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Q15E

Expert-verifiedFound in: Page 1

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

** 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 \(\left\{ {\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\4\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{12}\\{ - 2}\\{ - 6}\end{array}} \right]} \right\}\).

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:

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

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

** **

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

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.

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

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

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

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

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

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

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

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 \(\left\{ {\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\4\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{12}\\{ - 2}\\{ - 6}\end{array}} \right]} \right\}\).

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