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Chapter 1: Q23E (page 1)

Let \({\bf{T}}:{\mathbb{R}^{\bf{2}}} \to {\mathbb{R}^{\bf{2}}}\) be the linear transformation that reflects each point through the \({{\bf{x}}_{\bf{1}}}\)-axis. That is, \({\bf{T}}\left( {\bf{x}} \right) = \left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{ - {\bf{1}}}\end{array}} \right]{\bf{x}}\). Make two sketches that illustrate properties (i) \({\bf{T}}\left( {u + {\bf{v}}} \right) = {\bf{T}}\left( {\bf{u}} \right) + {\bf{T}}\left( {\bf{v}} \right)\) and (ii) \({\bf{T}}\left( {{\bf{cw}}} \right) = {\bf{cT}}\left( {\bf{w}} \right)\)of a linear transformation.

Short Answer

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i)

ii)

Step by step solution

01

Determine the first property

Let \(u = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right],v = \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right] \in {\mathbb{R}^2}\). Then,

\(\begin{array}{c}T\left( u \right) = T\left( {\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]} \right)\\ = \left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\\T\left( u \right) = \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right]\end{array}\)

\(\begin{array}{c}T\left( v \right) = T\left( {\left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]} \right)\\ = \left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]\\T\left( v \right) = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\end{array}\)

And \(u + v = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\3\end{array}} \right]\). Therefore,

\(\begin{array}{c}T\left( {u + v} \right) = T\left( {\left[ {\begin{array}{*{20}{c}}3\\3\end{array}} \right]} \right)\\ = \left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3\\3\end{array}} \right]\\T\left( {u + v} \right) = \left[ {\begin{array}{*{20}{c}}3\\{ - 3}\end{array}} \right]\end{array}\)

02

Provide the image that illustrates property (i)

Note that \(T\) reflects about the \({x_1}\)-axis.

03

Determine the second property

Let \(c = 3 \in \mathbb{R}\) and \(w = \left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right] \in {\mathbb{R}^2}\). Then,

\(cw = 3\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\3\end{array}} \right]\)

Its image is given by:

\(\begin{array}{c}T\left( {cw} \right) = T\left( {\left[ {\begin{array}{*{20}{c}}3\\3\end{array}} \right]} \right)\\ = \left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3\\3\end{array}} \right]\\T\left( {cw} \right) = \left[ {\begin{array}{*{20}{c}}3\\{ - 3}\end{array}} \right]\end{array}\)

04

Provide the image that illustrates property (ii)

Note that \(T\) reflects about the \({x_1}\)-axis.

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Most popular questions from this chapter

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let Sand U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that Thas a unique inverse, as asserted in theorem 9. (Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)).

Suppose \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}\) are distinct points on one line in \({\mathbb{R}^3}\). The line need not pass through the origin. Show that \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly dependent.

Let \({{\bf{a}}_1}\) \({{\bf{a}}_2}\), and b be the vectors in \({\mathbb{R}^{\bf{2}}}\) shown in the figure, and let \(A = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}}&{{{\bf{a}}_2}}\end{aligned}} \right)\). Does the equation \(A{\bf{x}} = {\bf{b}}\) have a solution? If so, is the solution unique? Explain.

Solve the linear system of equations. You may use technology.

|3x+5y+3z=257X+9y+19z=654X+5y+11z=5|

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