## reduction of quadratic form (principal axis)

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
$$\left(\begin{array}{ccc} x_{1} & x_{2} & x_{3} \end{array}\right) \left(\begin{array}{ccc} 2 & 1 & 0 \cr 1 & 2 & 1 \cr 0 & 1 & 1 \end{array} \right) \left(\begin{array}{c} x_{1} \cr x_{2} \cr x_{3} \end{array} \right)$$
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.

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 Recognitions: Science Advisor 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.

 Quote by 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.
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.

Recognitions:
 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.