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Math Help - trace

  1. #1
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    trace

    let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the coeff of x^(n-1) in the characteristic polynomial is -tr(A)
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the coeff of x^(n-1) in the characteristic polynomial is -tr(A)
    We have

    \begin{vmatrix} x-a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} &x-a_{22} & \ldots & a_{2n} \\ \vdots&&&\vdots \\ a_{n1} & a_{n2} &\ldots & x-a_{nn}\end{vmatrix}=\\<br />
(x-a_{11})(x-a_{22})\cdot\ldots\cdot(x- a_{nn})+\ldots

    and the terms we haven't written have degree \leq n-2.

    Regards.

    Fernando Revilla
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the coeff of x^(n-1) in the characteristic polynomial is -tr(A)
    Another method (the first is similar to that of Dr. Revilla's) is that you can first recall that p_A(x) is monic and it's roots are precisely the mulitiset spectrum \sigma\left(A\right) of A. and so


    \displaystyle p_A\left(x\right)=\prod_{\lambda\in\sigma\left(A\r  ight)}(x-\lambda)

    We can then recall Vieta's formula so that (remembering that p_A(x) is monic) that -\text{coeff}(x^{n-1})=\Pi_1(\lambda_1,\cdots,\lambda_n) where \Pi_1(\lambda_1,\cdots,\lambda_n) is the first symmetric polynomial. Thus,

    \displaystyle -\text{coeff}(x^{n-1})=\Pi_1(\lambda_1,\cdots,\lambda_n)=\sum_{\lambd  a\in\sigma(A)}\lambda=\text{tr}(A)
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