Q.61E

$61 .$

Let

c

and

d

be a scalars and let

v

be a vector in

ℝ^{3}

. Show that the following distributive property holds:

$(c + d) v = c v + d v .$

Verified

Answer

What do we mean when we say $S$ is linearly independent., $S$ is closed in both addition and scalar multiplication.

1Step 1: Introduction.

Consider the vector $v$ in $R^{3}$ and $c$ and $d$ are any two scalars.

The objective is to prove that $(c + d) v = c v + d v .$

If there are two vectors $u = <x_{1}, y_{1}, z_{1}>$

and

c

is any scalar, then given by

c u = <c x_{1}, c y_{1}, c z_{1}>

2Step 2: Given Information.

Now, let $v = <v_{1}, v_{2}, v_{3}> \dots \dots$ (1)

Therefore,

Now,

$(c + d) v = (c + d) ⟨ v 1, v 2, v 3 ⟩ = ⟨ (c + d) v 1, (c + d) v 2, (c + d) v 3 ⟩ = ⟨ c v 1 + d v 1, c v 2 + d v 2, c v 3 + d v 3 ⟩$

Therefore, $(c + d) v = ⟨ c v 1 + d v 1, c v 2 + d v 2, c v 3 + d v 3 ⟩$

3Step 3: Explanation (part a).

Now, rewrite $(c + d) v = <c v_{1} + d v_{1}, c v_{2} + d v_{2}, c v_{3} + d v_{3}>$

$(c + d) v = ⟨ c v 1, c v 2, c v 3 ⟩ + ⟨ d v 1, d v 2, d v 3 ⟩ = c ⟨ v 1, v 2, v 3 ⟩ + c ⟨ v 1, v 2, v 3 ⟩$

Hence, $(c + d) v = c <v_{1}, v_{2}, v_{3}> + c <v_{1}, v_{2}, v_{3}>$ ..........(2)

4Step 3: Explanation (part b).

Using $(1)$ in $(2)$

Therefore,

$(c + d) v = c v + d v$

Hence, it is proved that $(c + d) v = c v + d v$