Lines as Hermitian matrices

Now let's see if lines fit this pattern as well. A line is given as the set of solutions $x+iy$ where $$ax+by=d$$ for some real constants $a$, $b$ and $d$, where not both of $a,b$ are 0. Using complex conjugation, we have $x=(z+\bar{z})/2$ and $y=(z-\bar{z})/(2i)$. Then if we substitute these formulas into the equation for a line we get $$a (z+\bar{z}) -ib (z-\bar{z})=2d$$ or $$(a-ib) z + (a+ib)\bar{z} -2d = 0.$$ If we set $c=a+ib$, we can again realize this equation in matrix form as $$0 = \begin{bmatrix}\bar{z} & 1\end{bmatrix} \begin{bmatrix}0... ... \bar{c} & -2d \end{bmatrix}\begin{bmatrix}z\\ 1\end{bmatrix}.$$ The $2\times 2$ matrix in the middle is again Hermitian. The determinant is $-c\bar{c} =-\vert c\vert^2$ is again negative (remember that we assumed $c=a+ib\neq 0$).

Thus, circles and lines are both zero sets of Hermitian $2\times 2$ matrices with negative determinant. The feature that distinguishes circles from lines is that the upper left entry is 0 for lines.