# Calculating the determinant of a quadratic for Eigenvalues

1. Feb 26, 2008

### Bucky

1. The problem statement, all variables and given/known data
Find the eigenvalues and eigenvectors of

$A = \left(\begin{array}{ccc}5&1&1\\1&3&1\\1&1&3\end{array}\right)$

3. The attempt at a solution

The problem I'm having is finding the eigenvalues for the matrix. In 2d matricies it's not too bad, but in 3d the determinant is quite big, and I end up with
$-\lambda^3 - \lambda^2 - 20\lambda + 34 = 0$
(note - very possible this is wrong, theres enough numbers being pushed around for something to have went wrong.)

How do I get three roots from this? Taking the values given in the book I get the rest of the question, it's just this step thats giving me trouble.

Thanks.

Last edited: Feb 27, 2008
2. Feb 26, 2008

### olgranpappy

...yeah, I think something went wrong. You can try again and be very careful about the numbers, or you could use a computer program like mathematica to do the grunt work for you...

3. Feb 26, 2008

### olgranpappy

P.S. MMA finds that the eigenvalues given by
$$-32 + 35 \lambda - 11 \lambda^2 + \lambda^3$$

4. Feb 26, 2008

### olgranpappy

P.P.S... are you sure that the "2" in your matrix is not supposed to be a "1"

In the latter case the eigenvalues are much nicer...

5. Feb 27, 2008

### Bucky

sorry yeah that 2 should be a 1 (changed it in origonal post).

either way, how do i take this expression and get three roots from it?

book gives 2, 3, 6 as the results.

6. Feb 27, 2008

### olgranpappy

your equation is wrong. try to work out the secular equation again and be careful. you will end up with a cubic equation which is hard in general to solve, but usually in homework problems you can guess one of the roots which you can then factor out anf you only have to solve a quadradic equation which is easier.

7. Feb 29, 2008

### Bucky

ok I spoke to my tutor about it and he explained how to do these, which made things a lot clearer. I've moved onto doin the next question, and I get one root but i can't get the other two..

$A = \left(\begin{array}{ccc}-1&1&-1\\1&-1&-1\\-1&-1&-1\end{ar ray}\right)$

$D(\lambda) = \left(\begin{array}{ccc}-1- \lambda&1&-1\\1&-1 - \lambda&-1\\-1&-1&-1 - \lambda\end{ar ray}\right)$

Expand determinant to give:

$= (-1 - \lambda) ((-1-\lambda)^2 - 1) - 1((-1 - \lambda) - 1) - (( -1 - \lambda) + 1)$

$= (-1 - \lambda) \{ ((-1-\lambda)^2 - 1) - 1(1-1) - (1+1) \}$

$= (-1 - \lambda) \{ (\lambda^2 + \lambda) - 0 - 2 \}$

$= (-1 - \lambda) \{ \lambda^2 + 2 \lambda - 2 \}$

I'm sure I've dropped something somewhere (since the quadratic doesn't work out), but i've read this over and over and can't see where I've went wrong.

8. Feb 29, 2008

### olgranpappy

the third term is incorrect. it should be
$$-(-1+(-1-\lambda))$$