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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.

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  • $\begingroup$ How are you defining vectors and scalar multiplication? What are you allowed to assume in the proof? $\endgroup$ Jul 23, 2015 at 15:08
  • $\begingroup$ 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? $\endgroup$ Jul 23, 2015 at 15:10
  • $\begingroup$ 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. $\endgroup$ Jul 23, 2015 at 15:23

1 Answer 1

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

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