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Subspaces, R^n How to visualize?

  1. Dec 11, 2005 #1
    Hey,

    I have no problems dealing with vectors in space, R^3. But I am having a lot of trouble with vectors in R^n. One of my basic questions is what is R^n. I mean doesn't the vector space already encompass everything? How do I visualize R^n vectors? Can you reccomend any good online tutorials that explain how to approach problems in the subspace units?

    Any help would be great!!

    Thanks!!!
     
  2. jcsd
  3. Dec 11, 2005 #2

    matt grime

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    Don't visualize it. Period. These are not spaces that have any physically visible manifestation really, so don't bother - it is unnecssary to do so, and unhelpful in the long run if you only attempt mathematics that is 'visualizable'.

    R^n is just the set of n-tuples of numbers, and that is all you need.
     
  4. Dec 11, 2005 #3

    JasonRox

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    I can't even visualize R^3 half the time. :grumpy:

    Look at the bright side of R^n, you'll never be asked to draw a graph.
     
  5. Dec 11, 2005 #4

    Hurkyl

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    I dunno, I've seen people draw diagrams in much uglier spaces than than R^4. That's where the real fun begins. :smile:
     
  6. Dec 12, 2005 #5
    Thnx! I think Im starting to get the hang of this stuff, after constantly and constantly reading over notes and trying examples. Quick question. In one example it says,

    Let U = { X in R^n | AX = 0 (Zero Vector)}
    Is U a subspace?

    Am I correct in reading the question as, X is in R^n if and only if AX = 0. That is, when checking all three conditions, I should get the zero vector?
     
  7. Dec 12, 2005 #6
    It is the definition of a set U. It says X is in U iff X is a vector in R^n such that AX=0. Now you have to determine whether U satisfies the axioms of a vector space.
     
  8. Dec 12, 2005 #7

    matt grime

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    that is how we all learned mathematics.

    not quite, you should read it as X is in U if AX=0, where A is some fixed matrix, or that U is the subset of vectors x in R^n such that Ax=0. You need to show that U satisfies the axioms of a subspace, ie 0 is in U, and if x and y are in U that x+y is in U, and that if x is in U and t is a real number that tx is in U.



    no, that doesn't follow at all
     
  9. Dec 12, 2005 #8

    JasonRox

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    Are you talking about the all the homogeneous solutions for the matrix A?

    If that is the case, why not try yourself and see if it is a subspace. All you need to do is check if it is closed under addition and scalar multiplication.

    In order words, show that if x,y is in U, then x+y is in U, and if k is any scalar, then kx is in U.

    Note: If it has only one solution for the homegeneous system, then we know it is the trivial solution. The trivial solution is the zero vector itself, which creates a subspace all on it's own.
     
    Last edited: Dec 12, 2005
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