Finding eigenbasis and diagonalization

[TheFerruccio]

Homework Statement

Find the eigenbasis and diagonalize.

Homework Equations

$$\mathbf{A} = \left[ {\begin{array}{ccc} 5& \frac{8}{3} & \frac{-2}{3} \\ 2 & \frac{2}{3}& \frac{4}{3} \\ -4 & \frac{-4}{3} & \frac{-8}{3}\\ \end{array} } \right]$$

The Attempt at a Solution

I find the characteristic equation by finding the determinant of $$\mathbf{A} - \lambda \mathbf{I}$$$$\left|\mathbf{A} - \lambda \mathbf{I}\right| = \left| {\begin{array}{ccc} 5 - \lambda & \frac{8}{3} & \frac{-2}{3} \\ 2 & \frac{2}{3} - \lambda & \frac{4}{3} \\ -4 & \frac{-4}{3} & \frac{-8}{3} - \lambda\\ \end{array} } \right|$$ = 0

This gets me the cubic equation:
$$-\lambda^3 + 3\lambda^2 + 18\lambda = 0$$

So, here's the question: Are there any nice, fast ways to get the roots of the cubic equation? Furthermore, is there any faster way to find the characteristic polynomial that doesn't include such a high probability of arithmetic error when doing it by hand?

All of this is done by hand, no calculator, pencil and paper.

 

[Dick]

You are doing ok so far. You've got the right characteristic equation. Now just factor it. It's really not so bad for a cubic. You can factor out lambda right away and that leaves you with a quadratic. I don't think there is any easier way to find the characteristic polynomial in general than what you did. And there's no easier way to find cubic roots than to factor. If it doesn't factor, it's really hard.

 

[TheFerruccio]

You are doing ok so far. You've got the right characteristic equation. Now just factor it. It's really not so bad for a cubic. You can factor out lambda right away and that leaves you with a quadratic. I don't think there is any easier way to find the characteristic polynomial in general than what you did. And there's no easier way to find cubic roots than to factor. If it doesn't factor, it's really hard.

Oh, duh! I wasn't thinking. I was looking at the equation too fast and didn't see the obvious factorization right there.

Thanks for the help. I managed to figure this out after a few pages of number crunching.