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Math Help - Sum and product of eigenvalues

  1. #1
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    Sum and product of eigenvalues

    How to prove that
    - the sum of the eigenvalues of a matrix equals to the trace of the matrix?
    - the product of the eigenvalues equals to the determinant?
    (Of course the matrix is a square matrix)

    I would be very grateful if you could help me.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Consider A\in{\mathbb{C}^{n\times n}} then, A is similar to its canonical form of Jordan J\in{\mathbb{C}^{n\times n}}.

    Now use that similar matrices have the same trace and the same determinant.

    Regards.

    Fernando Revilla
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zadir View Post
    How to prove that
    - the sum of the eigenvalues of a matrix equals to the trace of the matrix?
    - the product of the eigenvalues equals to the determinant?
    (Of course the matrix is a square matrix)

    I would be very grateful if you could help me.
    Or you could notice that if A=[a_{i,j}] the the characteristic polynomial is given by


    \displaystyle p_A(x)=\prod_{j=1}^{n}(x-\lambda_j)


    Thus, recalling Vieta's formulas and noticing that the characteristic polynomial is monic we can see the the constant term of the above is \displaystyle (-1)^n\prod_{j=1}^{n}\lambda_j. That said, we know that for any polynomial q(x) the constant term is merely q(0). Thus, with equal validity we may conclude that the constant term of p_A(x) is


    p_A(0)=\det\left(I0-A\right)=\det\left(-A\right)=(-1)^n\det\left(A\right)


    comparing these two gives \displaystyle \det(A)=\prod_{j=1}^{n}\lambda_j as you required. A similar technique works for the trace.

    Remark: Vieta's formula (for the above, not the trace proof) is a bit of an overkill. It's easy to see that what I claimed was the constant term is, in fact, the constant term.
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