5.2.1: Exercises 5.2
- Page ID
- 71114
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In Exercises \(\PageIndex{1}\) – \(\PageIndex{5}\), a transformation \(T\) is given. Determine whether or not \(T\) is linear; if not, state why not.
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\end{array}\right]\right)=\left[\begin{array}{c}{x_{1}+x_{2}}\\{3x_{1}-x_{2}}\end{array}\right]\)
- Answer
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Yes
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\end{array}\right]\right)=\left[\begin{array}{c}{x_{1}+x_{2}^{2}}\\{x_{1}-x_{2}}\end{array}\right]\)
- Answer
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No; cannot have a squared term.
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\end{array}\right]\right)=\left[\begin{array}{c}{x_{1}+1}\\{x_{2}+1}\end{array}\right]\)
- Answer
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No; cannot add a constant.
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\end{array}\right]\right)=\left[\begin{array}{c}{1}\\{1}\end{array}\right]\)
- Answer
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No; cannot add a constant.
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\end{array}\right]\right)=\left[\begin{array}{c}{0}\\{0}\end{array}\right]\)
- Answer
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Yes
In Exercises \(\PageIndex{6}\) - \(\PageIndex{11}\), a linear transformation \(T\) is given. Find \([T]\).
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\end{array}\right]\right)=\left[\begin{array}{c}{x_{1}+x_{2}}\\{x_{1}-x_{2}}\end{array}\right]\)
- Answer
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\([T]=\left[\begin{array}{cc}{1}&{1}\\{1}&{-1}\end{array}\right]\)
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\end{array}\right]\right)=\left[\begin{array}{c}{x_{1}+2x_{2}}\\{3x_{1}-5x_{2}}\\{2x_{2}}\end{array}\right]\)
- Answer
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\([T]=\left[\begin{array}{cc}{1}&{2}\\{3}&{-5}\\{0}&{2}\end{array}\right]\)
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\\{x_{3}}\end{array}\right]\right)=\left[\begin{array}{c}{x_{1}+2x_{2}-3x_{3}}\\{0}\\{x_{1}+4x_{3}}\\{5x_{2}+x_{3}}\end{array}\right]\)
- Answer
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\([T]=\left[\begin{array}{ccc}{1}&{2}&{-3}\\{0}&{0}&{0}\\{1}&{0}&{4}\\{0}&{5}&{1}\end{array}\right]\)
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\\{x_{3}}\end{array}\right]\right)=\left[\begin{array}{c}{x_{1}+3x_{3}}\\{x_{1}-x_{3}}\\{x_{1}+x_{3}}\end{array}\right]\)
- Answer
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\([T]=\left[\begin{array}{ccc}{1}&{0}&{3}\\{1}&{0}&{-1}\\{1}&{0}&{1}\end{array}\right]\)
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\end{array}\right]\right)=\left[\begin{array}{c}{0}\\{0}\end{array}\right]\)
- Answer
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\([T]=\left[\begin{array}{cc}{0}&{0}\\{0}&{0}\end{array}\right]\)
\(T\left(\left[\begin{array}{c}{x_{1}}\\{x_{2}}\\{x_{3}}\\{x_{4}}\end{array}\right]\right)=[x_{1}+2x_{2}+3x_{3}+4x_{4}]\)
- Answer
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\([T]=[1\:2\:3\:4]\)