# Confused on finding Eigenvalues and Eigenvectors

mr_coffee
confused on finding Eigenvalues and Eigenvectors!!

hello everyone, i can't understand this example, how did they find the Eigen value of 3?! Aslo an Eigen vector of 1 1? http://img438.imageshack.us/img438/1466/lastscan1oc.jpg [Broken]
thanks.

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iNCREDiBLE
See http://mathworld.wolfram.com/CharacteristicEquation.html" [Broken].

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mr_coffee
thanks, i did that and I didn't get the right answer, look when i try to solve...
http://img442.imageshack.us/img442/4810/lastscan1fp.jpg [Broken]

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Muzza
You've factored it incorrectly.

mr_coffee
i know, i can't factor that and get a nice number, i'd have to use the quadtract equation, but that can't be right because the book got a nice answer of 3.

Homework Helper
As Muzza said, only the factoring was wrong!

$$\lambda ^2 - 4\lambda + 3 = \left( {\lambda - 1} \right)\left( {\lambda - 3} \right) \ne \left( {\lambda - 4} \right)\left( {\lambda + 1} \right)$$

mr_coffee
ohhh wow i suck hah, thank u so much!! why did they only use $$\lambda = 3$$ when it can also equal 1?

Homework Helper
mr_coffee said:
i know, i can't factor that and get a nice number, i'd have to use the quadtract equation, but that can't be right because the book got a nice answer of 3.
Even knowing that one solution was 3, so one factor must be x- 3 you couldn't factor it??

mr_coffee said:
ohhh wow i suck hah, thank u so much!! why did they only use $$\lambda = 3$$ when it can also equal 1?
Read it carefully! It specifically says "an eigenvalue", not the eigenvalue. And immediately below states that there is another and solves for it.

mr_coffee
Even knowing that one solution was 3, so one factor must be x- 3 you couldn't factor it??
This is what we witnessed today. :surprised
To think i have a 3.77 GPA. What is the world coming too?
Anywho, thanks for the explanation everyone.

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