## anonymous 4 years ago Can someone help me with eigenvalues and eigenvectors?

1. anonymous

2. anonymous

I need help with problem number 6

3. anonymous

Use: $det(A-\lambda I) = 0$ to find the eigenvalues. For two by two matrices I think you will get two values of lambra (as a quadratic in lambda will come out).

4. anonymous

I basically need help with part C

5. anonymous

Lambda=0 and 4

6. anonymous

Is that correct?

7. anonymous

You have to find a vector X such that $AX=\lambda X$

8. TuringTest

$\det(A-I\lambda)=\det\left[\begin{matrix}2-\lambda & 2 \\ 2 & 2-\lambda\end{matrix}\right]=4-2\lambda+\lambda^2-4$sorry gotta eat!

9. anonymous

LOL Go Eat:D

10. anonymous

For the corresponding eigenvectors, I think it's a case of finding X in $A-\lambda I = X$ when you know the two values of lambda...

11. anonymous

Turing its the opposite

12. TuringTest

you have I*Iambda-A I'm guessing, it's the same it is the same

13. anonymous

oh sorry $(A-\lambda I)X = 0$ (the ero matrix)

14. anonymous

zero*

15. anonymous

ohhh ya that is the one i have in my textbook

16. anonymous

Check out this example : http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors#Worked_example

17. anonymous

Thanks I figured it out :D

18. mathmate

For $$\lambda=0$$ reduced matrix is $\left[\begin{matrix}1 & 1 \\ 0 & 0\end{matrix}\right]$which gives an eigenvector of (1,-1). For $$\lambda=4$$, the reduced matrix is$\left[\begin{matrix}-1 & 1 \\ 0 & 0\end{matrix}\right]$whose eigenvector is (1,1).

19. anonymous

Thanks Guys :D I really appreciate ur help

20. mathmate

You're welcome! :)