Question

(M) Generate random vectors \(x\), \(y\), and \(v\) in \({\mathbb{R}^4}\) with integer

entries (and \(v \ne 0\)), and compute the quantities

\(\left( {\frac{{x \cdot v}}{{v \cdot v}}} \right)v,\left( {\frac{{y \cdot v}}{{v \cdot v}}} \right)v,\frac{{\left( {x + y} \right) \cdot v}}{{v \cdot v}}v,\frac{{\left( {10x} \right) \cdot v}}{{v \cdot v}}v\)

Repeat the computations with new random vectors x and

y. What do you conjecture about the mapping \(x \mapsto T\left( x \right) = \frac{{x \cdot v}}{{v \cdot v}}v\)

(for \(v \ne 0\))? Verify your conjecture algebraically.

Short answer

\(T\) is a linear transformation and it is verified algebraically.

Step-by-step solution

Orthogonal set

A set of vectors \(\left\{ {{v_1},......,{v_n}} \right\}\) is said to be orthogonal if \({v_i} \cdot {v_j} = 0\), for \(i \ne j\).

Compute the required quantity

We have to generate the random vectors in \({\mathbb{R}^4}\) with integer entries and compute \(\left( {\frac{{x \cdot v}}{{v \cdot v}}} \right)v,\left( {\frac{{y \cdot v}}{{v \cdot v}}} \right)v,\frac{{\left( {x + y} \right) \cdot v}}{{v \cdot v}}v,\frac{{\left( {10x} \right) \cdot v}}{{v \cdot v}}v\)

The random matrix

>> x= randi((-5,5),4,1)

>> y= randi((-5,5),4,1)

>> v= randi((-5,5),4,1)

Compute the Entries:

>> p=(dot(x,v)/dot(v,v))*v

>>q= (dot(y,v)/dot(v,v))*v

>> r=(dot(x+y,v)/dot(v,v))*v

>>s= (dot(10*x,v)/dot(v,v))*v

>> p+q

>> 10*p

We find that:

\(p = \left( {\begin{aligned}{*{20}{c}}{ - 1.1429}\\{0.2857}\\0\\{ - 1.4286}\end{aligned}} \right)\)

\(q = \left( {\begin{aligned}{*{20}{c}}{0.4762}\\{ - 0.1190}\\0\\{0.5952}\end{aligned}} \right)\)

\(r = \left( {\begin{aligned}{*{20}{c}}{ - 0.6667}\\{0.1667}\\0\\{ - 0.8333}\end{aligned}} \right)\)

\(s = \left( {\begin{aligned}{*{20}{c}}{ - 11.4286}\\{2.8571}\\0\\{ - 14.2857}\end{aligned}} \right)\)

Also

\(p + q = \left( {\begin{aligned}{*{20}{c}}{ - 0.6667}\\{0.1667}\\0\\{ - 0.8333}\end{aligned}} \right) = r\)

\(10p = \left( {\begin{aligned}{*{20}{c}}{ - 11.4286}\\{2.8571}\\0\\{ - 14.2857}\end{aligned}} \right) = s\)

The mapping \({\bf{x}} \mapsto T\left( x \right)\) is a linear transformation.

Verify the obtained result algebraically

To check \(T\) is linear or not,

\(\begin{aligned}{c}T\left( {x + y} \right) = \frac{{\left( {x + y} \right) \cdot v}}{{v \cdot v}}v\\ = \frac{{x \cdot v + y \cdot v}}{{v \cdot v}}v\\ = \frac{{x \cdot v}}{{v \cdot v}}v + \frac{{y \cdot v}}{{v \cdot v}}v\\ = T\left( x \right) + T\left( y \right)\end{aligned}\)

and

\(\begin{aligned}{c}T\left( {cx} \right) &= \frac{{\left( {cx} \right) \cdot v}}{{v \cdot v}}v\\ &= c\frac{{x \cdot v}}{{v \cdot v}}v\\ &= cT\left( x \right)\end{aligned}\)

Hence, \(T\) is a linear transformation.