Showing a 6x6 matrix has at least one positive eigenvalue

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Homework Statement



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.

Homework Equations


Characteristic polynomial: (-\lambda)^n + (\text{tr}A)(-\lambda)^{n-1} + ... \text{det} A


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 \lambda ^6, 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?
 
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What is the value of the polynomial at λ=0? For large positive λ?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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