# Thread: Find eigenvalues and eigenvectors of matrices

1. ## Find eigenvalues and eigenvectors of matrices

Find eigenvalues and eigenvectors of this matrix

I found the eigenvalue λ1=-3 and λ2=6
When I added λ1=-3 to the matrix, I have the equation system like this

x1-2x2+2x3=0
-2x1+4x2-4x3=0
2x1-4x2+4x3=0

I don't know how to solve this equation system to figure out x1,x2,x3 and eigenvector. Please help me!

2. ## Re: Find eigenvalues and eigenvectors of matrices

Hey math88.

You should eliminate two rows (since they are all linearly dependent) and you will get x1 - 2x2 + 2x3 = 0.

Now you need to convert that to a vector with a free parameter (call it t) by letting one variable equal to t and solving for the other parameters (in terms of t) and your vector will be (x1(t),x2(t),x3(t)) and this is your eigenvector. You can then normalize this to have length 1 (this is typically done for eigenvectors).

3. ## Re: Find eigenvalues and eigenvectors of matrices

I'm not sure, but
if x3=t and x2=-t so x1= -4t so the eigenvectors is t(-4,-1,1)
or if I choose x1=t, x2=2t so x3=3/2 t -> the eigenvectors is t(1,2,3/2)
I don't know which one is correct.

If x1=t -> x2-x3=t/2
What should I do next to find x2, x3 in term of t?

4. ## Re: Find eigenvalues and eigenvectors of matrices

Originally Posted by math88
Find eigenvalues and eigenvectors of this matrix

I found the eigenvalue λ1=-3 and λ2=6
When I added λ1=-3 to the matrix, I have the equation system like this

x1-2x2+2x3=0
-2x1+4x2-4x3=0
2x1-4x2+4x3=0

I don't know how to solve this equation system to figure out x1,x2,x3 and eigenvector. Please help me!
An nXn matrix always has n eigenvalues. But some of them may be repeated. What are your three eigenvalues?

5. ## Re: Find eigenvalues and eigenvectors of matrices

There are, as I am sure you know, to solve three equations in three variables. Here, an obvious solution is $x_1= x_2= x_3= 0$. Since you used an eigenvalue to get them, there must be another, in fact, infinitely many, non-zero ("non-trivial") solutions.

Here, you can immediately see that the third equation is just 2 times the first. And the second equation is -1 times the third. That is, we really only have one equation here. From the first equation, $x_1- 2x_2+ 2x_3= 0$ we can solve for x_1= 2x_2- 2x_3. Then any eigenvector can be written in the form [tex]<x_1, x_2, x_3>= <2x_2- 2x_3, x_2, x_3>= <2x_2, x_2, 0>+ <-2x_3, 0, x_3>= x_2<2, 1, 0>+ x_3<2, 0, 1>. That is, this "eigenspace" the space of all eigenvectors corresponding to eigenvalue -3 is two-dimensional and is spanned by <2, 1, 0> and <2, 0, 1>.