Confused on finding Eigenvalues and Eigenvectors

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SUMMARY

The discussion centers on the confusion surrounding the calculation of eigenvalues and eigenvectors, specifically regarding the eigenvalue of 3 and the eigenvector of (1, 1). Participants clarify that the characteristic equation λ² - 4λ + 3 = (λ - 1)(λ - 3) was incorrectly factored, leading to misunderstandings. The correct identification of eigenvalues is emphasized, noting that multiple eigenvalues can exist, as indicated by the phrase "an eigenvalue" in the problem statement. The conversation highlights the importance of careful reading and accurate factoring in solving eigenvalue problems.

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  • Understanding of eigenvalues and eigenvectors
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  • Knowledge of factoring polynomials
  • Basic proficiency in quadratic equations
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Students studying linear algebra, educators teaching eigenvalue concepts, and anyone seeking to improve their understanding of eigenvalues and eigenvectors in mathematical contexts.

mr_coffee
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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
thanks.
 
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See http://mathworld.wolfram.com/CharacteristicEquation.html" .
 
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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
 
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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.
 
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?
 
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.
To think i have a 3.77 GPA. What is the world coming too?
Anywho, thanks for the explanation everyone.
 
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