# Finding the eigenvectors of a 3x3 matrix - help please

1. Apr 14, 2010

### Major_Disaster

1. The problem statement, all variables and given/known data
Determine the eigenvalues and eigenvectors of the matric, A:

$$A=\left[\begin{array}{ccc}1 & 1 & 0\\ 1 & -2 & 0\\ 0 & 0 & 1\end{array}$$

2. Relevant equations
I think i understand what is going on. I have found the matrix equation to be:

$$\left(1-\lambda\right)\left[\left(-2-\lambda)(1-\lambda)-1\right]=0$$

So:

$$\lambda_{1}=1$$
$$\lambda_{2}=\frac{-1-\sqrt{13}}{2}$$
$$\lambda_{3}=\frac{-1-\sqrt{13}}{2}$$

3. The attempt at a solution
I have gotten some solutions (but this being my first attempt at Tex i think would take me an age to write out!) but im confused as to which is the best method.

In lectures, our lecturer seemed to say that when you get to

$$V_{1}=\left[\begin{array}{ccc}x_{1}\\x_{2}\\x_{3}\end{array}$$

You just set

$$x_{3}=1$$

To make life easier and go from there. But i dont see how or why?

I always tried to use the normalisation condition that the square of the three components of the vectors equals one and find another relation from there (or sub in something else)

EDIT: Sorry for my failed attempt at Tex - only took half an hour!

2. Apr 14, 2010

Hint: plug the eigenvalues into the equation (A - lambda*I)x = 0 and see what happens.

3. Apr 14, 2010

### vela

Staff Emeritus
If a vector v is an eigenvector of A, then any non-zero multiple of v is also an eigenvector of A for the same eigenvalue. In particular, unless x3=0, you can find some multiple of v where x3=1. So your lecturer is saying just set x3=1, just to keep things simple, and then find the corresponding x1 and x2. You can then choose to normalize the resulting eigenvector if you want.

(If you're having trouble actually solving for an eigenvector, do what radou suggested.)

4. Apr 14, 2010

### Major_Disaster

THANK YOU!

That was exactly what i have been looking for and have spent the better half of the past 4 hours trying to find. In text books i see "and now we set this to one" and no explanation as to why. It all makes sense now!

Im not having any troubles generally with finding eigenvectors - but for future reference then, which "method" is better / reccomended? The (in my eyes) "standard" method that radou posted or by exploiting the fact that you said? Surely setting x3 as one makes the whole process quicker and easier?

Thanks again

5. Apr 14, 2010

### vela

Staff Emeritus
It's not an either-or choice. You do what radou said to get the equations the eigenvector must satisfy. Those equations will have more than one solution. As long as you find a non-zero solution to the equations, you have an eigenvector. Sometimes you can just see what a solution is by inspection, and you can just write down what the answer is. More generally, though, you can manipulate the equation so that some of the variables are expressed in term of the rest. For example, x1 and x2 might be written in terms of x3. Then to get one particular solution, you just set x3=1 to make the arithmetic easy and calculate the values for x1 and x2.

6. Apr 14, 2010

### Major_Disaster

Thank you for your continued help.

I know see that was a somewhat silly question.

I have been going through the original question trying the various methods avaliable.

Using http://www.arndt-bruenner.de/mathe/scripts/engl_eigenwert.htm" [Broken] I know that:

Code (Text):
Characteristic polynomial:
x^3 - 4x + 3

Real eigenvalues:
{-2.302775637731995, 1, 1.3027756377319946}

Eigenvector of eigenvalue -2.302775637731995:
(-0.28978414868843005, 0.9570920264890529, 0)

Eigenvector of eigenvalue 1:
(0, 0, 1)

Eigenvector of eigenvalue 1.3027756377319946:
(0.9570920264890529, 0.28978414868843, 0)
First of all, i have found the eigenvalues (in agreement with the above). I know tried to calculate the eigenvectors.

I first used my original method:

$$[\begin{array}{ccc}1 & 1 & 0\\ 1 & -2 & 0\\ 0 & 0 & 1\end{array}] [\begin{array}{ccc}x_{1}\\x_{2}\\x_{3}\end{array}] =\lambda_{1} [\begin{array}{ccc}x_{1}\\x_{2}\\x_{3}\end{array}]$$

I then got some messy equations involving horrible fractions of $$x_{1}$$ and $$x_{2}$$. I was able to use the normalsiation condition to separate them both and eventually got eignvectors that match the link i posted.

I then tried doing the same as above, but also setting $$x_{3}=1$$, which gave me the following (normalised) eigenvector:

$$\frac{1}{\sqrt{11}}[\begin{array}{ccc}-3\\-1\\1\end{array}]$$

Which does not seem to be in agreement at all?

Finally i tried radou's form and once again set $$x_{3}=1$$. This worked great getting me straight to $$v_{1}$$. However, for $$v_{2}$$ $$x_{3}$$ is actually equal to 0 so the whole thing fell apart. Then what?

Where did i go wrong?

Thanks a lot

Last edited by a moderator: May 4, 2017
7. Apr 14, 2010

### vela

Staff Emeritus
For λ2, you get

$$(A-\lambda_2 I)\vec{x} = \begin{bmatrix}1-\frac{-1-\sqrt{13}}{2} & 1 & 0 \\ 1 & -2-\frac{-1-\sqrt{13}}{2} & 0 \\ 0 & 0 & 1-\frac{-1-\sqrt{13}}{2}\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = 0$$

which gives you the equations

$$\frac{3+\sqrt{13}}{2}x_1 + x_2 = 0$$

$$x_1 + \frac{-3+\sqrt{13}}{2}x_2 = 0$$

$${3+\sqrt{13}\over 2}x_3 = 0$$

From the last equation, you get x3=0. The first two equations are multiples of each others, so you only need to use one of them. So take, say, the second one and solve for x1. If you now set x2=1, you find

$$x_1 = \frac{3-\sqrt{13}}{2}x_2 = \frac{3-\sqrt{13}}{2}$$

So an eigenvector for λ2 is

$$V_2 = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}\frac{3-\sqrt{13}}{2} \\ 1 \\ 0\end{bmatrix}$$

Now, if you want to, you can normalize it.

8. Apr 15, 2010

### Major_Disaster

Thanks again vela for your continued support - i really appreciate it.

Ive had another go this morning on this problem and i see last night what i was doing was inserting x2=1 prematurely (in the original vector before i had actually expanded out the equations).

Everything works well with the method i was originally using:

$$[\begin{bmatrix}1 & 1 & 0\\ 1 & -2 & 0\\ 0 & 0 & 1\end{bmatrix}] [\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}] =\lambda_{1} [\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}]$$

I then tried again with your rearraged version:

$$(A-\lambda_2 I)\vec{x} = \begin{bmatrix}1-\frac{-1-\sqrt{13}}{2} & 1 & 0 \\ 1 & -2-\frac{-1-\sqrt{13}}{2} & 0 \\ 0 & 0 & 1-\frac{-1-\sqrt{13}}{2}\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = 0$$

Which works great for lamba2 and 3 - but for lamba1 (which equals 1) the whole right column of the matrix becomes zero. So i just get 0 = 0 and not the (0 0 1) vector that i got with the other method? Any ideas how to get around this?

Thanks again, i just want to get both methods straight in my head so i fully understand whats going on.

EDIT: Thinking about in, in my other method i just get x3 = x3, and i just said x3 = 1 - but it could be anything - so i guess this goes to what you originally said about any non zero multiple of the vector?)

Last edited: Apr 15, 2010
9. Apr 15, 2010

### vela

Staff Emeritus
Right. You're x3=x3 equation tells you nothing about x3, so it can be anything. In the other method, x3 just drops completely out of the equations, so again, there are no constraints on x3, so it can be anything.

10. Apr 15, 2010

### Major_Disaster

Perfect. So i could just have said x3 = 5, but then when i normalised it id get "one over the square root of 5 squared = 1/5)" and so just back to the (0 0 1) vector. That makes sense!

Thanks alot. This was driving me mad but ive got it all straight in my head now.

Thanks again.

11. Apr 15, 2010

### Major_Disaster

One last question. The next part of the problem reads:

"Find a matrix R such that RAR$$^{T}$$ = D, where D is a diagonal matrix (i.e.,
a matrix whose non-zero entries are all on the main diagonal)."

The answers say its just a matrix of the eigenvectors ive just found - why?

Thanks

12. Apr 15, 2010

### lanedance

so for a handwaving explanation

composing a the matrix R of the eigenvctors of A and taking D = RART in effect represents a change of basis, from whatever you started with to the basis of eignevectors.

When an eignvector is multiplied by the matrix, the result will be the corresponding eigenvalue tiems the eigenvector.

In the new basisi teh first eigenvector is represented by (1,0,..,0)T, hence why the diagonalised matrix must have the eignevalues on the diagonal

otherwise work the math to understand why