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Showing a 6x6 matrix has at least one positive eigenvalue

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that if a 6x6 matrix A has a negative determinant, then A has at least one positive eigenvalue. Hint: Sketch the graph for the characteristic polynomial of A.

    2. Relevant equations
    Characteristic polynomial: [itex](-\lambda)^n + (\text{tr}A)(-\lambda)^{n-1} + ... \text{det} A[/itex]


    3. The attempt at a solution
    I'm not really sure what to do at all. I know that the characteristic polynomial for a 6x6 matrix is going to be proportional to [itex]\lambda ^6[/itex], and shifted by det(A), and that the roots of the polynomial are going to be the eigenvalues...But I don't see how this shows there will be at least one positive eigenvalue. Can anyone point me in the right direction here?
     
  2. jcsd
  3. Nov 9, 2012 #2

    haruspex

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    What is the value of the polynomial at λ=0? For large positive λ?
     
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