# Moving quadratic vector out of determinant

In $\mathbb R^3$ I've got the equation

$$\det(A\cdot v, v, w) = 0$$

where $v$ and $w$ are vectors and $A$ is a matrix. So I'm composing three vectors to form a $3\times3$ matrix and then taking its determinant. I'd like to rewrite this as a simple quadratic equation in $v$, i.e. some equation

$$v^T\cdot B\cdot v=0$$

where $B$ is a symmetric matrix which depends on $A$ and $w$. I know I can do this at the coordinate level: $B=$

$$\scriptsize\begin{pmatrix} 2(A_{31}w_2 - A_{21}w_3) & -A_{31}w_1 + A_{32}w_2 + (A_{11} - A_{22})w_3 & A_{21}w_1 - (A_{11} - A_{33})w_2 - A_{23}w_3 \\ -A_{31}w_1 + A_{32}w_2 + (A_{11} - A_{22})w_3 & 2(-A_{32}w_1 + A_{12}w_3) & (A_{22} - A_{33})w_1 - A_{12}w_2 + A{13}w_3 \\ A_{21}w_1 - (A_{11} - A_{33})w_2 - A_{23}w_3 & (A_{22} - A_{33})w_1 - A_{12}w_2 + A{13}w_3 & 2(A_{23}w_1 - A_{13}w_2) \end{pmatrix}$$

But it feels as though there should be a more elegant way of expressing this. Can you come up with one?

For background: I'm trying to compute the orthogonal projections of a point onto a conic section, in a setup of projective geometry. So $w$ would be the given point, $v$ would be its projection, and $A\cdot v$ would be the point at infinity in the direction orthogonal to the conic in the point $v$. The determinant would be zero if the points are collinear. $A$ can be computed easily from the matrix of the conic, but I need $B$ to reduce the whole problem to conic-conic intersection in my formalism.

The determinant is equal to the triple product, $$\det(Av,v,w) = \det(w,Av,v)=(w \times Av) \cdot v$$ (with dot denoting the scalar product), and in brackets you have the composition of the linear transformation $A$ and the linear transformation “cross product by $w$ from the left” (call it $L$), with the matrix $$L = \begin{pmatrix} 0 & -w_3 & w_2 \\ w_3 & 0 & -w_1 \\ -w_2 & w_1 & 0 \end{pmatrix} .$$ So the expression becomes $$(L A v) \cdot v = (L A v)^t v = v^t (L A)^t v ,$$ which implies that $B$ is the symmetric part of $(L A)^t$ (since only the symmetric part of $X$ contributes to the quadratic form $v^t X v$).
That is, $$B = \frac{(LA)^t + LA}{2} = \frac{A^t L^t + LA}{2} = \frac{LA-A^t L}{2} .$$ (I think your $B$ is missing a factor $\frac12$.)
• I scaled my whole equation by a factor $2$ since all I'm interested is the case where it equals zero. Thanks, this is a very clear explanation which integrates nicely with what I have in place. – MvG Nov 3 '16 at 6:58