Eigenvalues; help with calculating characteristic polynomial roots

I need to find the eigenvalues of the linear transformation that the following matrix represent:

$\displaystyle A=\begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ 4 & -17 & 8\end{bmatrix}$

so I calculated the characteristic polynomial of A:

$\displaystyle f_A(\lambda)=det(\lambda I-A)=\lambda \begin{vmatrix}\lambda & 1\\ -17 & \lambda-8\end{vmatrix}-1\begin{vmatrix}0 & \lambda -8\\ 4 & 1\end{vmatrix}+0\begin{vmatrix}0 & \lambda\\ 4 & -17 \end{vmatrix}$

$\displaystyle f_A(\lambda)=\lambda(\lambda(\lambda-8)+17)-1(-4)$

$\displaystyle f_A(\lambda)=\lambda^3-8\lambda^2+17\lambda+4$

but I have no idea how to find its roots...

any ideas would be greatly appreciated.

TIA!

**edit

Is there another way to find the eigenvalues of A?

Re: Eigenvalues; help with calculating characteristic polynomial roots

There is some methods that work on a 3rd degree polynomial. You could use a factoration like

$\displaystyle f_A(\lambda)=(\lambda - a_0)(\lambda - a_1)(\lambda - a_2) $ where $\displaystyle a_0, a_1, a_2$ are polynomial roots.

There is another method that is called Ruffini's method (or Briot-Ruffini's method)

Ruffini's rule - Wikipedia, the free encyclopedia, take a look!

Re: Eigenvalues; help with calculating characteristic polynomial roots

Hi bosabarbosa, thanks for the help.

I'm familiar with polynomial division, but it is not useful here, since i don't know what are the divisors to be begin with (to use the example from the wiki-link - I don't know what 'r' is).

Division by $\displaystyle (\lambda -r)$ when 'r' isn't a root will produce remainder.

I'm looking for a way to find the roots without dealing with the polynomial's standard form ($\displaystyle a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$).

(maybe calculating the determinant in some other way will produce an easier-to-handle polynomial?)

Re: Eigenvalues; help with calculating characteristic polynomial roots

Quote:

Originally Posted by

**Stormey** I need to find the eigenvalues of the linear transformation that the following matrix represent:

$\displaystyle A=\begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ 4 & -17 & 8\end{bmatrix}$

so I calculated the characteristic polynomial of A:

$\displaystyle f_A(\lambda)=det(\lambda I-A)=\lambda \begin{vmatrix}\lambda & 1\\ -17 & \lambda-8\end{vmatrix}-1\begin{vmatrix}0 & \lambda -8\\ 4 & 1\end{vmatrix}+0\begin{vmatrix}0 & \lambda\\ 4 & -17 \end{vmatrix}$

You made a mistake calculating the determinant.

$\displaystyle f_A(\lambda)=det(\lambda I-A)=\lambda \begin{vmatrix}\lambda & -1\\ 17 & \lambda-8\end{vmatrix}-1\begin{vmatrix}0 & -1\\ -4 & \lambda-8\end{vmatrix}+0\begin{vmatrix}0 & \lambda\\ -4 & 17 \end{vmatrix}$

Re: Eigenvalues; help with calculating characteristic polynomial roots

I tried is some manipulations that seems to work fine:

$\displaystyle \lambda^3-8\lambda^2+17\lambda-4=0$

$\displaystyle \lambda(\lambda^2-8\lambda)+17\lambda-4=0$

doing again, we get:

$\displaystyle \lambda[(\lambda^2-8\lambda)+17]-4=0$

Now, if you try to complete the square:

$\displaystyle \lambda[(\lambda-4)^2-16+17]-4=0$

$\displaystyle \lambda[(\lambda-4)^2+1]-4=0$

$\displaystyle \lambda(\lambda-4)^2+(\lambda-4)=0$

Factoring $\displaystyle \lambda-4$:

$\displaystyle (\lambda-4)[\lambda(\lambda-4)+1]=0$

$\displaystyle (\lambda-4)[\lambda^2-4\lambda+1]=0$

Now, you have a root, 4! Now, the rest is up to you. :D

Re: Eigenvalues; help with calculating characteristic polynomial roots

Thats brilliant man.

Thank you so much!