# Finding eigenvectors

1. May 2, 2008

### t_n_p

1. The problem statement, all variables and given/known data

I have a matrix A = [3 -2; -3 2],
then I peform det(A-λI) and find solutions λ=0 and λ=5.

For the first case λ=0 I perform A-0I, which of course is just A [3 -2; -3 2]. I then reduce to row echelon form and find the values V1 = 2 and V2 = 3, therefore for λ=0 my eigenvector is [2; 3]

For the second case λ=5, I perform A-5I, which gives [-2 -2; -3 -3]. Reducing to row echelon form yields [-6 -6; 0 0][V1; V2] = [0; 0]

In linear form my, the equation is as follows, -6V1 - 6V2 = 0

The answer provided is [V1; V2] = [-1; 1], my question is, how do you know for a case where the coeffients are the same, that the eigenvector will be [-1; 1] rather than [1;-1] as -6(1) - 6(-1) = 0 and still satifies the equation.

Thanks

2. May 2, 2008

### Defennder

An infinite number of eigenvectors exists for every eigenvalue, though they may not be linearly independent. Both (1,-1) and (-1,1) are linear combinations of each other; they differ by a factor of -1 alone. Any multiple of this particular eigenvector would also satisfy the equation because you can cancel out the contants on both side of the matrix equation $$Ax=\lambda x$$

3. May 2, 2008

### Dick

If v is an eigenvector, the c*v is also an eigenvector for ANY constant c. Since A(cv)=cA(v)=c*lambda*v=lambda*(cv). So [-1,1] and [1,-1] are both eigenvectors. You can choose either one. For the eigenvalue 0 you could also have found the eigenvector to be [-2,-3]. That's also fine.

4. May 2, 2008

Thanks.

5. May 3, 2008

### t_n_p

I have another question. I have a matrix A = [0 -1 0; -4 0 0; 0 0 2]. I have found the characteristic equation to be (-1)^3(λ+2)(λ-2)^2 and hence my eigenvectors and λ=-2 and 2. In the answers it says the eigenvectors are [1; 2; 0], [1; -2; 0] and [0; 0; 1].

How can you have three eigenvectors from two eigenvalues? Yes I realised there are two solutions λ=2, but surely they provide the same eigenvectors?

Note: For λ=2 I found my eigenvector to be [1; -2; 0] and hence I am suspicious about the [0; 0; 1] eigenvector

6. May 3, 2008

### Defennder

There can be more than one linearly independent eigenvector for each eigenvalue. There's nothing wrong with that.

7. May 3, 2008

### t_n_p

Can you explain how you get the eigenvector [0; 0; 1] ?

8. May 3, 2008

### HallsofIvy

TRY! exactly like you did before.

Although I prefer the more fundamental:
$$\left[\begin{array}{ccc}0 & -1 & 0 \\ -4 & 0 & 0\\ 0 & 0 & 2\end{array}\right]\left[\begin{array}{c} x \\ y \\ z\end{array}\right]= \left[\begin{array}{c} 2x \\ 2y \\ 2z \end{array}\right]$$
directly from the definition of "eigenvector".

That gives three equations: -y= 2x, -4x= 2y, 2z= 2z . The last equation is true for all x, y, and z. From the first, y= -2x so -4x= 2(-2x) = -4x is true for any x, y satisfying y= -2x. That is <x, -2x, z> is an eigenvalue for any x or z. In particular, if you take x= 1, z= 0, you get <1, -2, 0> and if you take x= 0, z= 1, you get <0, 0, 1>.

9. May 3, 2008

### tiny-tim

Hi t_n_p!

No. An n-repeated eigenvalue will have a whole n-dimensional subspace of eigenvectors.

For example the nxn matrix In has n eigenvalues = 1, but obviously any nx1 vector will be an eigenvector!
How did you get [1; -2; 0]? You'll probably find that the same method gave you [0; 0; 1], and you threw it away for some reason.

10. May 3, 2008

### t_n_p

Halls of Ivy - In the notes we don't learn this way

Tiny tim - using the same method as I did before, obviously I get the same eigenvector if I'm using the same eigenvalue.

Here's a recap of what I tried.
-Got the characteristic equation as shown above
-Subbed in the value of λ=2 into my matrix and setup the following..

[-2 -1 0; -4 -2 0; 0 0 0][V1; V2; V3] = [0; 0; 0]

-Then reduced to row echelon to form -4V1 - 2V2 + 0V3 = 0

This is how I got the eigenvector [1; -2; 0]. Obviously I can see that [0; 0; 1] satifies the equation, in fact [0; 0; x] where x is any number will satisfy the equation, but how you get it using the way I found eigenvector [1; -2; 0] is what I wish to know.

Thanks

11. May 3, 2008

### HallsofIvy

What definition of "eigenvector" did you learn?

Yes, That gives V2= -2V1. But do you see that V3 can be anything and still satisfy that equation? You can choose V1 and V3 to be anything and then V2= -2V1. That's why the "eigenspace" corrresponding to $\lambda= 2$ is two dimensional. As I said, you can choose V1 and V3 to be anything you like. "1" and "0" are particularly easy and if you chose one to be 0 and the other to be 1, you are sure to get independednt vectors. If you take V1= 1 and V3= 0 you get <1, -2, 0> and if you take V1= 0, V3= 1, you get <0, 0, 1>.

Last edited by a moderator: May 3, 2008
12. May 3, 2008

### HallsofIvy

tiny-tim, that's not necessarily true. For example, the matrix
$$\left[\begin{array}{cc}2 & 1 \\ 0 & 2\end{array}\right]$$
has 2 as a "double eigenvalue" but all eigenvectors are of the form <x, 0>: only one dimensional. That's why it cannot be "diagonalized".

13. May 3, 2008

### t_n_p

Yes I see what you mean, my next question is now whether there is a more "mathematical" basis for obtaining the eigenvector <0, 0, x>, where x can be any number. Yes I can see that the eigenvector <0, 0, x> satisfies the equation by "eye", I'm just wondering if this is a more "logical" reasoning rather than a "mathematical" one. I hope you understand what I'm trying to say :tongue2:

14. May 4, 2008

### HallsofIvy

I think what you are trying to say is exactly what I said before:

The definition of "eigenvector" is "v is an eigenvector of A, corresponding to eigenvector $\lambda$ if and only if $Av= \lambda v$

$$\left[\begin{array}{ccc}0 & -1 & 0 \\ -4 & 0 & 0\\ 0 & 0 & 2\end{array}\right]\left[\begin{array}{c} x \\ y \\ z\end{array}\right]= \left[\begin{array}{c} 2x \\ 2y \\ 2z \end{array}\right]$$