\[2x^2-6xy+2y^2=5\]
.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 \[det(A- \lambda I)=0\]
\[\begin{equation} \begin{aligned} det( \left( \begin{array}{cc}2 & -3 \\ -3 & 2 \end{array} \right) -\left( \begin{array}{cc} \lambda & 0 \\ & 0 & \lambda \end{array} \right)) &= det (\left( \begin{array}{cc} 2- \lambda & -3 \\ -3 & 2- \lambda \end{array} \right) ) \\ &=(2- \lambda)^2 -9 \\ &= \lambda^2 -4 \lambda -5=(\lambda -5)(\lambda +1) =0 \end{aligned} \end{equation}\]
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
\[(x', y';)\left( \begin{array}{cc} 5 & 0 \\ 0 & -1 \end{array} \right) \begin{pmatrix}x'\\y'\end{pmatrix}= 5x'^2-y'^2=5\]