Question

Finding the characteristic polynomial of A

A=5 3 -3
1 3 -1
3 3 -1

Answer 1

Have you found the eigenvalues?

Answer 2

Oh, and \(\LaTeX\) can help here: `\(A=\begin{bmatrix} 5 & 3 & -3 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{bmatrix}\)` makes:

\(A=\begin{bmatrix} 5 & 3 & -3 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{bmatrix}\)

Answer 3

Does \(det(A − \lambda \mathrm{I})\) or \(|A − \lambda \mathrm{I}|\) where \(\mathrm{I}\) is the identity matrix sound at all familiar?

Answer 4

yes I changed the main diagonal to 5-x, 3-x, and -1-x. Confused on where to go from there

Answer 5

Get the determinant.

Answer 6

can you show me how. would row reduction be easiest or laplace expansion

Answer 7

Well, what methods have you been using to do 3x3 determinants up to this point?

Answer 8

both

Answer 9

Whichever you feel most comfortable with should work.

Answer 10

Cofactor expansion can be faster if there are values in a column or row that are 0. But because this does not have that, I am not sure which would be faster. Generally, the larger you get, the better it is to use reduction to a triangular matrix.

Answer 11

okay thanks

Answer 12

So, did you get an answer yet?

Answer 13

not yet :/

Answer 14

K. Well, I am going to bed. I won't say the signs or which position they are coefficients to, but I got a 16, 12, 7, and 1 as the coefficients of the \(\lambda\)s. And those are not in order.

Answer 15

those are the eigenvalues?

Answer 16

No. They are the coefficients in the characteristic polynomial.

\(Ax^2+Bx+C\) where A, B, and C are coefficients. That sort of thing. You would still need to factor it to get the eigenvalues.

Just gives you something to check against when you do find things.

Answer 17

perfect thanks for your help!

Answer 18

np. Have fun!

And like I said, I did not give the sign \((\pm)\) or position, so expect to see some differences. But if you get those values in the poly in some form, you should have done it right.

Answer 19

okay great!