Diagonalizing a Quadratic Form: x^{2} + 2y^{2} + z^{2} + 2xy + 4xz + 6yz

[T-7]

Homework Statement

x^{2} + 2y^{2} + z^{2} + 2xy + 4xz + 6yz

Write down the symmetric matrix A for which the form is expressible as x^{t}Ax where t denotes transpose. Diagonalise each of the forms and in each case find a real non-singular matrix P for which the matrix P^{t}AP is diagonal with entries in {1,-1,0}.

The Attempt at a Solution

I first tried this by completing the square.

<br /> x^{2} + 2y^{2} + z^{2} + 2xy + 4xz + 6yz<br /> = (x + y + 2z)^{2} + y^{2} - 3z^{2} + 2yz<br /> = (x + y + 2z)^{2} + (y + z)^2 - 4z^2<br /> = x_{1}^{2} + x_{2}^{2} - x_{3}^{2}

where

<br /> x_{1} = x + y + 2z,<br /> x_{2} = y + z,<br /> x_{3} = 2z,<br />

However, I just can't seem to find the eigenvalues for this form.

The symmetric matrix A for this quadratic form is

<br /> \[ \left( \begin{array}{ccc}<br /> 1 &amp; 1 &amp; 2 \\<br /> 1 &amp; 2 &amp; 3 \\<br /> 2 &amp; 3 &amp; 1 \end{array} \right)\] <br />

and the characteristic polynomial is given by

<br /> \[ \chi(\lambda) = \left| \begin{array}{ccc}<br /> 1-\lambda &amp; 1 &amp; 2 \\<br /> 1 &amp; 2-\lambda &amp; 3\\<br /> 2 &amp; 3 &amp; 1-\lambda \end{array} \right|.\] <br />

I find this comes to

f(\lambda) = \lambda^{3} - 4(\lambda^2) + 9(\lambda) - 4

which does not factorise -- so I can't get the eigenvalues, and can't form a matrix P. However, I have shown that a diagonal form is possible by completing the square. So surely I ought to be able to find three eigenvalues? Can someone point out where I've gone wrong?

Cheers!

 

[HallsofIvy]

You've misplaced a couple of signs! The coefficient of \lambda is +4+ 1- 4= 1. The correct equation is \lambda^3- 4\lambda^2- \lambda+ 4= 0 which obviously has \lambda- 1 as a factor.

 

[T-7]

You've misplaced a couple of signs! The coefficient of \lambda is +4+ 1- 4= 1. The correct equation is \lambda^3- 4\lambda^2- \lambda+ 4= 0 which obviously has \lambda- 1 as a factor.

I am obviously brain-dead today. I'm still getting what I got before:

<br /> \[ \chi(\lambda) = \left| \begin{array}{ccc}<br /> 1-\lambda &amp; 1 &amp; 2 \\<br /> 1 &amp; 2-\lambda &amp; 3\\<br /> 2 &amp; 3 &amp; 1-\lambda \end{array} \right|\]<br />

= (1-\lambda)((2-\lambda)(1-\lambda)-9)-(1-\lambda-6)+2(3-2(2-\lambda))
= (1-\lambda)(2-3\lambda+\lambda^{2}-9)-(-\lambda-5)+2(2\lambda-1)
=(1-\lambda)(\lambda^{2}-3\lambda-7)+\lambda+5+4\lambda-2
=(1-\lambda)(\lambda^{2}-3\lambda-7)+5\lambda+3
=\lambda^{2}-3\lambda-7-\lambda^{3}+3\lambda^{2}+7\lambda+5\lambda+3
=-\lambda^{3}+4\lambda^{2}+9\lambda-4

 

[HallsofIvy]

Well, that's not exactly what you had before but I messed up also.

The part about "diagonal with entries in {1,-1,0}" implies that the eigenvalues are 0, 1, and -1. But that can't possibly be correct.