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Math Help - one prove question

  1. #1
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    one prove question

    Let A be an nxn matrix with characteristic polynomial

    f(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1}+...+ a_1\lambda +a_0

    Prove that a_0 = (-1)^ndetA
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  2. #2
    Member kjchauhan's Avatar
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    Every square matrix satisfies its characteristic equation.
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  3. #3
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    Quote Originally Posted by kjchauhan View Post
    Every square matrix satisfies its characteristic equation.
    That may be, but you haven't proven this. This kind of question calls for an equation proof, not just a statement.

    I second the request for help on this.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Recall that f(t)=\det (A-tI). Set t=0 on both sides! You should just get a_0. The factor (-1)^n could come into the picture if somehow you had defined f(t)=\det (tI-A) but that is certainly not standard. The answer is just a_0.
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by kjchauhan View Post
    Every square matrix satisfies its characteristic equation.
    I'm not sure how you expected him to use the Cayley-Hamilton theorem.
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  6. #6
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    Quote Originally Posted by Bruno J. View Post
    Recall that f(t)=\det (A-tI). Set t=0 on both sides! You should just get a_0. The factor (-1)^n could come into the picture if somehow you had defined f(t)=\det (tI-A) but that is certainly not standard. The answer is just a_0.
    EDIT: Another source told me that the term f(t)=\det (tI-A) is equal to the characteristic polynomial, so that should help in solving the problem.
    Last edited by Runty; February 26th 2010 at 11:39 AM. Reason: Another source helped
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