# Find the eigenvectors given the eigenvalues

by sharks
Tags: eigenvalues, eigenvectors
 PF Gold P: 836 The problem statement, all variables and given/known data This is the matrix A, which i need to find the eigenvalues and eigenvectors. 3x3 matrix 5 6 12 0 2 0 -1 -2 -2 The attempt at a solution I have found the eigenvalues to be: 1, 2, 2. So, the final eigenvalues are : 1 and 2. Now, i found the eigenvector for eigenvalue = 1, which is: 3x1 column matrix: [-3 0 1]^T But for the eigenvalue = 2, i am stuck, as these are the system equations that i have before me: 3x1 + 6x2 + 12x3 = 0 -x1 - 2x2 - 4x3 = 0 I made x1 the subject of formula: -2x2 - 4x3 And then i'm not sure how to proceed. But i'm going out on a limb here, so please correct me. Let x2 = 1 and x3 = 0 Then i get this 3x1 column matrix: x2[-2 1 0]^T Let x3 = 1 and x2 = 0 I get another 3x1 column matrix: x3[-4 0 1]^T So, all the eigenvectors in a 3x3 matrix P, are: -3 -2 -4 0 1 0 1 0 1 Is this correct?? Most importantly, is my method correct? Is there a better method?
Mentor
P: 16,554
 Quote by sharks The problem statement, all variables and given/known data This is the matrix A, which i need to find the eigenvalues and eigenvectors. 3x3 matrix 5 6 12 0 2 0 -1 -2 -2 The attempt at a solution I have found the eigenvalues to be: 1, 2, 2. So, the final eigenvalues are : 1 and 2. Now, i found the eigenvector for eigenvalue = 1, which is: 3x1 column matrix: [-3 0 1]^T But for the eigenvalue = 2, i am stuck, as these are the system equations that i have before me: 3x1 + 6x2 + 12x3 = 0 -x1 - 2x2 - 4x3 = 0 I made x1 the subject of formula: -2x2 - 4x3 And then i'm not sure how to proceed. But i'm going out on a limb here, so please correct me. Let x2 = 1 and x3 = 0 Then i get this 3x1 column matrix: x2[-2 1 0]^T Let x3 = 1 and x2 = 0 I get another 3x1 column matrix: x3[-4 0 1]^T Is this correct??
Your equations reduce to one equation:
$$x_1=-2x_2-4x_3$$

Try to set $x_2,x_3$ equal to something and see what you get. For example, set $(x_2,x_3)=(1,0)$ and $(x_2,x_3)=(0,1)$. This will give rise to two linear independent eigenvectors which span the eigenspace.
 PF Gold P: 836 Thanks for your help, micromass. Could you please check on my final solution which i edited at the end of my first post above. BTW, you really have the best degree in the world. :)
Mentor
P: 16,554

## Find the eigenvectors given the eigenvalues

Looks good!!
 PF Gold P: 836 Thanks again. Now, i might be pushing into some daring territory here, but might you (or someone else) be familiar with matlab? Anyway, here goes the eigenvector solution from matlab: v = 0.9701 -0.9487 -0.6963 0 0 0.6963 -0.2425 0.3162 -0.1741 Notice that the last column should correspond to: [-4 0 1]^T (the ratio is what matters here) and since there are no middle zero from the matlab solution, i'm a bit uneasy that i might have made a mistake somewhere in my manual calculations, although i have doubled checked everything.
 Mentor P: 20,971 It looks to me like the first column, not the third column, represents a vector that is a multiple of <-4, 0, 1>^T.

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