1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\frac12$.)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – MvG Nov 3 '16 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.