anonymous
  • anonymous
Is the set { (0,0) } a vector space in R^2?
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • 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
  • anonymous
no, I don't have the list
anonymous
  • anonymous
What do you have for a definition of a vector space?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I remember the 0 vector must in the set is one of the conditions
anonymous
  • 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
  • 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
  • anonymous
yes
anonymous
  • 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
  • anonymous
yeah
anonymous
  • 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
  • anonymous
yeah. the vector (0,0)
anonymous
  • 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
  • anonymous
yeah
anonymous
  • 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
  • anonymous
yes
anonymous
  • anonymous
Great. Scalar sums must be distributive, i.e. (r+s)X = rX + sX. This one OK?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Good. Two to go. Vector sums must be distributive, i.e. r(X +Y) = rX + rY. This one OK?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Super. Last one. There must exist a scalar multiplication identity, a scalar r such that rX = X. What do you think?
anonymous
  • anonymous
r = 1?
anonymous
  • anonymous
Perfect. All conditions are satisfied so you have proven that your set is a vector space. Lot of work. Well done!
anonymous
  • anonymous
oh... also. Do additive inverses have to be in the set?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
so the set { (x,y) : x,r >0 } isn't a vector space?
anonymous
  • anonymous
sorry, x,y > 0
anonymous
  • 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
  • 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
  • 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
  • anonymous
huhm... It could be the first site is wrong.
anonymous
  • 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
  • anonymous
o: Wikipedia says they don't have to be in the set.
anonymous
  • anonymous
https://en.wikipedia.org/wiki/Vector_space#Definition
anonymous
  • anonymous
I think it does it says "there exists an element -v in V..."
anonymous
  • anonymous
https://en.wikipedia.org/wiki/Vector_space
anonymous
  • 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
  • anonymous
That would be correct. Good discussion. Glad to have that sorted out.
anonymous
  • anonymous
awesome! Thank you for your time :')
anonymous
  • anonymous
Any time. Keep up the good work!
anonymous
  • 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
  • anonymous
@oldrin.bataku thanks. Proving the set is closed under addition and multiplication is definitely quicker than checking those axioms in the definition.

Looking for something else?

Not the answer you are looking for? Search for more explanations.