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
thanks.

 

[iNCREDiBLE]

See http://mathworld.wolfram.com/CharacteristicEquation.html" .

 

[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

 

[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.

 

[TD]

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?

 

[HallsofIvy]

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??

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.
To think i have a 3.77 GPA. What is the world coming too?
Anywho, thanks for the explanation everyone.