Find the eigenvectors given the eigenvalues

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Homework Help Overview

The discussion revolves around finding the eigenvectors of a given 3x3 matrix after determining its eigenvalues. The matrix in question is presented, and the eigenvalues identified are 1 and 2.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find eigenvectors corresponding to the eigenvalue of 2 but encounters difficulties in solving the resulting system of equations. They explore specific values for variables to derive eigenvectors and question the correctness of their method.

Discussion Status

Participants are actively engaging in the problem, with one offering guidance on setting variables to find independent eigenvectors. There is acknowledgment of the original poster's attempts, and some participants express uncertainty regarding the accuracy of manual calculations compared to MATLAB results.

Contextual Notes

There is mention of potential discrepancies between manual calculations and results obtained from MATLAB, leading to questions about the validity of the original poster's findings. The discussion includes a focus on the implications of ratios in eigenvector representation.

DryRun
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Homework Statement
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?
 
Last edited:
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sharks said:
Homework Statement
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.
 
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. :)
 
Looks good!
 
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.
 
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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