If I have the conic $C$: $$ 5x^2 - 4xy + 8y^2 = 36 $$ How would I express it as a matrix of the form:
$$ \begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}a&b/2\\b/2&c\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = k $$ Also, how to find a unitary matrix $P$ such that $$ P^* \begin{bmatrix}a&b/2\\b/2&c\end{bmatrix}P $$ is diagonal.
Thanks for your help.
Let $$ A\equiv\left[\begin{array}{cc} a & \frac{1}{2}b\\ \frac{1}{2}b & c \end{array}\right] $$ so that \begin{align*} \left[\begin{array}{cc} x & y\end{array}\right]A\left[\begin{array}{c} x\\ y \end{array}\right] & =\left[\begin{array}{cc} x & y\end{array}\right]\left[\begin{array}{cc} a & \frac{1}{2}b\\ \frac{1}{2}b & c \end{array}\right]\left[\begin{array}{c} x\\ y \end{array}\right]\\ & =\left[\begin{array}{cc} x & y\end{array}\right]\left[\begin{array}{c} ax+\frac{1}{2}by\\ \frac{1}{2}bx+cy \end{array}\right]\\ & =ax^{2}+\frac{1}{2}byx+\frac{1}{2}bxy+cy^{2}\\ & =ax^{2}+bxy+cy^{2}. \end{align*} Therefore,$a=5$, $b=-4$ and $c=8$. Naturally, $k=36$. This gives us $$ A\equiv\left[\begin{array}{cc} a & \frac{1}{2}b\\ \frac{1}{2}b & c \end{array}\right]=\left[\begin{array}{cc} 5 & -2\\ -2 & 8 \end{array}\right]. $$ We want to diagonalize $A$. Note that the eigenvalues of $A$ are $\lambda_{1}=4$ and $\lambda_{2}=9$ (you can verify this). Constructing $P$ by placing the corresponding eigenvectors in the columns, $$ P\approx\left[\begin{array}{cc} -0.89443 & -0.44721\\ -0.44721 & 0.89443 \end{array}\right]. $$ You can verify that $$ P^{\star}AP\approx\left[\begin{array}{cc} \lambda_{1}\\ & \lambda_{2} \end{array}\right]. $$