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Proving something to be a basis.

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Letting u=[3, 0, -5], v=[2, 1, 5] and w=[-1, 3, 4], how would I show that a general vector can be written as a linear combination of this 'basis?' Without using an augmented matrix and getting a really messy result by using arbitrary a, b, and c values as the solutions?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 26, 2013 #2

    jbunniii

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    Can you show that the three elements of the canonical basis, namely [1, 0, 0], [0, 1, 0], and [0, 0, 1], can be written as linear combinations of the given vectors? If so, can you see how that helps?
     
  4. Sep 26, 2013 #3

    Ray Vickson

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    So much depends on what you know and are allowed to use. For example, do you know the connection between linear independence and determinants? If so, you can use that.

    Otherwise, you can try to show directly that u, v and w are linearly independent, so form a basis; then you can quote a theorem to finish the problem. Or, do you really want to express an arbitrary vector x = [x1,x2,x3] as an explicit linear combination of u, v and w? The problem statement seemed to say otherwise, but only you know for sure.
     
  5. Sep 26, 2013 #4
    I have shown that they are linearly independent. But I also need to show that the Span(s)=V.
     
  6. Sep 26, 2013 #5

    jbunniii

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    If you have three linearly independent vectors, what can you say about the dimension of their span?
     
  7. Sep 26, 2013 #6
    I figured it out :) I was also wondering if I am given four polynomials, of which the highest degree is x^2, they cannot be a basis for the vector space P3(R)?
     
  8. Sep 26, 2013 #7

    jbunniii

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    What is P3(R)? The space of polynomials with degree <= 3? If so, consider whether it's possible to write ##x^3## as a linear combination of the four given polynomials.
     
  9. Sep 26, 2013 #8
    Well, it wouldn't be possible unless my scalar was the variable x, and not a real number.
     
  10. Sep 26, 2013 #9

    jbunniii

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    Can you prove why it wouldn't be possible?
     
  11. Sep 26, 2013 #10
    Well, I converted the polynomials to a matrix, and tried to put it into RREF to prove they were linearly independent. But the first column is all 0's, so it was pointless to go any further as they were obviously linearly dependent and thus not a basis I assumed.
     
  12. Sep 26, 2013 #11

    jbunniii

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    OK, that's reasonable but you can also make an argument without even knowing anything about the four polynomials ##p_1(x),\ldots,p_4(x)## except that none of their degrees are higher than 2.

    Then suppose ##x^3## can be written as a linear combination of those four polynomials. We would then have
    $$x^3 + \sum_{n=1}^{4}a_n p_n(x) = 0$$
    for some scalars ##a_1,\ldots,a_4##. But ##x^3 + \sum_{n=1}^{4}a_n p_n(x)## is a polynomial of degree 3 (why?), so it cannot equal the zero polynomial (why?)
     
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