Find the eigenvectors given the eigenvalues

  • Thread starter DryRun
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  • #1
DryRun
Gold Member
838
4
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:

Answers and Replies

  • #2
22,129
3,297
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:
[tex]x_1=-2x_2-4x_3[/tex]

Try to set [itex]x_2,x_3[/itex] equal to something and see what you get. For example, set [itex](x_2,x_3)=(1,0)[/itex] and [itex](x_2,x_3)=(0,1)[/itex]. This will give rise to two linear independent eigenvectors which span the eigenspace.
 
  • #3
DryRun
Gold Member
838
4
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. :)
 
  • #4
22,129
3,297
Looks good!!
 
  • #5
DryRun
Gold Member
838
4
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.
 
  • #6
35,129
6,876
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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