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Thread: 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; Jan 30th 2010 at 09:54 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; Jan 31st 2010 at 08:39 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 $\displaystyle \{v_1, v_2, \cdot\cdot\cdot, v_n\}$ is a basis for W, then $\displaystyle \{A^{-1}v_1, A^{-1}v_2, \cdot\cdot\cdot, A^{-1}v_n\}$ 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 $\displaystyle \{A^{-1}v_1, A^{-1}v_2, \cdot\cdot\cdot, A^{-1}v_n\}$ is independent. Suppose $\displaystyle \alpha_1 A^{-1}v_1+ \alpha_2 A^{-1}v_2+ \cdot\cdot\cdot+ \alpha_n A^{-1}v_j= 0$ for some scalars, $\displaystyle \alpha_1$, $\displaystyle \alpha_2$, $\displaystyle \cdot\cdot\cdot$, $\displaystyle \alpha_n$. 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; Jan 31st 2010 at 12:53 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 $\displaystyle A^{-1}\{\vec v_1,\vec v_2,\cdots,\vec v_k\}$ is a basis for $\displaystyle U$, not $\displaystyle A^{-1}(\vec v_1+\vec v_2+\cdots+\vec v_k)$. That said....

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

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

    Multiplying both sides by $\displaystyle A$, we have

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

    But since $\displaystyle \mathcal{B}$ is linearly independent, then $\displaystyle \alpha_i=0$ for each $\displaystyle i$, which is to say $\displaystyle A^{-1}\mathcal{B}$ is linearly independent, and the conclusion follows.
    Last edited by hatsoff; Jan 31st 2010 at 01:44 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 $\displaystyle A^{-1}v_1$, $\displaystyle A^{-1}v_2$, ..., $\displaystyle A^{-1}v_n$ (NOT their sum) are independent and so form a basis for U.
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