Finding Eigenvectors for Distinct Real Eigenvalues

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Homework Help Overview

The discussion revolves around finding eigenvectors for a given matrix with distinct real eigenvalues. The original poster expresses confusion regarding the nature of eigenvectors, particularly the misconception that the zero vector could be an eigenvector.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the requirement that eigenvectors must be nonzero and explore the implications of this rule. There is an examination of the calculations related to the eigenvector associated with the eigenvalue of 1.

Discussion Status

Participants are actively clarifying the concept of eigenvectors and addressing the original poster's misunderstanding. Some guidance has been provided regarding the calculation of eigenvectors, and there is acknowledgment of the importance of attending lectures for understanding these concepts.

Contextual Notes

The original poster mentions a preference for independent learning and notes a potential oversight in their textbook regarding the nature of eigenvectors. There is a reference to the matrix involved in the problem, but specific details about the matrix are not fully explored in the discussion.

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Homework Statement


I don't know how to put matrices in, so I'll just link an http://forum.bodybuilding.com/attachment.php?attachmentid=3339921&d=1305058219"
Basically find the solution for that matrix.

Homework Equations


The Attempt at a Solution


This was the http://img560.imageshack.us/img560/9672/scr1305068624.png" I came up with. when I use an eigenvector calculator though, for c2 it gives me 0,1,0 instead of 0,0,0 which is what I got. I was wondering why I was wrong. Thanks
 
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An eigenvector can't be the zero vector.

Here's your matrix in LaTeX. They're not hard to do.

[tex]\begin{bmatrix}1&0&1 \\ 0&1&0 \\ 1&0&1 \end{bmatrix}[/tex]

Click the matrix to see my LaTeX script.

Check your calculation for the eigenvector associated with the eigenvalue of 1.
 
Mark44 said:
An eigenvector can't be the zero vector.

Here's your matrix in LaTeX. They're not hard to do.

[tex]\begin{bmatrix}1&0&1 \\ 0&1&0 \\ 1&0&1 \end{bmatrix}[/tex]

Click the matrix to see my LaTeX script.

Check your calculation for the eigenvector associated with the eigenvalue of 1.

I didn't know that, damnit. So the 1 in the center is an arbitrary number? The matric after you plug in 1 is
[tex] \begin{bmatrix}0&0&1 \\ 0&0&0 \\ 1&0&0 \end{bmatrix}[/tex]

I don't go to lectures and I learn off of textbooks. I only go to class to take the exams, i don't remember reading that it can't be the zero vector. So because it can't be the zero vector I have to make up a number that will make it work? Since the k2 for all of them is 0, i can just basically put anything in right?
 
To find your eigenvector, which is assumed to be nonzero (I'll bet that this is in your text and was stated in class), you're going to be finding solutions to the equation (A - 1I)x = 0.

Of course, x = 0 is a solution of this equation, but we want nonzero x.

Your matrix A - 1I says that x3 = 0, x1 = 0, and x2 is arbitrary, so the vector <0, 1, 0> is an eigenvector for the eigenvalue 1.

You ought to reconsider going to class...
 
Mark44 said:
To find your eigenvector, which is assumed to be nonzero (I'll bet that this is in your text and was stated in class), you're going to be finding solutions to the equation (A - 1I)x = 0.

Of course, x = 0 is a solution of this equation, but we want nonzero x.

Your matrix A - 1I says that x3 = 0, x1 = 0, and x2 is arbitrary, so the vector <0, 1, 0> is an eigenvector for the eigenvalue 1.

You ought to reconsider going to class...

Well now that I know that it can't be a non zero vector it all makes sense. I don't like going to lectures, it's not my thing. I prefer independent learning, which is hard for a lot of professors to understand. This was basically the only thing bothering me. Everything else is really easy. I probably skipped over that part in the text when it said it. I already knew about arbitrary numbers, but I didn't know about solutions not being able to be zero vectors. Thanks.
 

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