In Exercises 15 and 16, list five vectors in Span \(\left\{ {{v_1},{v_2}} \right\}\). For each vector, show the weights on \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) used to generate the vector and list the three entries of the vector. Do not make a sketch.

16. \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right]\)

Short Answer

Expert verified

The five vectors in span \(\left\{ {{v_1},{v_2}} \right\}\) are \(\left\{ {\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1\\0\\5\end{array}} \right],\left[ {\begin{array}{*{20}{c}}5\\0\\{ - 1}\end{array}} \right]} \right\}\).

Step by step solution

01

Choose the five sets of weights to generate the vector

The vector\({\mathop{\rm y}\nolimits} \)defined by\(y = {c_1}{v_1} + .... + {c_p}{v_p}\)is called alinear combination of\({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\)with weights\({c_1},{c_2},...,{c_p}\).

Choose the five sets of weights as:

\({\rm{w}} = \left\{ {0,0} \right\},\left\{ {1,0} \right\},\left\{ {0,1} \right\},\left\{ {1,1} \right\},\left\{ {1, - 1} \right\}\)

02

Generate the weight of the first vector

In\({\mathbb{R}^2}\), the sum of two vectors\({\mathop{\rm u}\nolimits} \)and\({\mathop{\rm v}\nolimits} \)is thevector addition \({\mathop{\rm u}\nolimits} + v\), which is obtained by adding the corresponding entries of\({\mathop{\rm u}\nolimits} \)and\({\mathop{\rm v}\nolimits} \).

Thescalar multipleof a vector \({\mathop{\rm u}\nolimits} \) by real number \(c\) is the vector \(c{\mathop{\rm u}\nolimits} \) obtained by multiplying each entry in \({\mathop{\rm u}\nolimits} \) by \(c\).

Use scalar multiplication and vector addition to generate the weight of the vector

\(\begin{aligned}{c}{V_1} &= 0{v_1} + 0{v_2}\\ &= 0\left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right] + 0\left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\end{aligned}\)

03

Generate the weight of the second vector

If \({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\) in \({\mathbb{R}^n}\), then the set of all linear combinations \({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\) is denoted by span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}} \right\}\) and is called the subset of \({\mathbb{R}^n}\) spanned \({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\). Span is the collection of all vectors that can be written in the form \({c_1}{v_1} + {c_2}{v_2} + .... + {c_p}{v_p}\) of \({c_1},...,{c_p}\) scalars.

Use scalar multiplication and vector addition to generate the weight of the vector

\(\begin{aligned}{c}{V_2} &= 1{v_1} + 0{v_2}\\ &= 1\left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right] + 0\left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{3 + 0}\\{0 + 0}\\{2 + 0}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right]\end{aligned}\)

04

Generate the weight of the third vector

Use scalar multiplication and vector addition to generate the weight of the vector

\(\begin{aligned}{c}{V_3} &= 0{v_1} + 1{v_2}\\ &= 0\left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right] + 1\left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right]\end{aligned}\)

05

Generate the weight of the fourth vector

Use scalar multiplication and vector addition to generate the weight of the vector

\(\begin{aligned}{c}{v_4} &= 1{v_1} + 1{v_2}\\ &= 1\left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right] + 1\left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{3 - 2}\\{0 + 0}\\{2 + 3}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}1\\0\\5\end{array}} \right]\end{aligned}\)

06

Generate the weight of the fifth vector

Use scalar multiplication and vector addition to generate the weight of the vector

\(\begin{aligned}{c}{V_5} &= 1{v_1} - 1{v_2}\\ &= 1\left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right] - 1\left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{3 + 2}\\{0 + 0}\\{2 - 3}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}5\\0\\{ - 1}\end{array}} \right]\end{aligned}\)

07

List the three entries of the vector

The weights on \({v_1}\) and \({v_2}\) to generate the vector are \(\left\{ {0,0} \right\},\left\{ {1,0} \right\},\left\{ {2,3} \right\},\left\{ {1,1} \right\},\left\{ {1, - 1} \right\}\)

The three entries of the vector are \(\left\{ {\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right],\left[ {\begin{array}{*{20}{c}}0\\0\\{13}\end{array}} \right]} \right\}\)

Hence, the five vectors in span \(\left\{ {{v_1},{v_2}} \right\}\) are \(\left\{ {\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1\\0\\5\end{array}} \right],\left[ {\begin{array}{*{20}{c}}5\\0\\{ - 1}\end{array}} \right]} \right\}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 5, write a system of equations that is equivalent to the given vector equation.

5. \({x_1}\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\5\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\)

Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. (Hint: Given u, v in \({\mathbb{R}^n}\), let \({\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\). Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \(T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \). Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).)

Give a geometric description of span \(\left\{ {{v_1},{v_2}} \right\}\) for the vectors \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}8\\2\\{ - 6}\end{array}} \right]\) and \({{\mathop{\rm v}\nolimits} _2} = \left[ {\begin{array}{*{20}{c}}{12}\\3\\{ - 9}\end{array}} \right]\).

Solve each system in Exercises 1–4 by using elementary row operations on the equations or on the augmented matrix. Follow the systematic elimination procedure.

  1. \(\begin{aligned}{c}{x_1} + 5{x_2} = 7\\ - 2{x_1} - 7{x_2} = - 5\end{aligned}\)

In Exercises 32, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

32. \(\left[ {\begin{array}{*{20}{c}}1&2&{ - 5}&0\\0&1&{ - 3}&{ - 2}\\0&{ - 3}&9&5\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&2&{ - 5}&0\\0&1&{ - 3}&{ - 2}\\0&0&0&{ - 1}\end{array}} \right]\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free