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Thread: Eigenvalues question

  1. #1
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    Eigenvalues question

    Let $\displaystyle A \in M_{n \times n}(\mathbb{R})$. Show that if A has n eigenvalues $\displaystyle \lambda_1,\lambda_2,...,\lambda_n$ in $\displaystyle \mathbb{R}$ then $\displaystyle \lambda_1+\lambda_2+...+\lambda_n = traceA$ and $\displaystyle \lambda_1 \lambda_2 ... \lambda_n = detA$.

    Not really sure where i'm going with this question. I am thinking of using the fact that for the polynomial equation obtained from $\displaystyle det(\lambda I_n - A)=0$ has a coefficient of trace(A) for the $\displaystyle \lambda^{n-1}$ term, but not really sure how this is useful.
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  2. #2
    A Plied Mathematician
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    The trace and determinant are unchanged under an orthogonal transformation. If you have, as I suspect, n distinct eigenvalues, then you're guaranteed to be able to orthogonally diagonalize A. If you can do that, and you know that the trace and determinant are unchanged by such a transformation, then you're done.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    An alternative:

    As $\displaystyle A\in \mathbb{R}^{n\times n}$ has $\displaystyle n$ real eigenvalues (including multiplicities), then $\displaystyle A$ is similar to its canonical form of Jordan $\displaystyle J$ . As similar matrices have the same trace and the same determinant, then:

    $\displaystyle \textrm{tr}A=\textrm{tr}J=\lambda_1+\ldots+\lambda _n$

    $\displaystyle \det A=\det J=\lambda_1\cdot \ldots \cdot \lambda_n$


    Fernando Revilla
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