# Thread:

1. ## 3x3 Matrix - Eigenvectors

Okey, im trying to find the eigenvectors of a 3x3 matrix.
(I have found the eigenvalues)

I'm going to use an example so I can solve my own matrix myself;

In this example the eigenvectors are found, but how?
So my question is How are the calculations done to recieve the values of x, y, and z (eigenvectors)?

Eg. take example 1 with the eigenvalue of 0

Any help appriciated!

2. in order to find the eigenvector for the eigenvalue r , you first subtract r from the diagonal of the matrix and then find the kernel of that matrix. this is because the eigenvector needs to solve the equation Av=rv, or 0 = Av-rv=Av-rIv=(A-rI)v. the best way to do it is to find its reduced row echelon form and then solve the system.
for example with the 0 eigenvalue you get
$<br /> \begin{bmatrix}<br /> 3-0 & 0 & 0 \\-4 & 6-0 & 2 \\16 & -15 & -5-0<br /> \end{bmatrix} \rightarrow \begin{bmatrix}<br /> 1 & 0 & 0 \\-2 & 3 & 1 \\3.2 & -3 & -1<br /> \end{bmatrix} \rightarrow \begin{bmatrix}<br /> 1 & 0 & 0 \\0 & 3 & 1 \\0 & 0 & 0<br /> \end{bmatrix}<br />$ (almost reduced form...)
now, a vector is in the kernel of that matrix only if it has the form (0,-x,3x), so the base for the kernel is the vector (0,-1,3) which is the eigenvector. if the base is of higher dimension then you have more than 1 eigenvector for that specified eigenvalue

3. Thanks for you answere, buw I didn't get a whole lot wiser.

What do you mean by "kernel". And why/how do you get to the form (0, -x, 3x) from the reduced matrix? :-P

Can it as well be solved by finding the equation = 0? (Or something similar?)

eg.

3Xn = 0 = Xn
-4Xn + 6Yn + 2Zn = 0 = Yn
16Xn - 15Yn - 5Zn = 0 = Zn

?

Thanks yet again!

Here is the matrix im trying to find the eigenvectors for;

I have found that the eigenvalues has to be (-0,8 , - 0,2 , 1) (Is this correct?)

Could you show me how you would find the eigenvectors?

4. The kernel for matrix A is the subspace of vectors that solves the equation Av=0
in the example I gave, you have
$<br /> \begin{bmatrix}<br /> 1 & 0 & 0 \\ 0 & 3 & 1 \\ 0&0&0<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> x \\ y \\ z<br /> \end{bmatrix} = <br /> \begin{bmatrix}<br /> 0 \\ 0 \\ 0<br /> \end{bmatrix}<br />$
or in other words you have the equations 1*x=0 from the first row, and 3y+z=0 from the second row, so x must be zero and the relation between y and z is -3y=z, so the vectors we're looking for are (0,y,-3y)

you can try to solve these equations without changing to the reduced echelon form but then you get a more difficult set of equations to solve. actually, the process of solving them is exactly the process of reducing the matrix to its echelon form, but I think that using the matrix notation is easier (specially, if the matrix is bigger than 3x3)

for example, with your matrix and the eigenvalue -0.8 you get
$<br /> \begin{bmatrix}<br /> 0-(-0.8) & 1.05 & 0.25 \\ 0.8 & 0-(-0.8) & 0 \\ 0&0.8&0-(-0.8)<br /> \end{bmatrix} = \begin{bmatrix}<br /> 0.8 & 1.05 & 0.25 \\ 0.8 & 0.8 & 0 \\ 0 & 0.8 & 0.8<br /> \end{bmatrix}<br />$ $<br /> \rightarrow \begin{bmatrix}<br /> 0 & 0 & 0 \\ 0.8 & 0.8 & 0 \\ 0 & 0.8 & 0.8<br /> \end{bmatrix} \rightarrow \begin{bmatrix}<br /> 0 & 0 & 0 \\ 0.8 & 0 & -0.8 \\ 0 & 0.8 & 0.8<br /> \end{bmatrix} \rightarrow \begin{bmatrix}<br /> 0 & 0 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & 1<br /> \end{bmatrix}<br />$
now to find the kernel you solve
$\begin{bmatrix}<br /> 0 & 0 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & 1<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> x \\ y \\ z<br /> \end{bmatrix} = <br /> \begin{bmatrix}<br /> 0 \\ 0 \\ 0<br /> \end{bmatrix}$
and you get x-z=0 and y+z=0 so x=z and y=-z, so the vectors are of the form (z,-z,z) and the base is the vector (1,-1,1). and just to check that this is the correct answer
$<br /> \begin{bmatrix}<br /> 0 & 1.05 & 0.25 \\ 0.8 & 0 & 0 \\ 0&0.8&0<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1 \\ -1 \\ 1<br /> \end{bmatrix} = <br /> \begin{bmatrix}<br /> -1.05+0.25 \\ 0.8 \\ -0.8<br /> \end{bmatrix} = <br /> -0.8*<br /> \begin{bmatrix}<br /> 1 \\ -1 \\ 1<br /> \end{bmatrix}<br />$

5. I can see how reduced echelon form helps. Thanks!

I'll take a deeper look on how to form a matrix on reduced echelon, if you dont mind explaining that bit a tiny more? (Since I have your attention... )
Or point me in a direction where its some "easy reading".

This part: In particular how you managed to get row 1 to 0 0 0 (Im assuming you have used Gauss-elimination)

6. after subtracting the second row from the first you get (0, 0.25, 0.25)
now multiply it by 0.8/0.25 you get (0, 0.8, 0.8) which is the same as the third row, so you can delete one of them.

finding the row reduced form of a matrix, is just the result of Gaussian elimination (if that what you meant by that question).

7. I see! Surley nice of you to explain it to me so thorough! I appriciate it!

So as I can se you have

R1 - R2
R1 * 0.8/0.25 (Which becomes the same as R3
Delete R1 (R1 - R3? )
Then R2 - R3
Correct?

Thanks yet again! You are very helpful indeed!

8. Im trying to solve the eigenvectors for the eigenvalues -0,2 and 1.

I can't seem to find any method on how to reduce those to find an answere.

I understand if you are getting sick of me asking, but if you could give me a hint/explanation on how to solve them, I sure would appriciate it...

9. well, the technique is simple. first go to the left most column that is not all zeros (in this case its the first column). multiply all the rows that have nonzero scalar on that column, so they will all have the same scalar.
for example with the eigenvalue -0.2, the first column has two nonzero scalars -0.2 and 0.8, so you multiply the first row by 4 (or the second by 1/4 or any other way you like).
$\begin{bmatrix} 0.2 & 1.05 & 0.25 \\ 0.8 & 0.2 & 0 \\ 0 & 0.8 & 0.2 \end{bmatrix} \rightarrow \begin{bmatrix} 0.8 & 4.2 & 1 \\ 0.8 & 0.2 & 0 \\ 0 & 0.8 & 0.2 \end{bmatrix}$
now you choose a row with nonzero scalar on the column and switch it with the first row. because in this case you already have a nonzero scalar on the first row you don't need to do the switch. now subtract the first row from the other rows so they'll all have zero (except for the first row) on that column.
$\begin{bmatrix} 0.8 & 4.2 & 1 \\ 0 & -4 & -1 \\ 0 & 0.8 & 0.2 \end{bmatrix}$
now you do the same thing with the bottom left 2x2 sub matrix
$\rightarrow \begin{bmatrix} 0.8 & 4.2 & 1 \\ 0 & -4 & -1 \\ 0 & 4 & 1 \end{bmatrix} \rightarrow \begin{bmatrix} 0.8 & 4.2 & 1 \\ 0 & -4 & -1 \\ 0 & 0 & 0 \end{bmatrix}$
now use the leading coefficients (the left most nonzero numbers in each row) to zero all the numbers above them. in this case you only have the -4 and you want to zero the 4.2 above it.
$\rightarrow \begin{bmatrix} 4 & 21 & 5 \\ 0 & -4 & -1 \\ 0 & 0 & 0 \end{bmatrix}\rightarrow \begin{bmatrix} 4 & 0 & -0.25 \\ 0 & -4 & -1 \\ 0 & 0 & 0 \end{bmatrix}$
you can multiply by scalars so that the leading coefficient would be 1's but this is not necessary. finally you get the 4x-z/4=0 and -4y-z=0 which give you the eigenvector (1/16,-1/4,1) or (1,-4,16)

this works the same for matrices of different size. anyway, I advise you to find an introductory book (or a web page) for linear algebra to see how it works in the general case.

10. Man! You saved my day!

I really appriciate the effort, thanks yet again! I understand a great deal now on how the process is done at least!

11. A quick question if you know how to explain it;

If all of the eigenvalues are less than 1 of a dynamic population matrix(linear equation) it would die out, and if all of the eigenvalues are bigger than 1 it would grow. Why is this?
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