Given a vector $a$, is there a matrix $A$ such that $Ab = a \times b$?

Question

If we let $$a = \begin{pmatrix} a_1 \\ a_2\\ a_3\\ \end{pmatrix}$$

Is there a matrix $A$ such that $Ab = a \times b$ for all $b\in\mathbb{R}^3$?

Answer 1

If you want to know whether there is or not, the function $X \mapsto a\times X$ is linear application from $\mathbb{R}^3$ into $\mathbb{R}^3.$ Then it can be represented by a matrix $A \in \mathcal{M}_{3\times 3}(\mathbb{R}).$ To compute $A,$ let $b = (x,y,z)^t.$ Then we have

$$ (a_1,b_2,a_3)^t \times (x,y,z)^t = \left( \begin{matrix} a_2z-a_3y\\ -a_1z+a_3x \\ a_1y -a_2x\end{matrix}\right) = \left( \begin{matrix} 0&-a_3& a_2\\ a_3&0&-a_1\\ -a_2&a_1&0\end{matrix}\right) b$$

Answer 2

Using the Levi-Civita ($\epsilon$) tensor you can write the cross product as $$\eqalign{ c &= a\times b \cr c_i &= (\epsilon_{ijk} a_j) b_k = A_{ik} b_k \cr }$$ So we just need a matrix whose components are given by $$\eqalign{ A_{ik} &= \epsilon_{ijk} a_j \cr }$$ For example, the matrix elements corresponding to $a_1$ are $$\eqalign{ A_{23} &= \epsilon_{213}\,a_1 = -a_1 \cr A_{32} &= \epsilon_{312}\,a_1 = +a_1 \cr }$$ since $\epsilon_{ijk}$ equals $+1$ for an even permutation of its indices, $-1$ for an odd permutation, and $0$ when any index is repeated.

You can also use the Levi-Civita tensor to reverse the process, i.e. given a skew-symmetric matrix $A$ find the vector $a$ from which it was constructed $$\eqalign{ a_i &= -\frac{1}{2}\epsilon_{ijk}\,A_{jk}\cr }$$