Showing a 6x6 matrix has at least one positive eigenvalue

[macaholic]

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?

 

[haruspex]

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