Question

Use the vector \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\) to verify the following algebraic properties of \({\mathbb{R}^n}\).

a. \({\bf{u}} + \left( { - {\bf{u}}} \right) = \left( { - {\bf{u}}} \right) + {\bf{u}} = {\bf{0}}\)

b. \(c\left( {d{\bf{u}}} \right) = \left( {cd} \right){\bf{u}}\) for all scalars \(c\) and \(d\).

Short answer

The algebraic properties are verified.

Step-by-step solution

(a) Step 1: Simplify the first part of the algebraic property

Consider the algebraic property \({\bf{u}} + \left( { - {\bf{u}}} \right) = \left( { - {\bf{u}}} \right) + {\bf{u}} = 0\).

Take the algebraic expression as \({\bf{u}} + \left( { - {\bf{u}}} \right)\). Here, \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\).

To obtain \( - {\bf{u}}\), take \( - 1\) as a scalar multiple, which is shown below:

\(\begin{aligned}{c} - {\bf{u}} &= - \left( {{u_1},...,{u_n}} \right)\\ &= - {u_1},..., - {u_n}\end{aligned}\)

Simplify \({\bf{u}} + \left( { - {\bf{u}}} \right)\) by using \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\), and \( - {\bf{u}} = - {u_1},..., - {u_n}\) as shown below:

\(\begin{aligned}{c}{\bf{u}} + \left( { - {\bf{u}}} \right) &= \left( {{u_1},...,{u_n}} \right) + \left( { - {u_1},..., - {u_n}} \right)\\ &= \left[ {{u_1} + \left( { - {u_1}} \right),...,{u_n} + \left( { - {u_n}} \right)} \right]\\ &= \left( {{u_1} - {u_1},...,{u_n} - {u_n}} \right)\\ &= \left( {0,...,0} \right)\\ &= {\bf{0}}\end{aligned}\)

Thus, \({\bf{u}} + \left( { - {\bf{u}}} \right) = 0\).

Simplify another part of the algebraic property

To show that \({\bf{u}} + \left( { - {\bf{u}}} \right) = \left( { - {\bf{u}}} \right) + {\bf{u}} = 0\), simplify another part \(\left( { - {\bf{u}}} \right) + {\bf{u}}\) as shown below:

\(\begin{aligned}{c}\left( { - {\bf{u}}} \right) + {\bf{u}} &= \left( { - {u_1},..., - {u_n}} \right) + \left( {{u_1},...,{u_n}} \right)\\ &= \left[ {\left( { - {u_1}} \right) + {u_1},...,\left( { - {u_n}} \right) + {u_1}} \right]\\ &= \left( { - {u_1} + {u_1},..., - {u_n} + {u_n}} \right)\\ &= \left( {0,...,0} \right)\\ &= {\bf{0}}\end{aligned}\)

Hence, it is proved that \({\bf{u}} + \left( { - {\bf{u}}} \right) = \left( { - {\bf{u}}} \right) + {\bf{u}} = 0\).

(b) Step 3: Simplify the left-hand side of the algebraic property

Consider the algebraic property \(c\left( {d{\bf{u}}} \right) = \left( {cd} \right){\bf{u}}\).

Take the left-hand side of the algebraic property as \(c\left( {d{\bf{u}}} \right)\). Here, \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\)and \(c\), and \(d\) are scalars.

Simplify \(d{\bf{u}}\)by using \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\)as shown below:

\(\begin{aligned}{c}d{\bf{u}} &= d\left( {{u_1},...,{u_n}} \right)\\ &= \left( {d{u_1},...,d{u_n}} \right)\end{aligned}\)

Simplify the left-hand side of the algebraic property further

Simplify \(c\left( {d{\bf{u}}} \right)\)by using \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\)as shown below:

\(\begin{aligned}{c}c\left( {d{\bf{u}}} \right) &= c\left( {d{u_1},...,d{u_n}} \right)\\ &= c\left[ {d\left( {{u_1},...,{u_n}} \right)} \right]\\ &= cd\left( {{u_1},...,{u_n}} \right)\\ &= \left( {cd} \right)\left( {{u_1},...,{u_n}} \right)\\ &= \left( {cd} \right){\bf{u}}\end{aligned}\)

Hence, it is proved that \(c\left( {d{\bf{u}}} \right) = \left( {cd} \right){\bf{u}}\).