Prove some properties of vector algebra in Rn

Prove that for all vectors in $\mathbb{R}^n$ we have commutativity of vector addition,
$A+ B = B + A$

associativity of vector addition,

$A + (B + C) = (A+B) + C$

associativity of scalar multiplication,

$c(dA) = (cd)A$

and distributivity,

$c(A+B) = cA + cB, \qquad \text{and} \qquad (c+d)A = cA + dA.$
Illustrate the geometric meaning of the distributive laws
$(c+d)A = cA + dA \qquad \text{and} \qquad c(A+B) = cA+ cB.$

Proof. Let $A = (a_1, \ldots, a_n)$ , $B = (b_1, \ldots, b_n)$ , and $C = (c_1, \ldots, c_n)$ be vectors in $\mathbb{R}^n$ and let $c,d \in \mathbb{R}$ be scalars. Then by commutativity of addition in $\mathbb{R}$ we have
$\begin{align*} A + B &= (a_1, \ldots, a_n) + (b_1, \ldots, b_n) \\ &= (a_1 + b_1, \ldots, a_n +b_n) = (b_1 + a_1, \ldots, b_n +a_n) \\ &= (b_1, \ldots, b_n) + (a_1, \ldots, a_n) \\ &= B+A. \end{align*}$

By associativity of addition in $\mathbb{R}$ we have

$\begin{align*} A+(B+C) &= (a_1, \ldots, a_n) + \big( (b_1, \ldots, b_n) + (c_1, \ldots, c_n) \big) \\ &= (a_1, \ldots, a_n) + (b_1 +c_1, \ldots, b_n + c_n) \\ &= (a_1 + (b_1 + c_1), \ldots, a_n + (b_n + c_n)) \\ &= ((a_1 + b_1) + c_1), \ldots, (a_n + b_n)+ c_n) \\ &= (a_1 + b_1, \ldots, a_n + b_n) + (c_1, \ldots, c_n) \\ &= \big( (a_1, \ldots, a_n)+ (b_1, \ldots, b_n) \big) + (c_1, \ldots, c_n) \\ &= (A+B) + C. \end{align*}$

For associativity of scalar multiplication we have

$\begin{align*} c(dA) &= c(d (a_1, \ldots, a_n)) \\ &= c(da_1, \ldots, d a_n) \\ &= (c(da_1), \ldots, c(da_n)) \\ &= ((cd)a_1, \ldots, (cd) a_n) \\ &= (cd)(a_1, \ldots, a_n) \\ &= (cd)A. \end{align*}$

For the first distributivity law we have

$\begin{align*} c(A+B) &= c \big( (a_1, \ldots, a_n) + (b_1, \ldots, b_n) \big) \\ &= c (a_1 + b_1, \ldots, a_n + b_n ) \\ &= (c(a_1 + b_1), \ldots, c(a_n + b_n)) \\ &= (ca_1 + cb_1, \ldots, ca_n + cb_n) \\ &= (ca_1, \ldots, ca_n) + (cb_1, \ldots, cb_n) \\ &= c(a_1, \ldots, a_n) + c(b_1, \ldots, b_n) \\ &= cA + cB. \end{align*}$

For the second distributivity law we have

$\begin{align*} (c+d)A &= (c+d)(a_1, \ldots, a_n) \\ &= \big( (c+d)a_1, \ldots, (c+d)a_n \big) \\ &= ( ca_1 + da_1, \ldots, ca_n + da_n ) \\ &= (ca_1, \ldots, ca_n) + (da_1, \ldots, da_n) \\ &= c(a_1, \ldots, a_n) + d(a_1, \ldots, a_n) \\ &= cA + dA. \qquad \blacksquare \end{align*}$
The distributive law $(c+d)A = cA+dA$ means the vector $(c+d)A$ is obtained by adding the arrow $dA$ to the end of the arrow $cA$ .
The distributive law $c(A+B) = cA+cB$ means that the vector $c(A+B)$ is the vertex of the parallelogram formed by $cA$ and $cB$ .