# How to find the eigenvectors

1. Jan 25, 2006

### stunner5000pt

find the eigenvectors for this matrix
$$\left( \begin{array}{ccc}3&0&6\\0&-3&0\\5&0&2 \end{array} \right)$$

easy to find the eigenvectors which are -3,-3 and 8
now how to find the eigenvectors
am i supposed to do
$$\left| \lambda I - A \right| X = 3 X$$?
then
$$\left( \begin{array}{ccc} -6&0&6\\0&0&0\\5&0&-5 \end{array} \right) \left( \begin{array}{c} X_{1}\\X_{2}\\X_{3} \end{array} \right) = 0$$ ? an find the solution for X1 x2 and X3?
it comes 1,0,1 then, which is not correct...

pleas help!

Last edited: Jan 25, 2006
2. Jan 25, 2006

### 0rthodontist

Well, you set

$$\left( \begin{array}{ccc} -6&0&6\\0&0&0\\5&0&-5 \end{array} \right) \left( \begin{array}{c} X_{1}\\X_{2}\\X_{3} \end{array} \right) = 0$$

up right but

$$\left| \lambda I - A \right| X = 3 X$$ is wrong. The equation you need to solve is
$$( \lambda I - A ) X = 0$$ for $$\lambda$$ = -3, and then 8. You only take the determinant if you're trying to find the eigenvalues. You can derive it because you're trying to solve

$$A X = \lambda X = \lambda I X$$
$$A X - \lambda I X = 0$$
$$(A - \lambda I) X = 0$$
or
$$(\lambda I - A) X = 0$$

But the final matrix equation you gave is not solved by (1, 0, 1). Probably just an arithmetic error.

Last edited: Jan 25, 2006
3. Jan 25, 2006

### stunner5000pt

this is the matrix for the characteristic polynomai lright ...
$$\left( \begin{array}{ccc}\lambda -3&0&6\\0&\lambda+3&0\\5&0&\lambda -2 \end{array} \right)$$

when u substitute -3 in there you get
$$\left( \begin{array}{ccc} -6&0&6\\0&0&0\\5&0&-5 \end{array} \right)$$
isnt htat correct?
Please point out my math error here...

4. Jan 25, 2006

### 0rthodontist

D'oh, I was confused. The error is the 6 and the 5 in the top right and bottom left corners. They should be negative.

5. Jan 25, 2006

### stunner5000pt

OF COURSE! I was completely disregarding that everything lese in the matrix A will be negated

thank you

6. Jan 25, 2006

### stunner5000pt

so from the -3 one of the eigenvectors is [-1 0 1] but why not [1 0 -1]??

since there are only 2 eigenvalues how would you get the third eigenvalue?

7. Jan 25, 2006

### 0rthodontist

I don't know what your question is. (-1, 0, 1) is an eigenvector for -3 and (1, 0, -1) is another eigenvector for -3, though they are not independent. Use row reduction to find the general form that an eigenvector for -3 must take, and you can get 2 independent eigenvectors.

8. Jan 25, 2006

### stunner5000pt

the real question is to diagonalize this matrix
and to diagonalizei need to find P where P = [X1 X2 X3 ... Xn]
thus i need to find THREE eigenvectors

9. Jan 25, 2006

### 0rthodontist

You don't need just any three eigenvectors, you need three independent eigenvectors. The third comes from 8. Can you give the general form that an eigenvector for -3 must take?

10. Jan 25, 2006

### stunner5000pt

however the text (and mathematica) say that hte eigenvectors are
[-1 0 1]
[0 1 0]
[6 0 5]
how how?

11. Jan 25, 2006

### 0rthodontist

Can you give the general form that an eigenvector for -3 must take?

12. Jan 25, 2006

### stunner5000pt

nevermind...
i found the last eigenvector when i used the eigenvalue 8 and it yielded 2 independant eigenvectors

13. Jan 25, 2006

### 0rthodontist

8 yields only 1 independent eigenvector. It has algebraic multiplicity 1 so it can't yield more than that.