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

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The discussion revolves around the calculation of eigenvectors for the matrix (1, 2; 2, 4) and the discrepancy between the user's result and WolframAlpha's output. The user initially calculated the eigenvector for the eigenvalue of 0 and derived (-1, 2), which was incorrect. It was clarified that the correct eigenvector should be (2, -1) or its scalar multiples. The confusion stemmed from misidentifying the eigenvector associated with the eigenvalue of 5 instead of 0. The conversation emphasizes the importance of correctly applying the definitions and methods for finding eigenvectors.
estro
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Given the matrix:
\left( \begin{array}{cc}<br /> 1 &amp; 2 \\<br /> 2 &amp; 4 \end{array} \right)

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

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

Then I do R_2-&gt;R_2-2R_1 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?
 
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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:
A v = \lambda v
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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