4.4E: Linear Operators on R³ Exercises
- Page ID
- 132818
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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 each case show that that \(T\) is either projection on a line, reflection in a line, or rotation through an angle, and find the line or angle.
- \(T\left[ \begin{array}{c} x\\ y \end{array} \right] = \frac{1}{5} \left[ \begin{array}{c} x + 2y\\ 2x + 4y \end{array} \right]\)
- \(T\left[ \begin{array}{c} x\\ y \end{array} \right] = \frac{1}{2} \left[ \begin{array}{c} x - y\\ y - x \end{array} \right]\)
- \(T\left[ \begin{array}{c} x\\ y \end{array} \right] = \frac{1}{\sqrt{2}} \left[ \begin{array}{c} -x - y\\ x - y \end{array} \right]\)
- \(T\left[ \begin{array}{c} x\\ y \end{array} \right] = \frac{1}{5} \left[ \begin{array}{c} -3x + 4y\\ 4x + 3y \end{array} \right]\)
- \(T\left[ \begin{array}{c} x\\ y \end{array} \right] = \left[ \begin{array}{c} -y\\ -x \end{array} \right]\)
- \(T\left[ \begin{array}{c} x\\ y \end{array} \right] = \frac{1}{2} \left[ \begin{array}{c} x - \sqrt{3}y\\ \sqrt{3}x + y \end{array} \right]\)
- Answer
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- \(A = \left[ \begin{array}{rr} 1 & -1\\ -1 & 1\\ \end{array} \right]\), projection on \(y = -x\).
- \(A = \frac{1}{5}\left[ \begin{array}{rr} -3 & 4\\ 4 & 3\\ \end{array} \right]\), reflection in \(y = 2x\).
- \(A = \frac{1}{2}\left[ \begin{array}{rr} 1 & -\sqrt{3}\\ \sqrt{3} & 1\\ \end{array} \right]\), rotation through \(\frac{\pi}{3}\).
Determine the effect of the following transformations.
- Rotation through \(\frac{\pi}{2}\), followed by projection on the \(y\) axis, followed by reflection in the line \(y = x\).
- Projection on the line \(y = x\) followed by projection on the line \(y = -x\).
- Projection on the \(x\) axis followed by reflection in the line \(y = x\).
- Answer
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- The zero transformation.
In each case solve the problem by finding the matrix of the operator.
- Find the projection of \(\mathbf{v} = \left[ \begin{array}{r} 1\\ -2\\ 3 \end{array} \right]\) on the plane with equation \(3x - 5y + 2z = 0\).
- Find the projection of \(\mathbf{v} = \left[ \begin{array}{r} 0\\ 1\\ -3 \end{array} \right]\) on the plane with equation \(2x - y + 4z = 0\).
- Find the reflection of \(\mathbf{v} = \left[ \begin{array}{r} 1\\ -2\\ 3 \end{array} \right]\) in the plane with equation \(x - y + 3z = 0\).
- Find the reflection of \(\mathbf{v} = \left[ \begin{array}{r} 0\\ 1\\ -3 \end{array} \right]\) in the plane with equation \(2x + y -5z = 0\).
- Find the reflection of \(\mathbf{v} = \left[ \begin{array}{r} 2\\ 5\\ -1 \end{array} \right]\) in the line with equation \(\left[ \begin{array}{r} x\\ y\\ z \end{array} \right] = t \left[ \begin{array}{r} 1\\ 1\\ -2 \end{array} \right]\).
- Find the projection of \(\mathbf{v} = \left[ \begin{array}{r} 1\\ -1\\ 7 \end{array} \right]\) on the line with equation \(\left[ \begin{array}{r} x\\ y\\ z \end{array} \right] = t \left[ \begin{array}{r} 3\\ 0\\ 4 \end{array} \right]\).
- Find the projection of \(\mathbf{v} = \left[ \begin{array}{r} 1\\ 1\\ -3 \end{array} \right]\) on the line with equation \(\left[ \begin{array}{r} x\\ y\\ z \end{array} \right] = t \left[ \begin{array}{r} 2\\ 0\\ -3 \end{array} \right]\).
- Find the reflection of \(\mathbf{v} = \left[ \begin{array}{r} 2\\ -5\\ 0 \end{array} \right]\) in the line with equation \(\left[ \begin{array}{r} x\\ y\\ z \end{array} \right] = t \left[ \begin{array}{r} 1\\ 1\\ -3 \end{array} \right]\).
- Answer
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- \(\frac{1}{21}\left[ \begin{array}{rrr} 17 & 2 & -8\\ 2 & 20 & 4\\ -8 & 4 & 5 \end{array} \right] \left[ \begin{array}{r} 0\\ 1\\ -3 \end{array} \right]\)
- \(\frac{1}{30}\left[ \begin{array}{rrr} 22 & -4 & 20\\ -4 & 28 & 10\\ 20 & 10 & -20 \end{array} \right] \left[ \begin{array}{r} 0\\ 1\\ -3 \end{array} \right]\)
- \(\frac{1}{25}\left[ \begin{array}{rrr} 9 & 0 & 12\\ 0 & 0 & 0\\ 12 & 0 & 16 \end{array} \right] \left[ \begin{array}{r} 1\\ -1\\ 7 \end{array} \right]\)
- \(\frac{1}{11}\left[ \begin{array}{rrr} -9 & 2 & -6\\ 2 & -9 & -6\\ -6 & -6 & 7 \end{array} \right] \left[ \begin{array}{r} 2\\ -5\\ 0 \end{array} \right]\)
- Find the rotation of \(\mathbf{v} = \left[ \begin{array}{r} 2\\ 3\\ -1 \end{array} \right]\) about the \(z\) axis through \(\theta = \frac{\pi}{4}\).
- Find the rotation of \(\mathbf{v} = \left[ \begin{array}{r} 1\\ 0\\ 3 \end{array} \right]\) about the \(z\) axis through \(\theta = \frac{\pi}{6}\).
- Answer
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- \(\frac{1}{2}\left[ \begin{array}{rrr} \sqrt{3} & -1 & 0\\ 1 & \sqrt{3} & 0\\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{r} 1\\ 0\\ 3 \end{array} \right]\)
Find the matrix of the rotation in \(\mathbb{R}^3\) about the \(x\) axis through the angle \(\theta\) (from the positive \(y\) axis to the positive \(z\) axis).
Find the matrix of the rotation about the \(y\) axis through the angle \(\theta\) (from the positive \(x\) axis to the positive \(z\) axis).
- Answer
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\(\left[ \begin{array}{ccc} \cos\theta & 0 & -\sin\theta\\ 0 & 1 & 0\\ \sin\theta & 0 & \cos\theta \end{array} \right]\)
If \(A\) is \(3 \times 3\), show that the image of the line in \(\mathbb{R}^3\) through \(\mathbf{p}_{0}\) with direction vector \(\mathbf{d}\) is the line through \(A\mathbf{p}_{0}\) with direction vector \(A\mathbf{d}\), assuming that \(A\mathbf{d} \neq \mathbf{0}\). What happens if \(A\mathbf{d} = \mathbf{0}\)?
If \(A\) is \(3 \times 3\) and invertible, show that the image of the plane through the origin with normal \(\mathbf{n}\) is the plane through the origin with normal \(\mathbf{n}_{1} = B\mathbf{n}\) where \(B = (A^{-1})^{T}\). [Hint: Use the fact that \(\mathbf{v} \cdot \mathbf{w} = \mathbf{v}^{T}\mathbf{w}\) to show that \(\mathbf{n}_{1} \cdot (A\mathbf{p}) = \mathbf{n} \cdot \mathbf{p}\) for each \(\mathbf{p}\) in \(\mathbb{R}^3\).]
Let \(L\) be the line through the origin in \(\mathbb{R}^2\) with direction vector \(\mathbf{d} = \left[ \begin{array}{r} a\\ b\\ \end{array} \right] \neq 0\).
- If \(P_{L}\) denotes projection on \(L\), show that \(P_{L}\) has matrix \(\dfrac{1}{a^2 + b^2}\left[ \begin{array}{cc} a^2 & ab\\ ab & b^2\\ \end{array}\right]\).
- If \(Q_{L}\) denotes reflection in \(L\), show that \(Q_{L}\) has matrix \(\dfrac{1}{a^2 + b^2}\left[ \begin{array}{cc} a^2 - b^2 & 2ab\\ 2ab & b^2 - a^2\\ \end{array}\right]\).
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- Write \(\mathbf{v} = \left[ \begin{array}{r} x\\ y \end{array} \right]\). \[\begin{aligned} P_{L}(\mathbf{v}) = \left(\frac{\mathbf{v} \cdot \mathbf{d}}{\| \mathbf{d} \|^2}\right)\mathbf{d} & = \frac{ax + by}{a^2 + b^2}\left[ \begin{array}{r} a\\ b \end{array} \right] \\ & = \frac{1}{a^2 + b^2}\left[ \begin{array}{c} a^2x + aby\\ abx + b^2y \end{array} \right] \\ & = \frac{1}{a^2 + b^2}\left[ \begin{array}{c} a^2 + ab\\ ab + b^2 \end{array} \right] \left[ \begin{array}{r} x\\ y \end{array} \right]\end{aligned} \nonumber \]
Let \(\mathbf{n}\) be a nonzero vector in \(\mathbb{R}^3\), let \(L\) be the line through the origin with direction vector \(\mathbf{n}\), and let \(M\) be the plane through the origin with normal \(\mathbf{n}\). Show that \(P_{L}(\mathbf{v}) = Q_{L}(\mathbf{v}) + P_{M}(\mathbf{v})\) for all \(\mathbf{v}\) in \(\mathbb{R}^3\). [In this case, we say that \(P_{L} = Q_{L} + P_{M}\).]
If \(M\) is the plane through the origin in \(\mathbb{R}^3\) with normal \(\mathbf{n} = \left[ \begin{array}{r} a\\ b\\ c \end{array} \right]\), show that \(Q_{M}\) has matrix
- Answer
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\[{\small \frac{1}{a^2 + b^2 + c^2}}{\footnotesize \left[ \begin{array}{ccc} b^2 + c^2 - a^2 & -2ab & -2ac \\ -2ab & a^2 + c^2 - b^2 & -2bc \\ -2ac & -2bc & a^2 + b^2 - c^2 \end{array} \right]} \nonumber \]