If I have a matrix represented by three column vectors like the one below, can I collect the cross product as I did below?
$$ [ \overrightarrow{a} \wedge \overrightarrow{b} \quad \overrightarrow{a} \wedge \overrightarrow{c} \quad \overrightarrow{a} \wedge \overrightarrow{d} ] = \overrightarrow{a} \wedge [ \overrightarrow{b} \quad \overrightarrow{c} \quad \overrightarrow{d} ] $$
Thank you for your time.
I assume that $\wedge$ means $\times$, i.e. cross-product (as in some countries like my one).
My answer : no and yes (!)
The answer, stricly speaking is "no" because on the RHS, you have an inhomogeneous cross product between a vector and a matrix.
But there is a way to say "yes" by rewriting this identity under the form of an ordinary matrix product (emphasized by a "*") which is :
$$[a_{\times}(b) \ , \ a_{\times}(c) \ , \ a_{\times}(d)] \ = \ a_{\times} \ * \ [b \ , \ c \ , \ d]$$
using the operator $v_{\times}$ associated with the cross product by vector v (with coordinates $v_x,v_y,v_z$) defined as the following skew-symmetrical matrix :
$$v_{\times}:=\begin{pmatrix}0&-v_z&v_y\\v_z&0&-v_x\\-v_y&v_x&0\end{pmatrix}.\tag{1}$$
Explanation of form (1). Let us apply this matrix to a vector $w$ with coordinates $w_x, w_y, w_z$ :
$$v_{\times}(w)=\begin{pmatrix}0&-v_z&v_y\\v_z&0&-v_x\\-v_y&v_x&0\end{pmatrix}\begin{pmatrix}w_x\\w_y\\w_z\end{pmatrix}=\begin{pmatrix}v_yw_z-v_zw_y\\ \cdots \\ \cdots\end{pmatrix} = v \times w \tag{2}$$
i.e., the classical formulas for the cross product. By the way, (2) explains the handiness of notation $v_{\times}$.
See for example Skew symmetric matrix of vector
