Further expanding the determinant that you have set up:
=-15 + 2\lambda + \lambda^2 - 9
So the characteristic polynomial is \det(A-\lambda\ I) = \lambda^2 + 2\lambda - 24. Setting this equal to zero to find the roots yields:
\lambda^2 + 2\lambda - 24 = 0
(\lambda + 6)(\lambda - 4) = 0...
I (and my TA) got eigenvalues with the associated eigenvectors as:
lambda = -6 with eigenvector <-1,3>
and
lambda = 4 with eigenvector <3,1>
EDIT: Yeah, just doublechecked with an eigenvector calculator online.
(Also, note how these two directions are perpendicular, which seems to contrast...
Homework Statement
http://imageshack.us/photo/my-images/861/screenshot20111211at928.png/
I am only concerned with part (a) of this problem.Homework Equations
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