To find the eigenvalues, find the roots of the polynomial .
Hiya,
Algebra exam coming up on Wednesday, and I'm stumped on how to get the egenvalues of a 3x3 matrix. I think I did them about 2 years ago, and they're gone completely out of my head.
Given:
I know the eigenvalues comes out as .
I know you subtract from the diagonal line at first, but then where do you go with it? I would really appreciate it if you could show step-by-step work. Thanks
Jo
So you are asking, basically, how to find those eigenvalues?
The number is an eigenvalue of A if and only if , which is equivalent to has "non-trivial solutions"- that is, a solution other than the obvious v= 0. If is invertible, then we could multiply on both sides by its inverse to get the unique solution. But on the right we have that matrix times the 0 vector which must be 0. That is, there exist other, non-zero solutions, if and only if is not invertible which means that its determinant must be 0:
You can expand that by the first row:
which has solutions , , and .