The task is to convert a quadric equation to a normal form.
$Q \equiv 3x_1^2 + 3x_2^2+3x_3^2+ 2x_1x_2+ 2x_1x_3 + 2x_2x_3+4x_1+4x_2+2x_3 +1=0$
What I've done is the following:
- Create a matrix A from the variables with degree 1 and 2
$$\begin{bmatrix}
x_1 & x_2 & x_3\\
3 & 1 & 1 & x_1\\
1 & 3 & 1& x_2\\
1 & 1 & 3& x_3\\
\end{bmatrix}$$
-
Determine the determinant of $A- \lambda I_3 = -\lambda^3 +9\lambda^2 -24\lambda+20= (\lambda-2)^2(-\lambda + 5) $
-
Determine the eigenvalues of $A- \lambda I_3$
$E_2= \operatorname{vect}\begin{pmatrix}
1\\
1\\
1\\
\end{pmatrix}$
$E_5= \operatorname{vect}\begin{pmatrix}
1\\
1\\
1\\
\end{pmatrix}$
but $\dim(E_2(A))+\dim(E_5(A)) \ne 3$. So a symmetrical matrix can't be constructed. Does this imply that there is no normal form? Can you maybe see if I made a mistake?
Best Answer
You are wrong about $E_2$ and $E_5$:$$E_5=\left\langle\begin{pmatrix}1\\1\\1\end{pmatrix}\right\rangle$$and$$E_2=\left\langle\begin{pmatrix}-1\\0\\1\end{pmatrix},\begin{pmatrix}-1\\1\\0\end{pmatrix}\right\rangle.$$So, $\dim E_5=1$ and $\dim E_2=2$.