## Standard Form of a Quadratic Curve

Suppose we have the quadratic curve
$2x^2-6xy+2y^2=5$
.
We can write this curve in matrix form, using
$ax^2+bx+c \rightarrow (x,y) \left( \begin{array}{cc} a & b/2 \\ b/2 & c & \end{array} \right) \begin{pmatrix}x\\y\end{pmatrix}$
as
$(x,y) \left( \begin{array}{cc} 2 & -3 \\ -3 & 2 & \end{array} \right) \begin{pmatrix}x\\y\end{pmatrix}$
(1)
Find the eigenvalues and eigenvectors of the above matrix.
The eigenvalues of
$A$
are the solutions to
$det(A- \lambda I)=0$

\begin{aligned} det( \left( \begin{array}{cc}2 & -3 \\ -3 & 2 \end{array} \right) -\left( \begin{array}{cc} \lambda & 0 \\ & 0 & \lambda \end{array} \right)) &= det (\left( \begin{array}{cc} 2- \lambda & -3 \\ -3 & 2- \lambda \end{array} \right) ) \\ &=(2- \lambda)^2 -9 \\ &= \lambda^2 -4 \lambda -5=(\lambda -5)(\lambda +1) =0 \end{aligned}

Hence
$\lambda = 5, \: -1$
.
If
$\lambda =5, \: (A- \lambda I) \mathbf{v}_1 = \left( \begin{array}{cc} -3 & -3 \\ -3 & -3 \end{array} \right) \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}-3x-3y\\-3x-3y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$

Hence
$x=1, \: y=-1 \rightarrow \mathbf{v}_1 = \begin{pmatrix}1\\-1\end{pmatrix}$

If
$\lambda =-1, \: (A- \lambda I) \mathbf{v}_1 = \left( \begin{array}{cc} 3 & -3 \\ -3 & 3 \end{array} \right) \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}3x-3y\\-3x+3y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$

Hence
$x = y=-1., \: \rightarrow \mathbf{v}_1 = \begin{pmatrix}1\\1\end{pmatrix}$

We can take
$P$
as the matrix with columns
$v_1 , v_2$
.
$P= \left( \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right)$

Then
$P^{-1}= \frac{1}{2} \left( \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right)$

Then
$D= \frac{1}{2} \left( \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right) \left( \begin{array}{cc}2 & -3 \\ -3 & 2 \end{array} \right) \left( \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right)= \frac{1}{2} \left( \begin{array}{cc} 5 & 0 \\ 0 & -1 \end{array} \right)$

Put
$\mathbf{x}'=P \mathbf{x}$
then (1) becomes
$(Px')^T P^{-1}DP \mathbf{x} =5$

$P=P^T$
so this simplifies to
$(\mathbf{x}')^T D \mathbf{x}'=5$
.
The quadratic form is
$(x', y';)\left( \begin{array}{cc} 5 & 0 \\ 0 & -1 \end{array} \right) \begin{pmatrix}x'\\y'\end{pmatrix}= 5x'^2-y'^2=5$

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