1
$\begingroup$

$$\begin{bmatrix} a\\\\b\end{bmatrix}\longmapsto \begin{bmatrix}-3&1\\\\0&2\end{bmatrix} \begin{bmatrix} a\\\\b\end{bmatrix}$$

Use the characteristic polynomail to find all eigenvalues for the transformation for each eigenvalues $\lambda$ find all eigenvectors with eigenvalues $\lambda$ and find a basis for $E_\lambda$

Here is what I have so far

$\chi_f(x)=\begin{bmatrix}-3-x&1\\\\0&2-x\end{bmatrix}$ $=(-3-x)(2-x)$ so there are two eigenvalues $\lambda=-3$ and $\lambda=2$

I'm not sure how to find a basis for these eigenvalues though

  • 0
    To find $E_\lambda$ you have to find a base for $ker(f-\lambda id)$ can you do that?2012-12-03
  • 0
    I think they might be of the form $\begin{bmatrix}a\\\\3a\end{bmatrix}$ for $E_0$ and $\begin{bmatrix}a\\\\2a\end{bmatrix}$ for $E_{-1}$2012-12-03
  • 0
    Your eigenvalues are wrong. $\chi_f(x)$ is $(-3-x)(2-x)$ which has roots $2,-3$.2012-12-03
  • 0
    @Stefan so a base of everything that maps char(f) to zero?2012-12-03
  • 0
    @PeterTamaroff Yes you are correct, for some reason I though 1 times 0 was 1 :(2012-12-03
  • 0
    @PeterTamaroff This helps for $E_2$ which is now $\begin{bmatrix}a\\\\5a\end{bmatrix}$ but I still get $-2b=-3b$ for $E_{-3}$2012-12-03
  • 0
    Well, that means $b=0$, clearly =) See my answer2012-12-03

1 Answers 1