Quadratic form and a matrix of a surface in $\mathbb{R}^2$

Question

If there is a surface in space $\mathbb{R}^2$ given by equation: $$2x^2+2y^2-5z^2+2xy-2x-4y-4z+2=0$$ what is its quadratic form and corresponding matrix?

If the quadratic form is $2x^2+2xy+2y^2-5z^2$ and the corresponding matrix is: $$A=\begin{pmatrix} 2 & 1 & 0\\ 1 & 2 & 0\\ 0 & 0 & -5 \end{pmatrix}$$

But this doesn't make much sense considering that the surface is in $\mathbb{R}^2$.
What would be the correct approach here? Is it possible that $z$ is just some variable from $\mathbb{R}$ not related to the quadratic form itself?

Answer 1

The $\mathbb R^2$ must be a typo and should surely read $\mathbb R^3$. Assuming that, I'll write $\mathbf x =(x,y,z)^T$ so that the equation is $f(\mathbf x)=0$ where $f(\mathbf x) = 2x^2+2xy+2z^2-5z^2-2x-4y-4z+2$. Normally if you want to understand the surface defined by an equation of this sort, there are essentially two steps:

1. First, one changes variables to $\mathbf x_1$ where $\mathbf x = \mathbf x_1+\mathbf d$ and $\mathbf d= (a,b,c)$, so that $f(\mathbf x) =f(\mathbf x_1+\mathbf d)$ has no terms which are linear in $\mathbf x_1$. That is, find $(a,b,c)$ such that

$$ \begin{split} f(x_1+a,y_1+b,z_1+c) &= 2(x_1+a)^2 +2(y_1+b)^2-5(z_1+c)^2 +2(x_1+a)(y_1+b) \\ &\quad -2(x_1+a)-4(y_1+b)-4(z_1+c)+2 \end{split} $$ $\quad$ becomes $f(x_1,y_1,z_1) = 2x_1^2+2x_1y_1+2y_1^2-5z_1^2+\alpha$ for some constant $\alpha\in \mathbb R$.

2. Now having found such $(a,b,c)$, we may write $f(\mathbf x_1)= \mathbf x_1^T C \mathbf x_1 + \alpha$ where $$ C=\left(\begin{array}{ccc}2 & 1 & 0\\ 1 & 2 & 0 \\ 0 & 0& -5\end{array}\right). $$ Now $C = \left(\begin{array}{cc} C_1 & 0 \\ 0 & -5 \end{array}\right)$ where $C_1 =\left(\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right)$, and since $C_1$ is symmetric, there is a rotation matrix $A_1 = \left(\begin{array}{cc}\cos(t) &-\sin(t)\\ \sin(t) & \cos(t)\end{array}\right)$ so that if $A = \left(\begin{array}{cc} A_1 & 0 \\ 0 & 1 \end{array}\right) $ then $$ \begin{split} AQA^T &= \left(\begin{array}{cc} A_1 & 0 \\ 0 & 1 \end{array}\right) \left(\begin{array}{cc} C_1 & 0 \\ 0 & -5\end{array}\right)\left(\begin{array}{cc} A_1^T & 0 \\ 0 & 1 \end{array}\right) \\ &= \left(\begin{array}{cc} A_1^TC_1A_1 & 0 \\ 0 & -5 \end{array}\right) %= %\left(\begin{array}{ccc}d_1 & 0 & 0\\ 0 & d_2 & 0 \\ 0 & 0 & -5 %\end{array}\right) \end{split} $$ where $A_1^TC_1A_1= \left(\begin{array}{cc} d_1 & 0 \\ 0 & d_2 \end{array}\right)$ is diagonal. But then if $\mathbf x_2 = A\mathbf x_1$, the equation $f(\mathbf x_1)=0$ becomes $f(\mathbf x_2) = d_1x_2^2+d_2y_2^2-5z_2^2+\alpha=0$. Thus by finding $\mathbf d$ and $A_1$, you can determine $d_1,d_2$ and $\alpha$, and hence determine what kind of quadric surface the original equation $f(x,y,z)=0$ describes.

Answer 2

A quadric hypersurface of dimension $D$ lies in a $(D+1)$-dimensional space. We can extend the usual matrix equation of a conic section to obtain the matrix-vector equation of the hypersurface. The matrix will be $(D+2)\times(D+2).$

The equation $$2x^2+2y^2-5z^2+2xy-2x-4y-4z+2=0$$ defines a $2$-dimensional quadric surface in the $3$-dimensional space. The matrix-vector equation can be written as

$$(x,y,z,1) \begin{pmatrix} \;\;2 & \;\;1 &\;\;0 & -1\\ \;\;1 & \;\;2 & \;\;0 & -2\\ \;\;0 & \;\;0 & -5 & -2\\ -1 & -2 & -2 & \;\; 2 \end{pmatrix} \begin{pmatrix} x\\y\\z\\1\end{pmatrix} =0$$

EDIT
We obtain the quadratic form when we replace $1$s in the vectors by a fourth variable, say $w,$ and expand.