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Math Help - Linear Algebra Stuff. Find a Basis, I Need Your Help!

  1. #1
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    Linear Algebra Stuff. Find a Basis, I Need Your Help!

    Hi, I have an assignment due and I have done most of the questions there is just one part of a question I have left, if someone can help that would be amazing



    Edit: I answered the second question, now all I need help with is 1B).

    I already did the first question part A), but I can't figure out B


    Thanks
    Last edited by mmmboh; January 30th 2010 at 10:54 AM.
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  2. #2
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    OK did the second one, can someone help me with 1B) please!
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  3. #3
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    For the first question part B, I know all the vectors in B are independent because B is a basis for W, so I know the inverse of all these vectors are also independent. And I know that A maps the vectors in U onto W, so the inverse of A will map the vectors in W, or B, onto U...but how should I write this out?
    Last edited by mmmboh; January 31st 2010 at 09:39 AM.
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  4. #4
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    Something obviously happened to your last post!

    The one you are having difficulty with is, I assume, 3b since there is no "problem 1" given. It says:

    If A is an n by n invertible matrix from U to W and \{v_1, v_2, \cdot\cdot\cdot, v_n\} is a basis for W, then \{A^{-1}v_1, A^{-1}v_2, \cdot\cdot\cdot, A^{-1}v_n\} is a basis for U. Since A is n by n, both U and W must be of dimension n so we need only show that \{A^{-1}v_1, A^{-1}v_2, \cdot\cdot\cdot, A^{-1}v_n\} is independent. Suppose \alpha_1 A^{-1}v_1+ \alpha_2 A^{-1}v_2+ \cdot\cdot\cdot+ \alpha_n A^{-1}v_j= 0 for some scalars, \alpha_1, \alpha_2, \cdot\cdot\cdot, \alpha_n. Apply to A to both sides of that equation. What does that, together with the fact that [tex]\{v_1, v_2, \cdot\cdot\cdot, v_n\}[tex] is a basis, tell you about the scalars?
    Last edited by HallsofIvy; January 31st 2010 at 01:53 PM.
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  5. #5
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    Sorry I didn't really understand your last sentence, but what I think you want to know is that the scalars are all zero, I think?
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  6. #6
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    I'm a little bit confused on how this proves that (A^-1)(V1+...Vk) is a basis for U though
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  7. #7
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    Quote Originally Posted by mmmboh View Post
    I'm a little bit confused on how this proves that (A^-1)(V1+...Vk) is a basis for U though
    EDIT: First of all, as HallsofIvy pointed out, your notation is off, here. You need to prove that A^{-1}\{\vec v_1,\vec v_2,\cdots,\vec v_k\} is a basis for U, not A^{-1}(\vec v_1+\vec v_2+\cdots+\vec v_k). That said....

    There are a couple of ways to do this. Going by HallsofIvy's method, we need to show that A^{-1}\mathcal{B} is linearly independent. To do this, let

    \sum_{i=1}^k\alpha_iA^{-1}\vec v_i=\textbf{0} for some scalars \alpha_1,\alpha_2,\cdots,\alpha_k\in F.

    Multiplying both sides by A, we have

    \sum_{i=1}^k\alpha_i\vec v_i=\textbf{0}.

    But since \mathcal{B} is linearly independent, then \alpha_i=0 for each i, which is to say A^{-1}\mathcal{B} is linearly independent, and the conclusion follows.
    Last edited by hatsoff; January 31st 2010 at 02:44 PM.
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  8. #8
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    Quote Originally Posted by mmmboh View Post
    I'm a little bit confused on how this proves that (A^-1)(V1+...Vk) is a basis for U though
    It doesn't. That's not true and not what you were asked to show!
    The point is that, since all the scalars are 0, the vectors A^{-1}v_1, A^{-1}v_2, ..., A^{-1}v_n (NOT their sum) are independent and so form a basis for U.
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