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Math Help - Lin Alg Proofs and Counterexamples

  1. #1
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    Lin Alg Proofs and Counterexamples

    I have compiled 57 prove or disprove lin alg questions for my final; however, these may be useful to all.

    Contributors to some of the solutions are HallsofIvy, Failure, Tikoloshe, jakncoke, tonio, and Defunkt.

    If you discover any errors in one of the solutions, then feel free to reply with the number and correction.


    Moderator Edit:
    1. If you want to thank dwsmith, please click on the Thanks button (do NOT post replies here unless you have a suggestion or erratum).
    2. The original thread can be viewed at http://www.mathhelpforum.com/math-he...tml#post511378.
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    Last edited by mr fantastic; May 12th 2010 at 01:55 AM.
    Thanks from x3bnm and Cesc1
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  2. #2
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    Here is a general proof for a Vector Space that shows when k+1 will be lin. ind. and lin. dep.

    Let \mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k be lin. ind. vectors in V. If we add a vector \mathbf{x}_{k+1}, do we still have a set of lin. ind. vectors?

    (i) Assume \mathbf{x}_{k+1}\in Span (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k)

    \displaystyle\mathbf{x}_{k+1}=c_1\mathbf{x}_{1}+..  .+c_k\mathbf{x}_{k}

    \displaystyle c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\ma  thbf{x}_{k+1}=0

    \displaystyle c_{k+1}=-1

    \displaystyle -1\neq 0; therefore, (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k, \mathbf{x}_{k+1}) are lin. dep.

    (ii) Assume \mathbf{x}_{k+1}\notin Span (\mathbf{x}_1, \mathbf{x}_2,  ..., \mathbf{x}_k)

    \displaystylec_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{  k}+c_{k+1}\mathbf{x}_{k+1}=0

    \displaystyle c_{k+1}=0 otherwise \displaystyle \mathbf{x}_{k+1}=\frac{-c_1}{c_{k+1}}\mathbf{x}+....+\frac{-c_k}{c_{k+1}}\mathbf{x}_k which is a contradiction.

    \displaystyle c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\ma  thbf{x}_{k+1}=0. Since (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k, \mathbf{x}_{k+1}) are lin ind., c_1=...c_{k+1}=0.
    Last edited by mash; March 6th 2012 at 01:08 AM. Reason: fixed latex
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    A is matrix n x n over field F, similar to an upper triangular matrix iff. the characteristic polynomial can be factored into an expression of the form \displaystyle (\lambda_1-\lambda)(\lambda_2-\lambda)...(\lambda_n-\lambda)

    \displaystyle det(A)=(a_{11}-\lambda)A_{11}+\sum_{i=2}^{n}a_{i1}A_{i1}

    \displaystyle (a_{11}-\lambda)A_{11}=(a_{11}-\lambda)(a_{22}-\lambda)...(a_{nn}-\lambda)

    \displaystyle =(-1)^n\lambda^n+...+(-1)^{n-1}\lambda^{n-1}

    \displaystyle p(0)=det(A)=\lambda_1\lambda_2...\lambda_n

    \displaystyle (-1)^{n-1}=tr(A)=\sum_{i=1}^{n}\lambda_i

    \displaystyle p(\lambda)=0 has exactly n solutions \lambda_1,...,\lambda_n

    \displaystyle p(\lambda)=(\lambda_1-\lambda)(\lambda_2-\lambda)...(\lambda_n-\lambda)

    \displaystyle p(0)=(\lambda_1)(\lambda_2)...(\lambda_n)=det(A)
    Last edited by mash; March 6th 2012 at 01:08 AM. Reason: fixed latex
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  4. #4
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    Let be A be a nxn matrix and B=I-2A+A^2. Show that if \mathbf{x} is an eigenvector of A belonging to an eigenvalue \lambda of A, then \mathbf{x} is also an eigenvector of B belonging to an eigenvalue \mu of B.

    B\mathbf{x}=(I-2A+A^2)\mathbf{x}=\mathbf{x}-2A\mathbf{x}+A^2\mathbf{x}=\mathbf{x}-2\lambda\mathbf{x}+A(\lambda\mathbf{x})=\mathbf{x}-2\lambda\mathbf{x}+(A\mathbf{x})\lambda =\mathbf{x}-2\lambda\mathbf{x}+\lambda^2\mathbf{x}=(1-2\lambda+\lambda^2)\mathbf{x}=\mu\mathbf{x}

    Hence, \mu=(1-2\lambda+\lambda^2)
    Last edited by mash; March 6th 2012 at 01:09 AM. Reason: fixed latex
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  5. #5
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    Let \lambda be an eigenvalue A and let \mathbf{x} be an eigenvector belonging to \lambda. Use math induction to show that, for m\geq 1, \lambda^m is an eigenvalue of A^m and \mathbf{x} is an eigenvector of A^m belonging to \lambda^m.

    A\mathbf{x}=\lambda\mathbf{x}

    p(k):=A^k\mathbf{x}=\lambda^k\mathbf{x}

    p(1):=A\mathbf{x}=\lambda\mathbf{x}

    p(k+1):=A^{k+1}\mathbf{x}=\lambda^{k+1}\mathbf{x}

    Assume p(k) is true.

    Since p(k) is true, then p(k+1):=A^{k+1}\mathbf{x}=\lambda^{k+1}\mathbf{x}.

    A^{k+1}\mathbf{x}=A^kA\mathbf{x}=A^k(\lambda\mathb  f{x})=\lambda(A^k\mathbf{x})=\lambda\lambda^k\math  bf{x}=\lambda^{k+1}\mathbf{x}

    By induction, A^k\mathbf{x}=\lambda^k\mathbf{x}.
    Last edited by mash; March 6th 2012 at 01:09 AM. Reason: fixed latex
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  6. #6
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    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!
    Last edited by mash; March 6th 2012 at 01:10 AM. Reason: fixed latex
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  7. #7
    Junior Member sorv1986's Avatar
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    Quote Originally Posted by dwsmith View Post
    A is matrix n x n over field F, similar to an upper triangular matrix iff. the characteristic polynomial can be factored into an expression of the form \displaystyle (\lambda_1-\lambda)(\lambda_2-\lambda)...(\lambda_n-\lambda)

    \displaystyle det(A)=(a_{11}-\lambda)A_{11}+\sum_{i=2}^{n}a_{i1}A_{i1}

    \displaystyle (a_{11}-\lambda)A_{11}=(a_{11}-\lambda)(a_{22}-\lambda)...(a_{nn}-\lambda)

    \displaystyle =(-1)^n\lambda^n+...+(-1)^{n-1}\lambda^{n-1}

    \displaystyle p(0)=det(A)=\lambda_1\lambda_2...\lambda_n

    \displaystyle (-1)^{n-1}=tr(A)=\sum_{i=1}^{n}\lambda_i

    \displaystyle p(\lambda)=0 has exactly n solutions \lambda_1,...,\lambda_n

    \displaystyle p(\lambda)=(\lambda_1-\lambda)(\lambda_2-\lambda)...(\lambda_n-\lambda)

    \displaystyle p(0)=(\lambda_1)(\lambda_2)...(\lambda_n)=det(A)
    A is matrix n x n
    over field F, similar
    to an upper
    triangular matrix
    iff. the
    minimal
    polynomial is a product of linear factors
    Last edited by mash; March 6th 2012 at 01:11 AM. Reason: fixed latex
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  8. #8
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    Re: Lin Alg Proofs and Counterexamples

    Great post thank you
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  9. #9
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    I have a workbook of 100 proofs I am typing up for Linear Alg. I am not sure if all the solutions are correct. However, I will post what I have done 15 problems so far for review and correct errors as they are found. Once they are all done, I will add them to other sticky of proofs I have up already.

    Deveno, Drexel, Pickslides, FernandoRevilla, and Ackbeet have helped with some of the problems already.

    Updated pdf with more solutions

    Updated pdf file.
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    Last edited by mr fantastic; December 20th 2011 at 04:31 PM. Reason: Merged on request by OP.
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