle 0,0\rangle$ look like a vector of the form $\displaystyle \langle x,0\rangle,$ for some $\displaystyle x?$
• Feb 25th 2011, 04:56 PM
evant8950
If x is 0 then if would be (0,0)
• Feb 25th 2011, 04:59 PM
Ackbeet
Right. So that tells you that the zero vector is in the set, and hence the set is nonempty. Next question:

If you take an arbitrary scalar $\displaystyle r\in\mathbb{R},$ and multiply it times an arbitrary vector $\displaystyle \mathbf{x}=\langle x,0\rangle$ in your set, is the result still in your set?
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