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Math Help - Minimal Polynomial

  1. #1
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    Minimal Polynomial

    Here is my problem:

    Let A be a complex n x n matrix with minimal polynomial q(x)=the sum from j=0 to m of \alpha_j x^j where m\leq n and \alpha_m = 1.

    Show: If A is non-singular then \alpha_0 does not equal 0.

    So, I get that 0=q(A)= \alpha_0 I_n + \alpha_1 A + \alpha_2 A^2 +...+A^m, but I'm not sure what to do here. I assume we will have to use the fact that A is non-singular, but I'm not sure how. Does it maybe involve multiplying both sides by x on the right side? Any hints would be much appreciated!
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  2. #2
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    If \alpha_0 is zero, then 0=\alpha_1 A + \alpha_2 A^2 +...+A^m=A(\alpha_1 I + \alpha_2 A +...+A^{m-1})<br />
. But since A is invertible...
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  3. #3
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    I have tried before to assume that \alpha_0 is 0, but I can't figure out what to do after. Because A is invertible, we can multiply both sides by A^{-1} so we get:

    0= \alpha_1+\alpha_2 A+...+A^{m-1}.

    Buth I haven't been able to understand how that poses a contradiction.

    Does it imply that all the alphas would have to be 0? Which would be aproblem because \alpha_m was given to be equal to 1.
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  4. #4
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    Oh!!! Does 0= imply that the there would have to be a polynomial of lower degree (m-1)? Which would be impossible because we were talking about the minimal polynomial to begin with.
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  5. #5
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    Quote Originally Posted by azdang View Post
    Oh!!! Does 0= imply that the there would have to be a polynomial of lower degree (m-1)? Which would be impossible because we were talking about the minimal polynomial to begin with.
    Yes.
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