How to prove that the cross product of two vectors is a linear transformation? I'm stuck, could I please have a hint? 
 A: (tacitly assuming that we are working with the traditional cross product of two vectors in $\mathbb{R}^3$)
Assuming that one of the vectors is held constant.  Let $v$ be the fixed vector.  We wish to prove that $L(x) = v\times x$ is a linear transformation of $x$ from $\mathbb{R}^3$ to $\mathbb{R}^3$.
Let $v = \begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}$.  Let $x = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$.
The cross product $v\times x = \det\left(\begin{bmatrix}\hat{i}&\hat{j}&\hat{k}\\
v_1&v_2&v_3\\x_1&x_2&x_3\end{bmatrix}\right) = \begin{bmatrix}v_2x_3-v_3x_2\\-v_1x_3+v_3x_1\\v_1x_2-v_2x_1\end{bmatrix} = x_1\begin{bmatrix}0\\v_3\\-v_2\end{bmatrix}+x_2\begin{bmatrix}-v_3\\0\\v_1\end{bmatrix}+x_3\begin{bmatrix}\star\\\star\\\star\end{bmatrix}$
Letting you fill in the $\star$'s above yourself.
How can we construct a matrix then such that $Ax = v\times x$?
If we successfully construct such a matrix, then that shows that the transformation is indeed a linear transformation.
A: For a function $L: \mathbb{R}^3 \to \mathbb{R}^3$ to be a linear transformation, you have to verify two properties:


*

*$L(a \overrightarrow{v})=aL(\overrightarrow{v})$ for any scalar $a$ and any vector $\overrightarrow{v}$

*$L(\overrightarrow{v}+\overrightarrow{w})=L(\overrightarrow{v})+L(\overrightarrow{w})$ for any two vectors $\overrightarrow{v}$ and $\overrightarrow{w}$


In this case, fix some vector $\overrightarrow{u}$, and consider the linear transformation defined by $L(\overrightarrow{v})= \overrightarrow{u} \times \overrightarrow{v}$.  To show that the cross-product is linear, you need to show that properties (1) and (2) above hold; in other words, you need to verify that:


*

*$\overrightarrow{u} \times (a\overrightarrow{v})=a ( \overrightarrow{u} \times \overrightarrow{v})$

*$\overrightarrow{u} \times (\overrightarrow{v} + \overrightarrow{w})=(\overrightarrow{u}\times \overrightarrow{v})+(\overrightarrow{u}\times \overrightarrow{w})$


Can you take it from there?
A: To prove that $\vec{a}\times(\vec{b} + \vec{c}) = \vec{a}\times\vec{b} + \vec{a}\times\vec{c}$, we have to make these vectors in Caertesian component using an orthonormal natural base $\{\hat{x}_j ~|~ j = 1, 2, 3\}$. I shall use Einstein Summation Convention, that is $\vec{a} = a_j\hat{x}_j$, $\vec{b} = b_j\hat{x}_j$, and $\vec{c} = c_j\hat{x}_j$, without summation symbol $\sum$.  Now, we get
\begin{align}
& \vec{a}\times(\vec{b} + \vec{c}) \\[6pt]
= {} & a_j\hat{x}_j\times\hat{x}_k(b_k + c_k) \\[6pt]
= {} & a_j(b_k + c_k)\hat{x}_j\times\hat{x}_k \\[6pt]
= {} & (a_jb_k + a_jc_k)\hat{x}_j\times\hat{x}_k \\[6pt]
= {} & a_jb_k\hat{x}_j\times\hat{x}_k + a_jc_k\hat{x}_j\times\hat{x}_k \\[6pt]
= {} & (a_j\hat{x}_j)\times(b_k\hat{x}_k) + (a_j\hat{x}_j)\times(c_k\hat{x}_k) \\[6pt]
= {} & \vec{a}\times\vec{b} + \vec{a}\times\vec{c}.
\end{align}
A: Cross product between vector $\vec{A}$ dan $\vec{B}$ is defined by
\begin{equation}
\vec{A}\times\vec{B} := |\vec{A}||\vec{B}|\sin\angle(\vec{A},\vec{B})\frac{\vec{A}\times\vec{B}}{|\vec{A}\times\vec{B}|}.
\end{equation}
Because for $\vec{A} := (A_x,A_y,A_z) \in \mathbb{R}^3$ and $\vec{B} := (B_x,B_y,B_z) \in \mathbb{R}^3$, it yields
\begin{eqnarray}
\sin\angle(\vec{A},\vec{B}) &=& \sqrt{1 - \cos^2\angle(\vec{A},\vec{B})} 
= \sqrt{1 - \left(\frac{\vec{A}\cdot\vec{B}}{|\vec{A}||\vec{B}|}\right)^2} 
= \frac{\sqrt{|\vec{A}|^2|\vec{B}|^2 - (\vec{A}\cdot\vec{B})^2}}{|\vec{A}||\vec{B}|} \nonumber\\
&=& \frac{\sqrt{(A_x^2 + A_y^2 + A_z^2)(B_x^2 + B_y^2 + B_z^2) - (A_xB_x + A_yB_y + A_zB_z)^2}}{|\vec{A}||\vec{B}|} \nonumber\\
&=& \frac{\sqrt{(A_yB_z - A_zB_y)^2 + (A_zB_x - A_xB_z)^2 + (A_xB_y - A_yB_x)^2}}{|\vec{A}||\vec{B}|},
\end{eqnarray}
so
\begin{equation}
\vec{A}\times\vec{B} = \sqrt{(A_yB_z - A_zB_y)^2 + (A_zB_x - A_xB_z)^2 + (A_xB_y - A_yB_x)^2}\frac{\vec{A}\times\vec{B}}{|\vec{A}\times\vec{B}|},
\end{equation}
so that
\begin{equation}
|\vec{A}\times\vec{B}| = \sqrt{(A_yB_z - A_zB_y)^2 + (A_zB_x - A_xB_z)^2 + (A_xB_y - A_yB_x)^2}.
\label{201810211741}
\end{equation}
There is many possibility of explicit form of $\vec{A}\times\vec{B}$ which satisfy equation (\ref{201810211741}).
Because $\hat{x}\times\hat{y} = \hat{z}$, $\hat{z}\times\hat{x} = \hat{y}$, and $\hat{y}\times\hat{z} = \hat{x}$ must be satisfied, so it must be
\begin{equation}
\vec{A}\times\vec{B} = (A_yB_z - A_zB_y)\hat{x} + (A_zB_x - A_xB_z)\hat{y} + (A_xB_y - A_yB_x)\hat{z}.
\end{equation}
