Is the set { (0,0) } a vector space in R^2?

- anonymous

Is the set { (0,0) } a vector space in R^2?

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- schrodinger

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- anonymous

In order to be a vector space, the set must satisfy about 7-8 conditions that prove it is closed under vector addition and scalar multiplication. Do you have the list of conditions?

- anonymous

no, I don't have the list

- anonymous

What do you have for a definition of a vector space?

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## More answers

- anonymous

I remember the 0 vector must in the set is one of the conditions

- anonymous

OK, here we go. For all of these conditions, assume X and Y are vectors belonging to the set. Since this set has only one member, X=Y=(0,0).

- anonymous

The first one is that the set must be commutative, i.e. X+Y = Y+X. Is this condition satisfied for this set?

- anonymous

yes

- anonymous

Great. The set must also be associative, i.e. (X+Y)+Z=X+(Y+Z). Is this satisfied? Oh, Z is also a member of the set.

- anonymous

yeah

- anonymous

Good. There also must exist an additive identity, a vector A such that A+X=X+A=X. Does such a vector exist?

- anonymous

yeah. the vector (0,0)

- anonymous

You're very good. There must be an additive inverse. For every X, there must be a -X such that X+(-X)=0. This satisfied?

- anonymous

yeah

- anonymous

We're getting there. The set must be associative under scalar multiplication. For scalars r and s, (rs)X=r(sX). This one OK?

- anonymous

yes

- anonymous

Great. Scalar sums must be distributive, i.e. (r+s)X = rX + sX. This one OK?

- anonymous

yes

- anonymous

Good. Two to go. Vector sums must be distributive, i.e. r(X +Y) = rX + rY. This one OK?

- anonymous

yes

- anonymous

Super. Last one. There must exist a scalar multiplication identity, a scalar r such that rX = X. What do you think?

- anonymous

r = 1?

- anonymous

Perfect. All conditions are satisfied so you have proven that your set is a vector space. Lot of work. Well done!

- anonymous

oh... also. Do additive inverses have to be in the set?

- anonymous

Yes.

- anonymous

so the set { (x,y) : x,r >0 } isn't a vector space?

- anonymous

sorry, x,y > 0

- anonymous

I was just checking my notes. I stand corrected. The additive inverse must exist, but there is no requirement I can see that it has to be in the set. Might want to confirm that with a Google search.

- anonymous

This site states that additive inverses must be in the set:
http://www.math.niu.edu/~beachy/courses/240/06spring/vectorspace.html
But this site doesn't seem to say so
http://www.math.ucla.edu/~tao/resource/general/121.1.00s/vector_axioms.html

- anonymous

The statement I have says "For every X in vector space V, there exists a -X such that X + (-X) = 0. If this is correct then there is no requirement for -X to be in the vector space. Wolfram Alpha seems to agree.

- anonymous

huhm... It could be the first site is wrong.

- anonymous

Just dug out my textbook. Yes, -X must be in the set. Your two references both confirm this. Must be an oversight on Wolfram.

- anonymous

o: Wikipedia says they don't have to be in the set.

- anonymous

https://en.wikipedia.org/wiki/Vector_space#Definition

- anonymous

I think it does it says "there exists an element -v in V..."

- anonymous

https://en.wikipedia.org/wiki/Vector_space

- anonymous

ahhhh! I overlooked! XD. Ok, so additive inverse do indeed have to be in the set. So the example I gave above isn't a vector space since (-x,-y) isn't in the set

- anonymous

That would be correct. Good discussion. Glad to have that sorted out.

- anonymous

awesome! Thank you for your time :')

- anonymous

Any time. Keep up the good work!

- anonymous

all you need to actually prove is, since we're dealing with a subset of \(\mathbb{R}^2\) which presumably you already know is a vector space, that we have closure under addition and scaling: $$k(0,0)=(0,0)\\(0,0)+(0,0)=(0,0)$$ so we're set -- the other properties are inherited from the parent space

- anonymous

@oldrin.bataku thanks. Proving the set is closed under addition and multiplication is definitely quicker than checking those axioms in the definition.

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