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  1. MHB Master

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    MHB Math Notes Award (Jan-June 2013)
    #1
    We used to have a bunch of problems and proofs that were in a pdf could be downloaded by anyone. Since we aren't able to upload pdf files of a certain size, I provided a link to google docs. If there is an error, typo, or something is just drastic wrong let me know.

    "https://docs.google.com/file/d/0B6UdM74FvKhPNWk3bWR3VHhod1U/edit"]Undgraduate Final Review[/URL] Practice problems with solutions
    Undergraduate level

    However, with this first link, I can't edit this document. It was created with Maple which I no longer have. So errors have to just be corrected in the thread and then consolidated for readability.

    This pdf has more advanced proofs in it.



    I have completed the second set. The only ones that need solutions are $A5$ part 2, $B7$ part2, $C4$ part 2 and 3, $C5$, $D4$, $F7$ needs to be checked, $H3$, $H10$, $I4$ part 3, $I5$, $I10$ part 2, $J7$, and $J10$.
    The rest of the problems I believe to be right but they should still be checked out.

    Comments and questions should be posted here:

    Last edited by MarkFL; April 7th, 2013 at 14:51. Reason: Added link to commentary topic

  2. MHB Master

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    MHB Math Notes Award (Jan-June 2013)
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    For the first document, here is what Ackbeet$\equiv $Ackbach suggested on MHF that needed to be adjusted
    Quote Originally Posted by Ackbeet View Post
    Very nice review! I just had a few comments:


    1. From Test 5, Problem 4, on page 4. I would say more than eigenvectors must be nonzero, by definition. It's not that the zero eigenvector case is trivial: it's that it's not allowed.


    2. Page 6, Problem 8: typo in problem statement. Change "I of -I" to "I or -I".


    3. Page 8, Problem 21: the answer is correct, but the reasoning is incorrect. It is not true that $\mathbf{x}$ and $\mathbf{y}$ are linearly independent if and only if $|\mathbf{x}^{T}\mathbf{y}|=0.$ That is the condition for orthogonality, which is a stronger condition than linear independence. Counterexample: $\mathbf{x}=(\sqrt{2}/2)(1,1),$ and $\mathbf{y}=(1,0).$ Both are unit vectors, as stipulated. We have that $|\mathbf{x}^{T}\mathbf{y}|=\sqrt{2}/2\not=0,$ and yet
    $a\mathbf{x}+b\mathbf{y}=\mathbf{0}$ requires$a=b=0,$ which implies linear independence.


    Instead, the argument should just produce a simple counterexample, such as $\mathbf{x}=\mathbf{y}=(1,0)$.


    Good work, though!

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