Given two vectors $A\in\mathbb{R}^{3}$ and $B\in\mathbb{R}^{3}$, does it hold that $A\cdot B = (-A)\cdot(-B)$ and $A\times B = (-A)\times(-B)?$ Following a space inversion, vector $\bf{A}$ goes to $\bf{A}^{'} = -\bf{A}$ and $\bf{B}$ to $\bf{B}^{'} = -\bf{B}$. How does $\bf{A}^{'} \cdot \bf{B}^{'}$ compare to $\bf{A} \cdot \bf{B}$ and $\bf{A}^{'}\times\bf{B}^{'}$ to $\bf{A} \times \bf{B}$?
My confusion in this question comes from not knowing how to express each operation for comparison. I suspect that because the cross-product and dot-product are linear operators, then each operation should yield equivalent results, but I am not entirely sure how to express this.
Any guidance is greatly appreciated.
 A: HINT
Suppose that $A = (a_{1},a_{2},a_{3})\in\mathbb{R}^{3}$ and $B = (b_
{1},b_{2},b_{3})\in\mathbb{R}^{3}$.
On the one hand, the dot product between $A = (a_{1},a_{2},a_{3})$ and $B = (b_{1},b_{2},b_{3})$ is calculated as:
\begin{align*}
A\cdot B = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}
\end{align*}
On the other hand, the cross product between $A$ and $B$ is calculated as follows:
\begin{align*}
A\times B =
\begin{vmatrix}
e_{1} & e_{2} & e_{3}\\
a_{1} & a_{2} & a_{3}\\
b_{1} & b_{2} & b_{3}
\end{vmatrix}
\end{align*}
where $e_{1} = (1,0,0)$, $e_{2} = (0,1,0)$ and $e_{3} = (0,0,1)$.
Can you take it from here?
A: In addition to @AtilaCorreia's good answer, ... you are certainly correct that the linearity in each argument is sufficient (without any further computations) to verify that
$$
(-A)\cdot (-B) \;=\; (-1)(-1) A\cdot B \;=\; A\cdot B
$$
and
$$
(-A)\times (-B) \;=\; (-1)(-1) A\times B \;=\; A\times B
$$
I think it's worthwhile to think about the fact that the property of these vector products, rather than a formulaic description, gives a more economical/conceptual way to understand further properties. I have to confess, when I was younger, I thought "formulas" were all we wanted, but, by now, I think "properties" are more relevant. :)
