I need help with a simple proof for the associative law of scalar
multiplication of a vectors. If
$$(rs)X =r (sX)$$
Define the elements belonging to $\mathbb{R}^2$ as $\{(a,b)|a,b\in\mathbb{R}\}$. Combining elements within this set under the operations of vector addition and scalar multiplication should use the following notation:
Vector Addition Example:
$$(–2,10)+(–5,0)=(–2–5,10+0)=(–7,10)$$
Scalar Multiplication Example:
$$–10×(1,–7)=(–10×1,–10×–7)=(–10,70)$$
where –10 is a scalar.
Best Answer
From your question, it appears you are only interested in $\mathbb{R}^2$, but in case not, we'll do the proof over $\mathbb{R}^n$. (If you only want $\mathbb{R}^2$, then set $n=2$ in what follows, or replace $(x_1, x_2, \ldots, x_n)$ by $(x,y)$.) (Of course, this law holds much more generally, but to keep things concrete we'll just be concerned with real numbers and $\mathbb{R}^n$.)
Let $X = (x_1, x_2, \ldots, x_n)$ be a vector, $r,s$ scalars. Then, \begin{align*} (rs)X &= (rs)(x_1, \ldots, x_n)\\ &= ((rs)x_1, (rs)x_2, \ldots, (rs)x_n) & (\text{Def. of scalar mult.})\\ &= (r(sx_1), r(sx_2), \ldots, r(sx_n)) & (\text{Assoc. law in } \mathbb{R})\\ &= r (sx_1, sx_2, \ldots, sx_n) & (\text{Def. of scalar mult. by } r) \\ &= r(s(x_1, x_2, \ldots, x_n) & (\text{Def. of scalar mult. by } s) \\ &= r(sX) & (\text{substituting in our def. of } X) \end{align*}
The key step (and really the only one that is not from the definition of scalar multiplication) is once you have $((rs)x_1, \ldots, (rs)x_n)$ you realize that each element $(rs)x_i$ is a product of three real numbers. Since you have the associative law in $\mathbb{R}$ you can use that to write $$ (rs)x_i = r(sx_i). $$