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

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Most popular questions from this chapter

Explain why a set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) in \({\mathbb{R}^5}\) must be linearly independent when \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly independent and \({{\mathop{\rm v}\nolimits} _4}\) is not in Span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\).

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

34. \(T\left( {{x_1},{x_2}} \right) = \left( {6{x_1} - 8{x_2}, - 5{x_1} + 7{x_2}} \right)\)

Consider a dynamical system x→(t+1)=Ax→(t) with two components. The accompanying sketch shows the initial state vector x→0and two eigen vectors υ1→  and  υ2→ of A (with eigen values λ1→andλ2→ respectively). For the given values of λ1→andλ2→, draw a rough trajectory. Consider the future and the past of the system.

λ1→=1,λ2→=0.9

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14. \(\left[ {\begin{array}{*{20}{c}}1&2&{ - 5}&{ - 6}&0&{ - 5}\\0&1&{ - 6}&{ - 3}&0&2\\0&0&0&0&1&0\\0&0&0&0&0&0\end{array}} \right]\).

Suppose Ais an \(n \times n\) matrix with the property that the equation \(A{\mathop{\rm x}\nolimits} = 0\) has at least one solution for each b in \({\mathbb{R}^n}\). Without using Theorem 5 or 8, explain why each equation Ax = b has in fact exactly one solution.

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