Calculating the determinant of a quadratic for Eigenvalues

In summary, the conversation is about finding the eigenvalues and eigenvectors of a 3x3 matrix. The problem is with finding the eigenvalues, and the conversation includes attempts at solving the equation and seeking help from a tutor. The conversation also involves double-checking the numbers and making corrections to the matrix given in the original problem.
  • #1
Bucky
82
0

Homework Statement


Find the eigenvalues and eigenvectors of


[itex]

A = \left(\begin{array}{ccc}5&1&1\\1&3&1\\1&1&3\end{array}\right)
[/itex]

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
[itex] -\lambda^3 - \lambda^2 - 20\lambda + 34 = 0 [/itex]
(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.
 
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  • #2
Bucky said:

Homework Statement


Find the eigenvalues and eigenvectors of [itex]

A = \left(\begin{array}{ccc}5&1&1\\1&3&2\\1&1&3\end{array}\right)
[/itex]

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
[itex] -\lambda^3 - \lambda^2 - 20\lambda + 34 = 0 [/itex]
(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...
 
  • #3
P.S. MMA finds that the eigenvalues given by
[tex]
-32 + 35 \lambda - 11 \lambda^2 + \lambda^3
[/tex]
 
  • #4
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
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
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
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..

[itex]A = \left(\begin{array}{ccc}-1&1&-1\\1&-1&-1\\-1&-1&-1\end{ar ray}\right)[/itex]

[itex]D(\lambda) = \left(\begin{array}{ccc}-1- \lambda&1&-1\\1&-1 - \lambda&-1\\-1&-1&-1 - \lambda\end{ar ray}\right)[/itex]

Expand determinant to give:

[itex]

= (-1 - \lambda) ((-1-\lambda)^2 - 1) - 1((-1 - \lambda) - 1) - (( -1 - \lambda) + 1) [/itex]

[itex]
= (-1 - \lambda) \{ ((-1-\lambda)^2 - 1) - 1(1-1) - (1+1) \} [/itex]

[itex]
= (-1 - \lambda) \{ (\lambda^2 + \lambda) - 0 - 2 \} [/itex]

[itex]
= (-1 - \lambda) \{ \lambda^2 + 2 \lambda - 2 \}
[/itex]

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
Bucky said:
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..

[itex]A = \left(\begin{array}{ccc}-1&1&-1\\1&-1&-1\\-1&-1&-1\end{ar ray}\right)[/itex]

[itex]D(\lambda) = \left(\begin{array}{ccc}-1- \lambda&1&-1\\1&-1 - \lambda&-1\\-1&-1&-1 - \lambda\end{ar ray}\right)[/itex]

Expand determinant to give:

[itex]

= (-1 - \lambda) ((-1-\lambda)^2 - 1) - 1((-1 - \lambda) - 1) - (( -1 - \lambda) + 1) [/itex]
the third term is incorrect. it should be
[tex]
-(-1+(-1-\lambda))
[/tex]
 

Related to Calculating the determinant of a quadratic for Eigenvalues

What is the determinant of a quadratic matrix?

The determinant of a 2x2 quadratic matrix is the product of the elements on the main diagonal minus the product of the elements on the other diagonal.

Why is calculating the determinant important for finding eigenvalues?

The determinant is necessary for finding eigenvalues because it is a key factor in determining the solutions to the characteristic equation, which is used to find eigenvalues.

What is the process for calculating the determinant of a quadratic matrix?

To calculate the determinant of a 2x2 quadratic matrix, you multiply the elements on the main diagonal and subtract the product of the elements on the other diagonal.

Can the determinant of a quadratic matrix be negative?

Yes, the determinant of a quadratic matrix can be negative. The sign of the determinant depends on the arrangement of the elements within the matrix.

What is the relationship between the determinant and the eigenvalues of a quadratic matrix?

The determinant is used to find the solutions to the characteristic equation, which is used to find the eigenvalues of a quadratic matrix. The eigenvalues are the solutions to the characteristic equation.

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