Confused on finding Eigenvalues and Eigenvectors

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Homework Help Overview

The discussion revolves around the process of finding eigenvalues and eigenvectors, specifically focusing on an example where the eigenvalue of 3 and an eigenvector of (1, 1) are mentioned. Participants express confusion regarding the calculations and the methods used to arrive at these values.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to understand the factorization of a characteristic polynomial and question the correctness of their calculations. Some express frustration over not obtaining the expected results and seek clarification on why only one eigenvalue is emphasized.

Discussion Status

The discussion is active, with participants providing feedback on each other's attempts. Some guidance has been offered regarding the factorization process, and there is an exploration of multiple eigenvalues, although no consensus has been reached on the best approach to solve the problem.

Contextual Notes

Participants mention the use of the quadratic equation and the expectation of obtaining "nice" answers, indicating a possible constraint in their understanding of the problem setup. There is also a reference to a textbook example that may influence their expectations.

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!

[tex]\lambda ^2 - 4\lambda + 3 = \left( {\lambda - 1} \right)\left( {\lambda - 3} \right) \ne \left( {\lambda - 4} \right)\left( {\lambda + 1} \right)[/tex]
 
ohhh wow i suck hah, thank u so much! why did they only use [tex]\lambda = 3[/tex] 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 [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.
 
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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