Calculating the determinant of a quadratic for Eigenvalues

[Bucky]

Homework Statement

Find the eigenvalues and eigenvectors of


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

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, there's 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 that's giving me trouble.

Thanks.

 

[olgranpappy]

Homework Statement

Find the eigenvalues and eigenvectors of $A = \left(\begin{array}{ccc}5&1&1\\1&3&2\\1&1&3\end{array}\right)$

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, there's enough numbers being pushed around for something to have went wrong.)

...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...

 

[olgranpappy]

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

 

[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...

 

[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.

 

[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.

 

[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 doing 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.

 

[olgranpappy]

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 doing 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:

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

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