Learn How to Find Eigenvalues of 3x3 Matrices | Eigenvalue Algorithm Explained

[sciencegirl1]

Homework Statement

find the eigenvalues of 3x3 matrix:

I have to learn how to find eigenvalues of 3x3 matrix

and this is the link, am I not supposed to do lamda-1 instead of 1-lamda like here?
http://en.wikipedia.org/wiki/Eigenvalue_algorithm
(the chapter name is "Eigenvalues of 3×3 matrices")


is there any simpler rule?

 

[Mark44]

Sadly, no simple rules.

For eigenvalues, you're solving the following equation for \lambda.
A\lambda = \lambdax
or equivalently,
(A - \lambdaI)x = 0

For the equation above to be true for arbitrary x (which turn out to be the eigenvectors),
(A - \lambdaI) has to be singular, which means that its determinant must be zero.

 

[sciencegirl1]

A= ( -1 -2 -2;
1 2 1;
-1 -1 0)

this is the matrix

I don´t understand how to get the eigenvalues which are supposed to be -1 1 1
because when I calculate it with the wikipedia rule I get -lamda^3+lamda^2-3lamda+1 and that does not make sense to me.

 

[Mark44]

You're taking the determinant of A - \lambdaI and setting it to zero, so you'll have a cubic polynomial in \lambda that is equal to zero.
You need to solve for \lambda in that polynomial.

 

[sciencegirl1]

What do you mean?

 

[Mark44]

Which part of what I said don't you understand?

 

[sciencegirl1]

so you'll have a cubic polynomial in LaTeX Code: \\lambda that is equal to zero.
You need to solve for LaTeX Code: \\lambda in that polynomial.

can you show me?

 

[Mark44]

Here's the equation you need to solve:
-\lambda^3+\lambda^2-3\lambda+1 = 0

I haven't checked your work, so it's possible that this isn't the right equation.

If (\lambda - a) is a factor, a has to be a divisor of 1, which severely limits the possibilities for a. By divisor, I mean a has to go into 1 a whole number of times.

Use polynomial long division to work this out. Once you get one factor, you'll be left with a quadratic in lambda, which should be pretty easy to factor or use the quadratic formula on.

 

[sciencegirl1]

thanks for your help
i just don´t know how to solve it :( I´be been trying but is there any rule for solving equations like this or...?

 

[Mark44]

Let me say it again.

Use polynomial long division to work this out. Once you get one factor, you'll be left with a quadratic in lambda, which should be pretty easy to factor or use the quadratic formula on.

Didn't you learn about polynomial long division back when you took elementary or intermediate algebra?

 

[lanedance]

A= ( -1 -2 -2;
1 2 1;
-1 -1 0)

this is the matrix

I don´t understand how to get the eigenvalues which are supposed to be -1 1 1
because when I calculate it with the wikipedia rule I get -lamda^3+lamda^2-3lamda+1 and that does not make sense to me.

It may be worth looking at your determinant calculation again as well, very easy to make a miscalulation in those steps.

if you know the eigenvalues are -1, 1, 1 the you should be able to factorise the determinant in term of these... so try subsitituting 1 or -1 into
-\lambda^3+\lambda^2-3\lambda+1=0
ie
-(1)^3+(1)^2-3.(1)+1=-2\neq 0
which makes me think something may have gone wrong in the det calc

and then
-\lambda^3+\lambda^2-3\lambda+1 \neq (\lambda-1)^2(\lambda+1)

 

[HallsofIvy]

Your matrix is
A= \begin{bmatrix} -1 & -2 & -2 \\ 1 & 2 & 1 \\ -1 & -1 & 0\end{bmatrix}

So the characteristic equation is
|A- \lambda I|= \left|\begin{array}{ccc}-1- \lambda & -2 & -2 \\ 1 & 2- \lambda & 1 \\ -1 & -1 & -\lambda\end{array}\right|= 0
expanding, on, say, the first row, that is
(-1-\lambda)\left|\begin{array}{cc}2-\lambda 1 \\ -1 & -\lambda\end{array}\right|+ 2\left|\begin{array}{cc}1 & 1 \\ -1 & -\lambda\end{array}\right|+ 2\left|\begin{array}{cc}1 & 2-\lambda \\ -1 & -1\end{array}\right|
= (-1-\lambda)(\lambda^2- 2\lambda+ 1)+ 2(-\lambda+ 1)-2(1-\lambda)
= -(1+\lambda)(\lambda- 1)^2- 2(\lambda-1)+ 2(\lambda-1)
and, in that form, it is obvious that the last two terms cancel leaving the already factored form
-(\lambda+ 1)(\lambda- 1)^2

 

[maze]

When you are learning all this keep in mind that no-one actually solves the characteristic polynomial to find eigenvalues in real-world problems. This is mostly useful for small toy problems, or to gain understanding of the theory.

For real problems you would use something like the QR algorithm or the Arnoldi iteration.

 

[CFDFEAGURU]

Here is a link to a great set of linear algebra notes.

http://tutorial.math.lamar.edu/Classes/LinAlg/LinAlg.aspx

Thanks
Matt

 

[sciencegirl1]

Thanks a lot everybody!