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

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In order to be a vector space, the set must satisfy about 78 conditions that prove it is closed under vector addition and scalar multiplication. Do you have the list of conditions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, I don't have the list

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do you have for a definition of a vector space?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I remember the 0 vector must in the set is one of the conditions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OK, 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
 one year ago
Best ResponseYou've already chosen the best response.0The first one is that the set must be commutative, i.e. X+Y = Y+X. Is this condition satisfied for this set?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great. 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
 one year ago
Best ResponseYou've already chosen the best response.0Good. There also must exist an additive identity, a vector A such that A+X=X+A=X. Does such a vector exist?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah. the vector (0,0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You'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
 one year ago
Best ResponseYou've already chosen the best response.0We're getting there. The set must be associative under scalar multiplication. For scalars r and s, (rs)X=r(sX). This one OK?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great. Scalar sums must be distributive, i.e. (r+s)X = rX + sX. This one OK?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good. Two to go. Vector sums must be distributive, i.e. r(X +Y) = rX + rY. This one OK?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Super. Last one. There must exist a scalar multiplication identity, a scalar r such that rX = X. What do you think?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Perfect. All conditions are satisfied so you have proven that your set is a vector space. Lot of work. Well done!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh... also. Do additive inverses have to be in the set?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the set { (x,y) : x,r >0 } isn't a vector space?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0huhm... It could be the first site is wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just dug out my textbook. Yes, X must be in the set. Your two references both confirm this. Must be an oversight on Wolfram.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0o: Wikipedia says they don't have to be in the set.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it does it says "there exists an element v in V..."

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ahhhh! 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
 one year ago
Best ResponseYou've already chosen the best response.0That would be correct. Good discussion. Glad to have that sorted out.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome! Thank you for your time :')

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Any time. Keep up the good work!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0all 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
 one year ago
Best ResponseYou've already chosen the best response.0@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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