# Characteristic polynomial and eigenvalues of a $3 \times3$ matrix.

Hi so I have to find the characteristic polynomials and the eigenvalues of the matrix: $$A = \begin{bmatrix}1 & 0 & 3\\2 & -2 & 2\\3 & 0 & 1\end{bmatrix}$$

So I know you use the formula $\det(A-\lambda\cdot I)$ so then you get the matrix: $$\det(A-\lambda \cdot I) = \begin{bmatrix}1-\lambda & 0 & 3\\2 & -2-\lambda & 2\\3 & 0 & 1-\lambda\end{bmatrix}$$

So using cofactor expansion on the first row I got: $$\ det(A-\lambda \cdot I) = (1-\lambda)\cdot\begin{bmatrix}-2-\lambda & 2\\0 & 1-\lambda\end{bmatrix} - (0)\cdot\begin{bmatrix}2 & 2\\3 & 1-\lambda\end{bmatrix}\\+ (3)\cdot\begin{bmatrix}2 & -2-\lambda\\3 & 0\end{bmatrix}$$

And then: $$\det(A-\lambda \cdot I) = (1-\lambda)[(-2-\lambda)(1-\lambda)]+(3)[-(-2-\lambda)(3)]$$

And basically once I factored this out I got the characteristic polynomial of $$\det(A-\lambda \cdot I) = -20-6\lambda-\lambda^3$$

But this doesn't really seem like it's supposed to be right and I'm not quite sure how to find the eigenvalues out of a polynomial that doesn't have a $x^2$ value.

Am I doing this right? Any help would be appreciated. Thanks.

• Looks like $\lambda=-2$ is a root of that polynomial (guess and check), then you can divide by $\lambda+2$ to get a quadratic. Jul 28, 2014 at 6:20
• There seems to be some arithmetic bugs in the last step. Jul 28, 2014 at 6:23
• According to Wolfram Alpha your calculation seems to be wrong Jul 28, 2014 at 6:26
• $(1-x)((-2-x)(1-x)+3(-(-2-x)3 =-(1-x)^2(2+x)+9(2+x)=(2+x)(9-(1-x)^2)$. Seems like this is easier to analyze w.r.t the roots. (I used $x$ instead of $\lambda$ because i was lazy.) Jul 28, 2014 at 6:31

$$\det(A-\lambda I) = (1-\lambda)(-2-\lambda)(1-\lambda) + 3(2+\lambda)3=\\ =(1-2\lambda + \lambda^2)(-2-\lambda) + 9(2+\lambda) = \\ =(-2-\lambda + 4\lambda + 2\lambda^2 - 2\lambda^2 - \lambda^3) + 18 + 9\lambda = \\ = -\lambda^3 + 3\lambda - 2 + 18 + 9\lambda = -\lambda^3 + 12\lambda + 16,$$
$$\det(A-\lambda \cdot I) = (\lambda-4)(\lambda+2)^2$$