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Math Help - Cayley Hamilton Theorem Question

  1. #1
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    Cayley Hamilton Theorem Question

    Hello, a tad stuck on an algebra assignment.

    The Cayley Hamilton theorem implies that, for any nxn matrix A over a field K, there is a polynomial p(x) with coefficents in K and degree n in x such that p(A)=0.
    By considering the matrices A^i for ,0 less than or eqaul to (i) less than or equal to (n^2), prove, without using the cayley hamilton theorem, that there is a non-zero polynomial p(x) of degree at most (n^2) in x, such that p(A)=0.

    Hints/pointers/answers will all be welcomed.

    Thankyou
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Look at the n^2 matrices:

    I,A,A^2,...,A^(n^2).

    Remember that the vector space M of matrices nxn is n^2, but you have n^2+1 matrices, so...?
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Look at the n^2 matrices:

    I,A,A^2,...,A^(n^2).
    Done this, but still stuck.
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  4. #4
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    Quote Originally Posted by Also sprach Zarathustra View Post
    but you have n^2+1 matrices
    why?
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by lordslytherin View Post
    why?

    I,A,A^2,...,A^(n^2) <==>A^0,A,A^2,...,A^(n^2).
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  6. #6
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    Quote Originally Posted by Also sprach Zarathustra View Post
    I,A,A^2,...,A^(n^2) <==>A^0,A,A^2,...,A^(n^2).
    Thought as much, when i read it at n^3, rather than (n^2)+1. Still at a loss to do after that though.
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    I will continue from my first comment.

    ...hence, n^2+1 matrices are linear dependence, in other words:

    there exist t_0,t_1,...t_{n^2+1} scalars that not all of them is zero, which for them:

    t_{n^2+1}A^{n^2+1}+...+t_1A+t_0I=0

    or in other writing...

    A is a root of f(x)=t_{n^2+1}x^{n^2+1}+...+t_1x+t_0I.
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  8. #8
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    Thanks alot Also sprach zarathustra, most helpful.
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