# Confused on finding Eigenvalues and Eigenvectors

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

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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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You've factored it incorrectly.

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.

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

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

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

HallsofIvy
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.

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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