1. Eigen value determination

I need to determine eigen values for the following matrix.

- 3 2 -1

2 2 -6

-1 -6 5

I get Det of A - lambda I =
I will use x instead of lambda

(-3 - x) (2 - x) (5 - x) - 36(-3 - x) - 4 (2 - x) + 10 + x = 0

I know my factoring skills need work, but is there another way besides trial and error to determine my eigenvalues??? I get that lambda (or x as I have used) = -2 is one of my eigen values. I also understand, I think, that I should have two more. Is there a method I am not aware of or a site on the internet that has a tutorial for this process. So far all the ones I have found just use 2 x 2 matrices which are so simple even I can do them!!! Thanks so much for looking at this. Frostking

2. You already know that one eigenvalue is -2.

Now, instead of factoring the polynomial, you can use these two facts to find the remaining two:

$det A = 60 = \prod_{i=1}^{3}\lambda_i = -2\lambda_2\lambda_3$

$trace(A) = 4 = \sum_{j=1}^{3}\lambda_j= \lambda_2 + \lambda_3 - 2$

Spoiler:
-4, 10

3. doublepost

4. Originally Posted by Frostking
I need to determine eigen values for the following matrix.

- 3 2 -1

2 2 -6

-1 -6 5

I get Det of A - lambda I =
I will use x instead of lambda

(-3 - x) (2 - x) (5 - x) - 36(-3 - x) - 4 (2 - x) + 10 + x = 0

I know my factoring skills need work, but is there another way besides trial and error to determine my eigenvalues??? I get that lambda (or x as I have used) = -2 is one of my eigen values. I also understand, I think, that I should have two more. Is there a method I am not aware of or a site on the internet that has a tutorial for this process. So far all the ones I have found just use 2 x 2 matrices which are so simple even I can do them!!! Thanks so much for looking at this. Frostking

Long division: -2 is a root of the pol. iff x+2 divides the pol., so divide your pol. by x+2 and get as quotient $x^2-6x-40=(x-10)(x+4)$ and the other two eigenvalus are 10 and -4

Tonio