When does the bilinear form $x^TAy,$ with $x,y$ are non-negative vectors, vanish?

Question

I have a map $F:\mathbb{R}^2\to\mathbb{R}^2$ and it's Jacobian turned out to be of the form: \begin{bmatrix} a^Tx & b^Tx\\ a^Ty & b^Ty \end{bmatrix}, with $x,y,a,b\in\mathbb{R^n}$ ($n=4$ to be specific) and $x,y$ have non-negative entries, and $a,b$ are smooth vector functions of two variables .

So its determinant is $|J| = x^Tab^Ty - x^Tba^Ty = x^T(ab^T - ba^T)y = x^TAy$, with $A = ab^T - ba^T$ is a skew-symmetric matrix. Is there any criteria on $A$ that ensures the non-zeroness of this bilinear form $B(x,y) = x^TAy $ ?

My google search reveals that any skew-symmetric bilinear form like this must have even rank but I am not sure if this can be useful.

Answer 1

Your function $R(x,y)$ is called a bilinear form, and there's a lot of literature on them. I think the most interesting thing about your matrix is the fact that it is rank 2. The zero set for $y$ will be the set $(a^T x)(b^T y) = (b^T x)(a^T y)$. Now, $(b^T y)$ and $(b^T x)$ are just scalars, so we can divide by them and get $\frac{a^T x}{b^T x}=\frac{a^T y}{b^T y}$.

There is another way to think about this. You can do a coordinate tranformation $\Phi:\mathbb{R}^n \to \mathbb{R}^n$ such that the first element of $\Phi(z)$ is $a^T z$ and the second element is $b^T z$, and take $u:=\Phi(x)$ and $v:=\Phi(y)$. In these coordinates, the zero set is the set $u_1v_2-u_2v_1=0$. This is the third coordinate of the cross product $\left(\begin{array}{c}u_1\\u_2\\0\end{array}\right)\times \left(\begin{array}{c}v_1\\v_2\\0\end{array}\right)$. When it is zero, the first two coordinates of $u$ and $v$ are on the same line through 0. All the other coordinates can be arbitrary.

Answer 2

The expression can be simplified as $\langle a,x\rangle \langle b,y \rangle - \langle a,y \rangle \langle b,x\rangle$ or $2 x^T (a \wedge b) y$, where $a \wedge b = \frac{1}{2}(ab^T - ba^T) $ is the exterior product. It can be proven that $2 x^T (a \wedge b) y = - 2 y^T (a \wedge b) x$.

Another form this can take is: $2 \langle x\wedge y , \;a \wedge b \rangle$ or $\frac{1}{2}\text{Tr}((ab^T - ba^T)(yx^T - xy^T))$. The object $(x \wedge y)$ can be interpreted as antisymetric matrix or as a $2$-form, which roughly associates with the plane spanned by $x$ and $y$ the value of the area of the parallelogram formed by $x$ and $y$.This suggests a geometric interpretation to the problem: $2 \langle x\wedge y , \;a \wedge b \rangle$ vanishes when:

1. $x$ and $y$ are collinear or
2. $a$ and $b$ are collinear or
3. The (orthogonal) projection of the plane/subspace spanned by $x,y$ onto the plane/subspace spanned by $a,b$ is not surjective. Another way to state this is the plane spanned by $x,y$ contains a non-zero vector that is orthogonal to the plane spanned by $a,b$. A third way is there exists a non-zero linear combination of $x$ and $y$ that is orthogonal to both $a$ and $b$. A fourth way to state this is the intersection of the subspace spanned by $a,b$ with the orthogonal complement of the subspace spanned by $x,y$ is not $\{0\}$.

The condition at 1 states $x\wedge y = 0$. The condition at 2 states $a\wedge b = 0$. The condition at 3 is a geometric way of saying that $(x\wedge y) \perp (a\wedge b)$ when $(x\wedge y)$ and $(a\wedge b)$ are taken as vectors over the space of $2$-forms.