Eigenvectors of a 3x3 Matrix

In summary, the problem is that I am not getting the correct remainder when calculating the determinant. I am assuming that I am doing something wrong, but I am not sure what it is. The only solution that I have found so far is to use row expansion to find the eigenvectors.
  • #1
Bkkkk
12
0

Homework Statement


Given Matrix B:

[ 1 2 1]
[-1 2 -1]
[ 2 -2 3]


and knowing that one of the Eigenvalues is 4, find one other value and its corresponding eigenvector

Homework Equations


Bx=Lx (The basic idea behind eigenvectors)
det(B-LI)=0

The Attempt at a Solution


Ive set up the above determinant


[ 1-L 2 1]
[-1 2-L -1]
[ 2 -2 3-L]


Equal to zero, the only way I could figure how to do this question was using long division after getting the characteristic equation, but I keep getting a remainder which I shouldn't get If I am not making a huge mistake.

For the determinant I get either -L^3 + L^2 -9L +3 or -L^3 + 6L^2 -11L + 6
but I checked both using a calculator and long division and both of them give me a remainder.

I don't want the answer flat out maybe point out where I am going wrong, this problem has been bugging me for ages and I really want to know what the hell is wrong.:grumpy:

Thanks
 
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  • #3
I don't get either of your polynomials. Maybe you should show your steps in calculating the determinant.
 
  • #4
He this is the determinant as I set it up

(1-X){[(2-X)(3-X)]-2}-2{[(-1)(3-X)]+2}+{[(-1)(-2)]-[(2)(2-X)]}

(1-X){[6-5X+X^2-2}-2{X-1}+{2-4-2X}

(1-X){[-5X+X^2+4}-2X+2-2-2X

(1-X){-5X+X^2+4}-2X+2-2-2X

-5X+X^2+4+5X^2-X^3-4X-4X

-13X+6X^2+4-X^3

Completely different but still wrong.
 
  • #5
{[(-1)(-2)]-[(2)(2-X)]} = 2 -4 -(-2x) = -2 + 2x
 
  • #6
Hey!

I just tried your problem twice. The first time, I did it the way you did, by calculating the determinant, and like you, I ended up with a factor left after long division. But then, I calculated det(LI-B)=0. It'd do it this way, if I were you, because then you don't get all the -L's in there, which just means there's less chance of numerical errors when you calculate the determinants.
 
  • #7
malawi_glenn said:
{[(-1)(-2)]-[(2)(2-X)]} = 2 -4 -(-2x) = -2 + 2x
Oh I hate when that happens!

SO then

The final solution is

(x-1)(x-1)(X-4)
 
  • #8
Bkkkk said:
Oh I hate when that happens!

SO then

The final solution is

(x-1)(x-1)(X-4)
Well, no, the "final solution" is not a polynomial! You still need to find the eigenvalues and corresponding eigenvectors.
 
  • #9
Yes, sorry I meant the polynomials solution.

So all I have to do now is put the value for lamba back into Matrix B, multiply by the column matrix, make that equal to zero and work out the values of X, Y and Z.
 
  • #10
Yes, finding the eigenvectors should be straightforward. Notice, however, that you have x=1 as a double root. There may be two independent eigenvectors corresponding to that.
 
  • #11
I would use row expansion.

I get:

[tex](1 - r)[(2 - r)·(3 - r) - 2] - 2[(-1)·(3 - r) + 2] + [2 - 2·(2 - r)] = 0[/tex]

After simplifying you get:

[tex](4 - r)·(r - 1)^2[/tex]

So one eigenvalue is 4, like you said, and the others would be 1 and 1 (repeated).

Finding the eigenvectors is straight forward from here. If you need anymore help I would be glad to go further :smile:

Good luck!
 
  • #12
Thanks everyone, I managed to get the other eigenvector, there is only one from the repeated L=1 so it was straight forward.

Thanks again.
 

1. What is an eigenvector?

An eigenvector is a vector that does not change direction when multiplied by a matrix. Instead, it only scales by a constant factor, known as the eigenvalue.

2. Why are eigenvectors important?

Eigenvectors are important because they help us understand the behavior of a matrix. They can be used to simplify calculations, identify patterns, and solve complex problems in fields such as physics, engineering, and computer science.

3. How do you find the eigenvectors of a 3x3 matrix?

To find the eigenvectors of a 3x3 matrix, you first need to find the eigenvalues by solving the characteristic equation. Then, for each eigenvalue, you can use the equation (A - λI)x = 0 to find the corresponding eigenvector, where A is the original matrix, λ is the eigenvalue, and I is the identity matrix.

4. Can a matrix have more than 3 eigenvectors?

Yes, a matrix can have up to n linearly independent eigenvectors, where n is the size of the matrix. In the case of a 3x3 matrix, it can have up to 3 eigenvectors.

5. What is the relationship between eigenvectors and eigenvalues?

Eigenvectors and eigenvalues are closely related. The eigenvalues represent the scaling factor of the corresponding eigenvector when multiplied by the matrix. In other words, the eigenvalues determine the direction and magnitude of the eigenvectors. Additionally, eigenvectors with different eigenvalues are orthogonal to each other.

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