0
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ Looks like $\lambda=-2$ is a root of that polynomial (guess and check), then you can divide by $\lambda+2$ to get a quadratic. $\endgroup$ Jul 28, 2014 at 6:20
  • $\begingroup$ There seems to be some arithmetic bugs in the last step. $\endgroup$ Jul 28, 2014 at 6:23
  • 1
    $\begingroup$ According to Wolfram Alpha your calculation seems to be wrong $\endgroup$
    – Thomas
    Jul 28, 2014 at 6:26
  • 1
    $\begingroup$ $(1-x)((-2-x)(1-x)+3(-(-2-x)3 =-(1-x)^2(2+x)+9(2+x)=(2+x)(9-(1-x)^2)$. Seems like this is easier to analyze w.r.t the roots. (I used $x$ instead of $\lambda$ because i was lazy.) $\endgroup$
    – Thomas
    Jul 28, 2014 at 6:31

2 Answers 2

1
$\begingroup$

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

not what you got...

$\endgroup$
1
  • $\begingroup$ Oh okay I see what I did. Forgot to multiply in the negative.. $\endgroup$
    – Sofia June
    Jul 28, 2014 at 6:46
1
$\begingroup$

you should treat it as a normal polynominal when you are trying to find the roots (which are eigenvalues).

$$\det(A-\lambda \cdot I) = (\lambda-4)(\lambda+2)^2$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.