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Q15E

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Linear Algebra and its Applications
Found in: Page 1
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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.

15. \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\)

The five vectors in span \(\left\{ {{v_1},{v_2}} \right\}\) are \(\left\{ {\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\4\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{12}\\{ - 2}\\{ - 6}\end{array}} \right]} \right\}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Choose the five sets of weights to generate the vector

If vectors \({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\) in \({\mathbb{R}^n}\) are given with scalars \({c_1},{c_2},...,{c_p}\), then the vector \({\mathop{\rm y}\nolimits} \) defined by \(y = {c_1}{v_1} + .... + {c_p}{v_p}\) is called a linear 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:

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

Step 2: 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 the vector addition \({\mathop{\rm u}\nolimits} + v\), which is obtained by adding the corresponding entries of \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \).

The scalar multiple of 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

\(\begin{aligned}{c}{V_1} &= 0{v_1} + 0{v_2}\\ &= 0\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right] + 0\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\end{aligned}\)

Step 3: 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 \({c_1}{v_1} + {c_2}{v_2} + .... + {c_p}{v_p}\) with \({c_1},...,{c_p}\) scalars.

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

\(\begin{aligned}{c}{V_2} &= 1{v_1} + 0{v_2}\\ &= 1\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right] + 0\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right]\end{aligned}\)

Step 4: Generate the weight of the third vector

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

\(\begin{aligned}{c}{V_3} &= 0{v_1} + 1{v_2}\\ &= 0\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right] + 1\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\end{aligned}\)

Step 5: Generate the weight of the fourth vector

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

\(\begin{aligned}{c}{V_4} &= 1{v_1} + 1{v_2}\\ &= 1\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right] + 1\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{7 - 5}\\{1 + 3}\\{ - 6 + 0}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}2\\4\\{ - 6}\end{array}} \right]\end{aligned}\)

Step 6: Generate the weight of the fifth vector

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

\(\begin{aligned}{c}{V_5} &= 1{v_1} - 1{v_2}\\ &= 1\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right] - 1\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{7 + 5}\\{1 - 3}\\{ - 6 - 0}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{12}\\{ - 2}\\{ - 6}\end{array}} \right]\end{aligned}\)

Step 7: List the three entries of the vector

The three entries of the vector are \(\left\{ {\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{12}\\{ - 2}\\{ - 6}\end{array}} \right]} \right\}\)

Hence, the five vectors in span \(\left\{ {{v_1},{v_2}} \right\}\) are \(\left\{ {\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\4\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{12}\\{ - 2}\\{ - 6}\end{array}} \right]} \right\}\).

Most popular questions for Math Textbooks

An important concern in the study of heat transfer is to determine the steady-state temperature distribution of a thin plate when the temperature around the boundary is known. Assume the plate shown in the figure represents a cross section of a metal beam, with negligible heat flow in the direction perpendicular to the plate. Let \({T_1},...,{T_4}\) denote the temperatures at the four interior nodes of the mesh in the figure. The temperature at a node is approximately equal to the average of the four nearest nodes—to the left, above, to the right, and below. For instance,

\({T_1} = \left( {10 + 20 + {T_2} + {T_4}} \right)/4\), or \(4{T_1} - {T_2} - {T_4} = 30\)

33. Write a system of four equations whose solution gives estimates

for the temperatures \({T_1},...,{T_4}\).

Suppose T and S satisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that S is 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 S to 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 \(T\left( {cx} \right) = cT\left( x \right)\).)

Suppose the system below is consistent for all possible values of \(f\) and \(g\). What can you say about the coefficients \(c\) and \(d\)? Justify your answer.

27. \(\begin{array}{l}{x_1} + 3{x_2} = f\\c{x_1} + d{x_2} = g\end{array}\)

If A is an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

Construct a \(3 \times 3\) matrix\(A\), with nonzero entries, and a vector \(b\) in \({\mathbb{R}^3}\) such that \(b\) is not in the set spanned by the columns of\(A\).
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