Question

$61.$

Let $c$and $d$be a scalars and let $\mathbf{v}$be a vector in ${\mathrm{ℝ}}^3$. Show that the following distributive property holds:

$(c+d)\mathbf{v}=c\mathbf{v}+d\mathbf{v}.$

Short answer

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

Step-by-step solution

Introduction.

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

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

If there are two vectors

and $c$is any scalar, then given by

Given Information.

Now, let (1)

Therefore,

Now,

$$\begin{gathered} (c+d)v=(c+d)⟨v1,v2,v3⟩ \\ =⟨(c+d)v1,(c+d)v2,(c+d)v3⟩ \\ =⟨cv1+dv1,cv2+dv2,cv3+dv3⟩ \end{gathered}$$

Therefore,$(c+d)v=⟨cv1+dv1,cv2+dv2,cv3+dv3⟩$

Explanation (part a).

Now, rewrite

$$\begin{gathered} (c+d)v=⟨cv1,cv2,cv3⟩+⟨dv1,dv2,dv3⟩ \\ =c⟨v1,v2,v3⟩+c⟨v1,v2,v3⟩ \end{gathered}$$

Hence,..........(2)

Explanation (part b).

Using $\left(1\right)$in$\left(2\right)$

Therefore,

$(c+d)\mathbf{v}=c\mathbf{v}+d\mathbf{v}$

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