# 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