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Expert-verified Found in: Page 1 ### 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 # 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 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 $$\left\{ {\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\4\\{ - 6}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{12}\\{ - 2}\\{ - 6}\end{array}} \right]} \right\}$$. ### Want to see more solutions like these? 