# I need help with a simple proof for the associative law of scalar multiplication of a vectors.

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.

• How are you defining vectors and scalar multiplication? What are you allowed to assume in the proof? Jul 23, 2015 at 15:08
• The proof depend from the vector space in which you are working, and the definition of the scalar multiplication in this space. So, what is your vector space? Jul 23, 2015 at 15:10
• ector spaces possess a collection of specific characteristics and properties. Use the definitions in the attached “Definitions” to complete this task. Define the elements belonging to R2 as {(a,b)|a,b∈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. Jul 23, 2015 at 15:23

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 $$(rs)x_i = r(sx_i).$$