Reduction of quadratic form (principal axis)

[xman]

i keep getting nonzero off diagonal elements when i try to reduce to simple sum of squares, of the equation
$$2 x_{1}^{2}+2x_{2}^{2}+x_{3}^{2}+2x_{1}x_{3}+2x_{2}x_{3}$$
what i have is
$$\left(\begin{array}{ccc} x_{1} & x_{2} & x_{3} \end{array}\right)<br /> \left(\begin{array}{ccc}<br /> 2 & 1 & 0 \cr<br /> 1 & 2 & 1 \cr<br /> 0 & 1 & 1 <br /> \end{array} \right) <br /> \left(\begin{array}{c} x_{1} \cr x_{2} \cr x_{3} \end{array} \right)$$
so my thought was to calculate the eigenvalues of the coefficient matrix above, which yield complex solutions from the characteristic equation
$$1-6 \lambda+5 \lambda^{2}-\lambda^{3}=0$$
From the complex eigenvalues I obtain complex eigenvectors, which i'll post if necessary, but are rather lengthy. From the eigenvectors I choose to use Gram-Schmidt orthogonalization to form an orthonormal basis set. From which I construct a matrix with the corresponding basis set, and use diagonalize the system I have the diagonalization matrix
$$D = \left(\mid n \rangle \langle m \mid \right)^{T} A \left( \mid n \rangle \langle m \mid \right)$$
where the matrix
$$\left(\mid n \rangle \langle m \mid \right)$$
is the orthonormal eigenvector matrix. When I'm done with all of this I'm not getting a diagonalized matrix. I was wondering if I am making a mistake in my approach, or if anyone else does get a diagonalized matrix equation.

 

[0rthodontist]

The equation 1-6x+5x^2-x^3=0 has all real solutions. You could have known this because your matrix is symmetric and symmetric matrices in the reals are orthogonally diagonalizable in the reals.

 

[xman]

The equation 1-6x+5x^2-x^3=0 has all real solutions. You could have known this because your matrix is symmetric and symmetric matrices in the reals are orthogonally diagonalizable in the reals.

That's interesting, I was bad and usinig Maxima to calculate the roots of the equation and getting imaginary components, interesting, when I plot and find the roots, you're correct the roots are all real. Thanks, for the tip, I didn't know that about symmetric matrices, either. Thanks again.

 

[0rthodontist]

That's interesting, I was bad and usinig Maxima to calculate the roots of the equation

I was using my TI-89.

 

[xman]

Perhaps the -89 is superior to my cas, even mathematica, maple given imaginary components, though on order of $$10^{-16}$$ or so. I wonder why that is. Not to mention I'm glad I'm not the only one whose cheats on algebra parts of problems...well that may get you into trouble as I found out today. Hey thanks again.