Finding eigenvalues of matrix

1. Jul 20, 2009

1. The problem statement, all variables and given/known data

Find the eigenvalues of the following matrix:

$$\left( \begin{array}{ccc} 1 & 0 & -3 \\ 1 & 2 & 1 \\ -3 & 0 & 1 \end{array} \right)$$

2. Relevant equations

3. The attempt at a solution

I think I'm forgetting a basic algebra rule or something. I know there are supposed to be 3 eigenvalues, but I am only getting 2.

$$0=\det (\lambda I-A)=\left| \begin{array}{ccc} \lambda -1 & 0 & 3 \\ -1 & \lambda -2 & -1 \\ 3 & 0 & \lambda -1 \end{array} \right|$$
$$=\frac{(\lambda -1)(\lambda -2)(\lambda -1)-3(3)(\lambda -2)}{(\lambda -2)}$$
$$=(\lambda -1)(\lambda -1)-9$$
$$=(\lambda -1)^2-9$$
$$(\lambda -1)^2=9$$
$$\lambda =\pm 3+1$$

2. Jul 20, 2009

AUMathTutor

Re: Eigenvalues

Where you cancel the (lambda - 2) term, you lose one solution: lambda = 2.

Remember, you have to factor. By canceling the term (lambda - 2), you're basically saying "lambda cannot under any circumstances equal 2". If you want to do that, you have to consider two separate cases: lambda=2 and lambda!=2.

So the three eigenvalues are 3+1=4, -3+1=-2, and 2.

3. Jul 20, 2009

Pengwuino

Re: Eigenvalues

Why did you divide by (lambda - 2) in the first place?

4. Jul 20, 2009

Re: Eigenvalues

So let me get this straight. Whenever I divide by a term to cancel it, I count that term as one of my solutions, right? Does this always work? I didn't quite understand what you meant by testing two separate cases.

5. Jul 20, 2009

Re: Eigenvalues

Because it was much easier to do that than it was to multiply everything out and try to factor a third-degree polynomial.

6. Jul 20, 2009

VeeEight

Re: Eigenvalues

Why did you divide by lambda-2 in the second line of your equations? Do you remember the formulas for a determinant of a 3x3 matrix?

7. Jul 20, 2009

Re: Eigenvalues

Oh, duh. Since there are two terms on the RHS and they are constants then the must both be equal to zero separately. Haha.

8. Jul 20, 2009

VeeEight

Re: Eigenvalues

You can just factor (x-2) to get an easily solvable equation

9. Jul 20, 2009

Re: Eigenvalues

I was just doing two steps at once. Bad form on my part.

10. Jul 20, 2009

Pengwuino

Re: Eigenvalues

Yes. Think of it this way, let's say you're trying to find the solutions of x for (x-1)(x-2)(x-3)=0. Obviously your solutions are x=1,2, and 3. If you divide out (x-1), you lose a solution and are left with (x-2)(x-3)=0.

11. Jul 20, 2009

Re: Eigenvalues

Very good point. Thanks!

12. Jul 20, 2009

Pengwuino

Re: Eigenvalues

Your welcome. Also, to clarify AU's point, when you divide by (lambda-2), your left with that (lambda-2) in the denominator of the left hand side (the 0). Now for any number other then 2, it's going to be 0 divided by some finite number which is 0. However, with lambda = 2, you get 0/0 which is undefined, not 0.

13. Jul 20, 2009