How to find eigenvectors of 2x2 by gauss jordan method

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SUMMARY

The discussion focuses on finding eigenvectors of the matrix A = [1 1; 2 2] using the Gauss-Jordan elimination method. The eigenvalues calculated are λ1 = 0 and λ2 = 3. To find the eigenvector corresponding to λ2 = 3, the equation (A - 3I)v = 0 is solved, leading to the matrix [[-2, 1], [2, -1]]. Row reduction reveals that the eigenvector is v = <1, 2>. The same process is suggested for the eigenvalue λ1 = 0.

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Homework Statement



how to find eigenvectors by using gauss jordan

A=[1 1; 2 2]

Homework Equations


I know how to use gauss jordan but don´t know how to use it to find eigenvectors


The Attempt at a Solution


First I find the eigenvalues: ((y-1)(y-2)-(1*2)=> y1=0 and y2=3

Then I don´t know how to use the G.J method to find the eigenvectors.
 
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If 3 is an eigenvalue for A then there exist a non-zero vector v such that Av= 3v or (A- 3I)v= 0. Use Gauss-Jordan to find a non-zero solution to that equation and that solution is an eigenvector.
Here, A- 3I is
[tex]\left[\begin{array}{cc}-2 & 1 \\ 2 & -1\end{array}\right][/tex]
(I haven't used the "augmented" matrix since the third column would always be 0.)
Row reduce that. Because 3 is an eigenvalue, the last row will become all 0s and then the other rows give the eigenvectors. Here just adding the first row to the second will do that and the first row remains -2, 1. That is the same as -2x+ y= 0 or y= 2x. Taking x= 1, an eigenvector is <1, 2>. See what happens: What is
[tex]\left[\begin{array}{cc} 1 & 1 \\ 2 & 2\end{array}\right]\left[\begin{array}{c}1 \\ 2\end{array}\right][/tex]?

Do the same thing with eigevalue 0.
 

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