## feldy90 Group Title How do I find the eigenvectors of the following matrix? 2 years ago 2 years ago

1. feldy90 Group Title

$A=\left[\begin{matrix}1 & 2 \\ 2 & -2\end{matrix}\right]$

2. Coolsector Group Title

$Av = \lambda v$

3. hesam Group Title

a=-6

4. Coolsector Group Title

first find eigenvalues :det(A - λI) = 0 λI - A = $\left[\begin{matrix}1-λ & 2 \\ 2 & -2-λ\end{matrix}\right]$ so Det : (1−λ)(−2−λ) - 4 = 0 λ^2 - λ + 2λ -2 -4 = 0 λ^2 +λ -6 = 0 λ^2 +3λ - 2λ - 6 = 0 λ(λ+3) - 2(λ+3) (λ-2)(λ+3) = 0 λ=-3,2 now Av=λv Av = 2v v1+2v2 = 2v1 2v1 -2v2 = 2v2 any vector of the form : 2v2 = v1 Av = -3v v1 + 2v2 = -3v1 2v1 - 2v2 = -3v2 2v2 = -4v1 any vector of the form : v2 = -2v1

5. Coolsector Group Title

by the way .. the eigenvector cant be zero.

6. Coolsector Group Title

example for the eigenvectors are : 2v2 = v1 -> (2,1) v2 = -2v1 ->(-1,2)

7. Coolsector Group Title

please tell me if it is fine..

8. feldy90 Group Title

Sorry - I have actually left my books for a moment... I will definitely check this when I get back to them - thanks so much!!

9. feldy90 Group Title

10. Coolsector Group Title

11. feldy90 Group Title

Haha, later is better than never, right? :P Quick question, for the eigenvalue -3, is the final eigenvector$v=\left(\begin{matrix}-2 \\ 1\end{matrix}\right)$ Since $v_2=-2v_1$??

12. Coolsector Group Title

if you take v1 = 1 then v2 = -2 (1,-2) its the opposite from yours

13. feldy90 Group Title

Oh I see, thanks!

14. feldy90 Group Title

So would the eigenspace of eigenvalue 2, be: $span \left\{ \left(\begin{matrix}2 \\ 0\end{matrix}\right) ,\left(\begin{matrix}0 \\ 1\end{matrix}\right)\right\}$ or just $span \left\{ \left(\begin{matrix}2 \\ 1\end{matrix}\right) \right\}$ Thanks for the ongoing help :)

15. feldy90 Group Title

Never mind, I got it :)