## Writing a Conic in Canonical Form

Given the equation of a conic of the form
$Ax_1^2+Bx_1x_2 +Cx^2_2=k$
we can write in matrix form as
$(x_1.x_2)\left( \begin{array}{cc} A & B/2 \\ B/2 & C \end{array} \right)\begin{pmatrix}x_1\\x_2\end{pmatrix}=k$

We can find the eigenvalues and eigenvectors of the matrix above and express the conic in the canonical form
$A'y_1^2+B'y_2^2=C'$
.
Example: Write the conic
$5x_1^2+4x_1x_2+8x_2^2=9$
in canonical form.
We find the eigenvalues and eigenvectors of the matrix
$A=\left( \begin{array}{cc} 5 & 4/2 \\ 4/2 & 8 \end{array} \right)=\left( \begin{array}{cc} 5 & 2 \\ 2 & 8 \end{array} \right)$

\begin{aligned} \left| \begin{array}{cc} 5- \lambda & 2 \\ 2 & 8- \lambda \end{array} \right| & =(5-- \lambda)(8- \lambda )-2^2 \\ &=\lambda^2-13 \lambda-36 \\ & =(\lambda-4)(\lambda-9)=0 \end{aligned}

Hence
$\lambda=4, \: 9$

$\lambda=4$
:
\begin{aligned}\left( \begin{array}{cc} 5-4 & 2 \\ 2 & 8-4 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix}&=\left( \begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix}\\ &=\begin{pmatrix}x_1+2x_2\\2x_1+4x_2\end{pmatrix} \\ &= \begin{pmatrix}0\\0\end{pmatrix} \end{aligned}

We can take
$x_1=-2, \: x_2=1$

Normalising gives
$x_1=-\frac{2}{\sqrt{5}}, \: x_2=\frac{1}{\sqrt{5}}$

The first eigenvector is
$\begin{pmatrix}-\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\end{pmatrix}$

$\lambda=9$
:
\begin{aligned} \left( \begin{array}{cc} 5-9 & 2 \\ 2 & 8-9 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix} &= \left( \begin{array}{cc} -4 & 2 \\ 2 & -1 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix} \\ &= \begin{pmatrix}-4x_1+2x_2\\2x_1-x_2\end{pmatrix} \\ &= \begin{pmatrix}0\\0\end{pmatrix}\end{aligned}

We can take
$x_1=1, \: x_2=2$

Normalising gives
$x_1=-\frac{1}{\sqrt{5}}, \: x_2=\frac{2}{\sqrt{5}}$

The second eigenvector is
$\begin{pmatrix}\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\end{pmatrix}$

The matrix of eigenvectors is
$\left( \begin{array}{cc} -\frac{2}{\sqrt{5}} & -\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{array} \right)$

Now define the transformation
$\mathbf{x}=P \mathbf{y}$
then
$\mathbf{x}^TA \mathbf{x}=9$
becomes
$(P \mathbf{y})^TA(P \mathbf{x})=\mathbf{y}^TP^TAP=\mathbf{y}D\mathbf{y}=9$

We have
$(y_1.y_2)\left( \begin{array}{cc} 4 & 0 \\ 0 & 9 \end{array} \right)\begin{pmatrix}y_1\\y_2\end{pmatrix}=y_1^2+9y_2^2=9$