Confused on finding Eigenvalues and Eigenvectors

  • Thread starter mr_coffee
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  • #1
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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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Answers and Replies

  • #2
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See http://mathworld.wolfram.com/CharacteristicEquation.html" [Broken].
 
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  • #3
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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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  • #4
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You've factored it incorrectly.
 
  • #5
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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.
 
  • #6
TD
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As Muzza said, only the factoring was wrong!

[tex]\lambda ^2 - 4\lambda + 3 = \left( {\lambda - 1} \right)\left( {\lambda - 3} \right) \ne \left( {\lambda - 4} \right)\left( {\lambda + 1} \right)[/tex]
 
  • #7
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ohhh wow i suck hah, thank u so much!! why did they only use [tex]\lambda = 3[/tex] when it can also equal 1?
 
  • #8
HallsofIvy
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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 [tex]\lambda = 3[/tex] 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.
 
  • #9
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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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