# Are my eigenvalues right?

1. Mar 4, 2004

### ilikephysics

Find the eigenvalues and eigenvectors of the general real symmetric 2 x 2 matrix A= a b
b c

The two eigenvalues that I got are a-b and c-b. I got these values from this:

(a-eigenvalue)(c-eigenvalue)-b^2=0
(a-eigenvalue)(c-eigenvalue) = b^2
(a-eigenvalue)= b = a-b
(c-eigenvalue)= b = c-b

Will there be 4 eigen values instead of the two that I have? Like, a-b and a+b and c-b and c+b?
I haven't gotten to the eigenvectors yet. I'll post what I have in a minute.

2. Mar 4, 2004

### ilikephysics

eigenvectors

Are these the eigenvectors?

For a-b,

-1
c-a-b

For c-b,

a-c-b
1

3. Mar 4, 2004

For the eigenvalues, you can't quite perform one of the steps you did. It's an incorrect assumption that

a - r = b = c - r
where r is an eigenvalue.

You've got the right equation up until that point, i.e.,

(a-r)(c-r) - b^2 = 0

but this is a quadratic equation in the r variable. Solve it the way you would normally solve a quadratic equation! Nothing too fancy.

4. Mar 5, 2004

### HallsofIvy

Staff Emeritus
If you are working with eigenvalues and eigenvectors it would be a really, really good idea not to mess up basic algebra!

You have:
(a-eigenvalue)(c-eigenvalue) = b^2
(a-eigenvalue)= b = a-b
(c-eigenvalue)= b = c-b

You appear to be thinking that if xy= b^2, then x= b and y= b.

That is not at all true!! (9*1= 32;5*(9/5)= 32, etc.)

Once you have the equation (a-eigenvalue)(c-eigenvalue) = b^2

(I'm going to use &lambda; for the eigenvalue)

(a- &lamda;)(c-&lamba;)- b2= 0
&lambda;2-(a+c)&lambda+ (ac- b2)= 0

Now you can use the quadratic formula to solve for the two values of &lambda;