Find the eigenvectors given the eigenvalues

In summary: But I am not familiar with MATLAB, so I could be wrong. In summary, the conversation discusses finding eigenvalues and eigenvectors for a 3x3 matrix. The eigenvalues are found to be 1 and 2, and the corresponding eigenvectors are [-3 0 1]^T and [-3 -2 -4]^T. The method used to find the eigenvectors involves setting x2 and x3 equal to 1 and 0, respectively, and then setting x2 and x3 equal to 0 and 1, respectively. However, the solution from MATLAB may suggest a mistake in the manual calculations.
  • #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:
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  • #2
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:
[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
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. :)
 
  • #5
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
It looks to me like the first column, not the third column, represents a vector that is a multiple of <-4, 0, 1>^T.
 

1. What is the purpose of finding eigenvectors given eigenvalues?

Eigenvectors and eigenvalues are important concepts in linear algebra, and they have various applications in fields such as physics, engineering, and computer science. Finding the eigenvectors corresponding to the given eigenvalues allows us to understand the behavior of a linear transformation or a matrix, and it also helps us to solve systems of linear equations more efficiently.

2. How do you find the eigenvectors given the eigenvalues?

To find the eigenvectors corresponding to a given eigenvalue, we first need to find the null space of the matrix A - λI, where A is the original matrix and λ is the eigenvalue. This can be done by solving the system of equations (A - λI)x = 0. Once we have the null space, we can choose any non-zero vector from it as the eigenvector.

3. Can a matrix have more than one eigenvector for a given eigenvalue?

Yes, a matrix can have multiple eigenvectors corresponding to the same eigenvalue. In fact, there can be infinitely many eigenvectors for a given eigenvalue. However, these eigenvectors must all be linearly independent, meaning they are not scalar multiples of each other.

4. What is the relationship between eigenvectors and eigenvalues?

Eigenvectors and eigenvalues are closely related. An eigenvector is a non-zero vector that, when multiplied by a matrix, results in a scalar multiple of itself. The corresponding scalar multiple is the eigenvalue. In other words, the eigenvalue represents the factor by which the eigenvector is scaled when multiplied by the matrix.

5. Can all matrices be diagonalized using eigenvectors?

No, not all matrices can be diagonalized using eigenvectors. Only square matrices with distinct eigenvalues can be diagonalized. If a matrix has repeated eigenvalues or if it is not square, it cannot be diagonalized using eigenvectors. In these cases, other methods must be used to find a diagonal or nearly diagonal form of the matrix.

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