Why Does My Eigenvector Calculation Differ from WolframAlpha's Result?

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Homework Help Overview

The discussion revolves around the calculation of eigenvectors for a given matrix, specifically addressing discrepancies between the original poster's results and those provided by WolframAlpha. The subject area is linear algebra, focusing on eigenvalues and eigenvectors.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the eigenspace for the eigenvalue of zero and presents their calculations. Some participants question the correctness of the eigenvector derived from the calculations, while others clarify definitions and concepts related to eigenvectors.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's calculations and definitions. There are multiple interpretations of the eigenvector results, and some guidance has been offered regarding the correct forms of eigenvectors associated with the eigenvalues.

Contextual Notes

Participants note that the matrix in question has two eigenvalues, 5 and 0, and there is a focus on the eigenvector corresponding to the eigenvalue of zero. The original poster acknowledges a mistake in terminology but maintains that they understand the concepts involved.

estro
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Given the matrix:
[tex]\left( \begin{array}{cc}<br /> 1 & 2 \\<br /> 2 & 4 \end{array} \right)[/tex]

I'll find eigenspace for the eigenvalue of t=0, So I have to solve:

[tex]\left( \begin{array}{cc}<br /> 1 & 2 \\<br /> 2 & 4 \end{array} \right) {(x,y)}^t=0[/tex]

Then I do [tex]R_2->R_2-2R_1[/tex] and get x+2y=0 => x=-2y => get the eigenvector (-1,2).

But wolframalpha tells me the eigenvector for this eigenvalue should be (1,2).

Where is my sin?
 
Last edited:
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I'n my first pose I did a mistake, I wrote eigenvalue but meant eigenvector.
Fixed it.
 
estro said:
I'n my first pose I did a mistake, I wrote eigenvalue but meant eigenvector.
Fixed it.

Still, an eigenvector of a matrix A is a vector v that satisfies:
[tex]A v = \lambda v[/tex]
for some number lambda, which is the eigenvalue.

To solve it you need to set up the so called characteristic quadratic equation and solve that...
 
Thank you for your response, but I know what is eigenvalue, eigenvector and characteristic polynomial, I left out the technical details and only showed how i calculate the eigen space for eigenvalue 0. [this matrix has 2 eigenvalues 5 and 0]
 
estro said:
Thank you for your response, but I know what is eigenvalue, eigenvector and characteristic polynomial, I left out the technical details and only showed how i calculate the eigen space for eigenvalue 0. [this matrix has 2 eigenvalues 5 and 0]

Sorry! My bad! I guess I didn't read carefully enough. :redface:

In that case, your solution is the wrong way around: it should be for instance (2, -1).

The solution you mentioned from WolframAlpha is the eigenvector belonging to eigenvalue 5.
 
Your sin is an incorrect solution: from x = -2y you get (-2,1) (by putting y = 1) or (2,-1) (by putting y = -1). There is no way to get (-1,2) or (1,-2).

RGV
 
Thanks!
 

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