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anonymous

  • 5 years ago

Can anyone direct me somewhere to understand vector subspaces? My teacher lacks the visual aspect. I think this would help understand best.

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  1. anonymous
    • 5 years ago
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    Vector subspaces are just subsets of vector spaces that are themselves, vector spaces. They are subsets that satisfy the axioms of a vector space. What is it you need to know? Have you tried YouTube?

  2. anonymous
    • 5 years ago
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    The part I'm having a problem with is proving the Axiom. Example problem I'm unsure with states this: Determine whether the set together with the indicated operations, is a vector space. The set {(x,y)| y=ax^2 . This is supposed to be in R2 space. I'm sure if fails some Axioms, but not sure how, we need to show our work for which ones don't pass. (I hope that all makes sense)

  3. anonymous
    • 5 years ago
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    I'm having a hard time figuring out a search term that targets exactly this type of question. This forum is really great, I was hoping to get some direction here!

  4. anonymous
    • 5 years ago
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    Oh, so you're not proving a subspace, you're proving a vector space entirely...that's a longer process.

  5. anonymous
    • 5 years ago
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    Yes, sorry for the confusion, that is how confused I am on this. I understand the space is R2 and I even understand that y=ax^2 would result a parabola in this space, I'm thinking the multiply scalar by -1 would be one failure, but I'm not sure if I'm thinking it correctly.

  6. anonymous
    • 5 years ago
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    It's actually easier than it looks. You basically have to test your set under the operations defined on your space (vector addition and scalar multiplication) for each one of the axioms. There are 10 of these to test.

  7. anonymous
    • 5 years ago
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    It's a tedious process where you just have to check your elements under the defined operations obey those 10 axioms. If they do, you have a vector space.

  8. anonymous
    • 5 years ago
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    Yes I get all of that. The way I'm looking at it that is confusing me is treating a parabola as a vector? It doesn't make sense.

  9. anonymous
    • 5 years ago
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    Oh! Forget about the parabola.

  10. anonymous
    • 5 years ago
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    Your element is (x,ax^2)...look at that, not the parabola.

  11. anonymous
    • 5 years ago
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    Are you still in the frame of mind of thinking as vectors as those 'things' in physics with 'magnitude' and 'direction'?

  12. anonymous
    • 5 years ago
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    Probably. I didn't do well with physics, but I have a tendency to think of it that way I guess. My teacher failed to explain this graphically, I think that would really help. I'm having a hard time seeing how I would add another vector to it or or multiply it.

  13. anonymous
    • 5 years ago
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    Well, this is what I'm saying - it all depends on how the addition and multiplication is *defined* on the set. In this instance, unless you've omitted something, addition would be like:

  14. anonymous
    • 5 years ago
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    \[(x_1,ax^2_12)+(x_2,ax^2_2)=(x_1+x_2,ax^2_1+ax^2_2)\]

  15. anonymous
    • 5 years ago
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    The X doesn't appear to be a problem with the Axiom tests, it's the ax^2, but does the a have a significance?

  16. anonymous
    • 5 years ago
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    Yes. I'm falling asleep, but from what I can see, one of the axioms is that, for EVERY element in your set, in order for your set to be a vector space, you must an 'additive inverse' - for example, if u is in V, then there must be some -u in V where u + (-u) = 0 (the zero vector). I think your set fails here.

  17. anonymous
    • 5 years ago
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    Is it because the x^2 is no longer P2?

  18. anonymous
    • 5 years ago
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    All you need to do is show a test case. From your set, you have the element (2,a(2)^2) which equals (2,4a). If you are to have an element that will be an additive inverse, you must have (x,ax^2) such that (x,ax^2)+(2,4a) = (0,0). That means, for the first co-ordinate, your x would have to be -2. But then because of the definition of your set, if x is -2, the other co-ordinate would be a(-2)^2 = 4a again! So all you could do is (2,4a) + (-2, 4a) = (0,8a) does not equal (0,0).

  19. anonymous
    • 5 years ago
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    i.e. you have an element in your set that does not have an additive inverse under your definition of vector addition. You therefore do not have a vector space here. Does that help?

  20. anonymous
    • 5 years ago
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    I can only stick around for a couple more minutes - I have to go to bed to be up again in five hours :)

  21. anonymous
    • 5 years ago
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    I guess I'm looking into it too deeply because I'm also looking at that a and trying to see if that has any bearing. So basically I ignore that a and rewrite the main vector as a second vector that is created by one of the Axioms to disprove it will work? Sorry to keep you up. That's why I figured an online resource would be more helpful. Thanks for your help!

  22. anonymous
    • 5 years ago
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    What I think is that you're probably not looking at it abstractly enough. People have a hard time letting go of the specific rules taught at the secondary level. Now, having said that, all you have to do with this testing of spaces thing is blindly follow the rules: take your elements, take your definitions of addition and scalar multiplication (whatever they happen to be) and just do what it says on the tin ;) Try reading this - http://tutorial.math.lamar.edu/Classes/LinAlg/VectorSpaces.aspx

  23. anonymous
    • 5 years ago
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    That is exactly what it is. I love math, but the abstract parts like this linear algebra class are really difficult. I'll look at the link. Thanks!

  24. anonymous
    • 5 years ago
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    If you have anymore questions, post them - either someone else will respond or if not, I will. Look for online lectures as well - MIT on YouTube might have something. PS - the 'a' in your ax^2 is assumed a non-zero constant (by me, anyway).

  25. anonymous
    • 5 years ago
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    No worries - 'fan' me - I need the points! ;)

  26. anonymous
    • 5 years ago
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    I actually watched the MIT and it helped, but not this specifically for these type of proofs in vector space type problems.

  27. anonymous
    • 5 years ago
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    I will definitely fan you, thanks a ton for the help. Hopefully my brain will let go and let me think more abstractly. ;-)

  28. anonymous
    • 5 years ago
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    Yeah, and it's good you love math - it will click eventually. If it were easy, we'd know everything already.

  29. anonymous
    • 5 years ago
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    Good luck!

  30. anonymous
    • 5 years ago
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    Bed...

  31. anonymous
    • 5 years ago
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    Thanks again and get some sleep!

  32. anonymous
    • 5 years ago
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    If anyone else out there has any tricks to understanding this better I'm all ears. Thanks!

  33. anonymous
    • 5 years ago
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    are you still trying to get this?

  34. anonymous
    • 5 years ago
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    is the subset of \[\mathbb{R}^2\] you are looking at \[H=\left\{(x,y)\in\mathbb{R}^2:y=ax^2, a\in \mathbb{R}\right\}\]

  35. anonymous
    • 5 years ago
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    Sounds like you are having the same problem I had with LA. No matter how much I studied, I just couldn't get the conceptual stuff. A friend referred me to this site. He's the Intro to LA prof. at MIT and what I had time to watch seemed very good. http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/

  36. anonymous
    • 5 years ago
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    I took my test and argued my reasoning with my professor and he gave me points back. I think he could have explained this better and he realized it too. This is a great resource, I have Calc II next semester, so I'll be a frequent flier here. I'm looking forward to getting away from the abstract.

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