# Finding eigenvalues to use in Cayley-Hamilton theorem problem

1. Oct 28, 2012

### phosgene

1. The problem statement, all variables and given/known data

Let C =

2,0,-2
1,1,2
-1,-1,-1

Use the Cayley-Hamilton theorem to compute C^3.

2. Relevant equations

Cayley-Hamilton theorem says that every square matrix satisfies its own characteristic equation.

$C^3=PD^3P^{-1}$

where P is the matrix formed from linearly independant eigenvectors of C and D is the diagonal matrix formed from the eigenvalues of C.

3. The attempt at a solution

I get the characteristic equation of C is

$\lambda^3 - 2\lambda^2 - \lambda - 2 = 0$

I get stuck because I can't factorise this and get the eigenvalues to proceed. Is there some trick to factorising cubics like this?

2. Oct 28, 2012

### hedipaldi

Attached

#### Attached Files:

• ###### 001.jpg
File size:
12.8 KB
Views:
54
3. Oct 28, 2012

### phosgene

Thanks, I was just going back over the lecture notes and realized that I was absurdly confused in that section (I'm embarassed I even asked this question!)...anyway, I get it now, thanks for that :)