Solved Problems for Chapter 6

Show that the function \(T_{\vec {u}}\) defined by \(T_{\vec {u}} \left ( \vec {v}\right ) = \vec {v}-\mbox {proj}_{\vec {u}}\left ( \vec {v}\right ) \) is also a linear transformation.

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\[ T_{\vec {u}}\left ( a\vec {v}+b\vec {w}\right ) =a\vec {v}+b\vec {w}-\frac {\left ( a\vec {v}+b\vec {w}\right )\dotp \vec {u} }{\norm { \vec {u} } ^{2}}\vec {u} =\]
\[ =a\vec {v}-a\frac {\left ( \vec {v}\dotp \vec {u}\right ) }{\norm { \vec {u} } ^{2}}\vec {u}+b\vec {w}-b\frac {\left ( \vec {w}\dotp \vec {u} \right ) }{\norm { \vec {u}} ^{2}}\vec {u} =aT_{\vec {u}}\left ( \vec {v}\right ) +bT_{\vec {u}}\left ( \vec {w} \right ) \]
Let \(\vec {u}\) be a fixed vector. The function \(T_{\vec {u}}\) defined by \(T_{\vec {u}}\vec {v}=\vec {u}+\vec {v}\) has the effect of translating all vectors by adding \(\vec {u}\neq \vec {0}\). Show that \(T_{\vec {u}}\) is not a linear transformation.

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Linear transformations take \(\vec {0}\) to \(\vec {0}\) which \(T\) does not. Also \(T_{\vec {a}}\left ( \vec {u}+\vec {v}\right ) \neq T_{\vec {a}}\vec {u}+T_{\vec {a}} \vec {v}\).
Find the matrix for the linear transformation which rotates every vector in \(\mathbb {R}^{2}\) through an angle of \(\pi /3.\)

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\[\left [ \begin{array}{cc} \cos \left ( \frac {\pi }{3}\right ) & -\sin \left ( \frac {\pi }{3}\right ) \\ \sin \left ( \frac {\pi }{3}\right ) & \cos \left ( \frac {\pi }{3}\right )\end{array} \right ] = \left [ \begin{array}{cc} \frac {1}{2} & -\frac {1}{2}\sqrt {3} \\ \frac {1}{2}\sqrt {3} & \frac {1}{2} \end{array} \right ] \]
Find the matrix for the linear transformation which rotates every vector in \(\mathbb {R}^{2}\) through an angle of \(2\pi /3\) and then reflects across the \(x\) axis.

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\[ \left [ \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \right ] \left [ \begin{array}{cc} \cos \left ( \frac {2\pi }{3}\right ) & -\sin \left ( \frac {2\pi }{3}\right ) \\ \sin \left ( \frac {2\pi }{3}\right ) & \cos \left ( \frac {2\pi }{3}\right ) \end{array} \right ] = \left [ \begin{array}{cc} -\frac {1}{2} & -\frac {1}{2}\sqrt {3} \\ -\frac {1}{2}\sqrt {3} & \frac {1}{2} \end{array} \right ] \]
Find the matrix for the linear transformation which rotates every vector in \(\mathbb {R}^{2}\) through an angle of \(\pi /6\) and then reflects across the \(x\) axis followed by a reflection across the \(y\) axis.

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\[ \left [ \begin{array}{rr} -1 & 0 \\ 0 & 1 \end{array} \right ] \left [ \begin{array}{cc} \cos \left ( \frac {\pi }{6}\right ) & -\sin \left ( \frac {\pi }{6}\right ) \\ \sin \left ( \frac {\pi }{6}\right ) & \cos \left ( \frac {\pi }{6}\right ) \end{array} \right ] = \left [ \begin{array}{cc} -\frac {1}{2}\sqrt {3} & \frac {1}{2} \\ \frac {1}{2} & \frac {1}{2}\sqrt {3} \end{array} \right ] \]
Let \(T\) be a linear transformation induced by the matrix \(A = \left [ \begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right ]\) and let \(S\) be a linear transformation induced by \(B = \left [ \begin{array}{rr} 0 & -2 \\ 4 & 2 \end{array}\right ]\). Find matrix of \(S \circ T\) and find \(\left ( S \circ T \right ) \left ( \vec {x} \right )\) for \(\vec {x} = \left [ \begin{array}{r} 2 \\ -1 \end{array} \right ]\).

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The matrix of \(S \circ T\) is given by \(BA\).
\[ \left [ \begin{array}{rr} 0 & -2 \\ 4 & 2 \end{array}\right ] \left [ \begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right ] = \left [ \begin{array}{rr} 2 & -4 \\ 10 & 8 \end{array} \right ] \]
Now, \(\left ( S \circ T \right ) \left ( \vec {x} \right ) = (BA) \vec {x}\).
\[ \left [ \begin{array}{rr} 2 & -4 \\ 10 & 8 \end{array} \right ] \left [ \begin{array}{r} 2 \\ -1 \end{array} \right ] = \left [ \begin{array}{r} 8 \\ 12 \end{array} \right ] \]
Let \(T\) be a linear transformation and suppose \(T \left ( \left [ \begin{array}{r} 1 \\ -4 \end{array} \right ] \right ) = \left [ \begin{array}{r} 2 \\ -3 \end{array} \right ]\). Suppose \(S\) is a linear transformation induced by the matrix \(B = \left [ \begin{array}{rr} 1 & 2 \\ -1 & 3 \end{array} \right ]\). Find \(\left ( S \circ T \right ) \left ( \vec {x} \right )\) for \(\vec {x} = \left [ \begin{array}{r} 1 \\ -4 \end{array} \right ]\).

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To find \(\left ( S \circ T \right ) \left ( \vec {x} \right )\) we compute \(S(T(\vec {x}))\).
\[ \left [ \begin{array}{rr} 1 & 2 \\ -1 & 3 \end{array} \right ] \left [ \begin{array}{r} 2 \\ -3 \end{array} \right ] = \left [ \begin{array}{r} -4 \\ -11 \end{array} \right ] \]
Let \(T\) be a linear transformation given by
\[ T \left [ \begin{array}{r} x\\ y \end{array}\right ] = \left [ \begin{array}{rrr} 1 &1 \\ 1 & 1 \end{array}\right ] \left [ \begin{array}{r} x\\ y \end{array}\right ] \]
Find a basis for \(\ker \left ( T\right )\) and a basis for \(\mbox {im} \left ( T\right ) \). Find the dimension of the kernel and the image of \(T\).

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A basis for \(\ker \left ( T\right )\) is \(\left \{ \left [ \begin{array}{r} 1 \\ -1 \end{array} \right ] \right \}\) and a basis for \(\mbox {im} \left ( T\right )\) is \(\left \{ \left [ \begin{array}{r} 1 \\ 1 \end{array} \right ] \right \}\).
There are many other possibilities for the specific bases. \(\dim \left ( \ker \left ( T\right ) \right )=1 \) and \(\dim \left ( \mbox {im} \left ( T\right ) \right )=1\).
Let \(T\) be a linear transformation given by
\[ T \left [ \begin{array}{r} x\\ y \end{array}\right ] = \left [ \begin{array}{rrr} 1 & 0 \\ 1 & 1 \end{array}\right ] \left [ \begin{array}{r} x\\ y \end{array}\right ] \]
Find a basis for \(\ker \left ( T\right )\) and a basis for \(\mbox {im} \left ( T\right ) \).

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In this case \(\ker \left ( T\right ) =\{0\}\) and \(\mbox {im} \left ( T\right ) = \mathbb {R}^2\) (pick any basis of \(\RR ^2\)).
Let \(T\) be a linear transformation given by
\[ T \left [ \begin{array}{r} x\\ y \\ z \end{array}\right ] = \left [ \begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right ] \left [ \begin{array}{r} x\\ y \\ z \end{array}\right ] \]
What is \(\dim ( \ker \left ( T \right ) )\)?

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We can easily see that \(\dim ( \mbox {im} \left ( T \right ) ) =1\), and thus \(\dim ( \ker \left ( T \right ) ) = 3 - \dim ( \mbox {im} \left ( T \right ) ) = 3- 1 = 2\).
Suppose \(T:\RR ^2\rightarrow \RR ^3\) is a linear transformation such that \(T\left (\begin{bmatrix}2\\-1\end{bmatrix}\right )=\begin{bmatrix}7\\-6\\1\end{bmatrix}\), and \(T\left (\begin{bmatrix}4\\1\end{bmatrix}\right )=\begin{bmatrix}-2\\5\\0\end{bmatrix}\). Find the image of \(\begin{bmatrix}-2\\-2\end{bmatrix}\) under \(T\).

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First, observe that \(\begin{bmatrix}-2\\-2\end{bmatrix}=\begin{bmatrix}2\\-1\end{bmatrix}-\begin{bmatrix}4\\1\end{bmatrix}\). Therefore,
\[T\left (\begin{bmatrix}-2\\-2\end{bmatrix}\right )=T\left (\begin{bmatrix}2\\-1\end{bmatrix}\right )-T\left (\begin{bmatrix}4\\1\end{bmatrix}\right )=\begin{bmatrix}7\\-6\\1\end{bmatrix}-\begin{bmatrix}-2\\5\\0\end{bmatrix}=\begin{bmatrix}9\\-11\\1\end{bmatrix}\]
Suppose \(T:\RR ^2\rightarrow \RR ^2\) is a linear transformation that maps \(\vec {i}\) to \(\begin{bmatrix}1\\1\end{bmatrix}\), and \(\vec {j}\) to \(\begin{bmatrix}-2\\3\end{bmatrix}\). Find vector \(\vec {v}\) such that \(T(\vec {v})=\begin{bmatrix}20\\-30\end{bmatrix}\).

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Observe that \(T\) is induced by the matrix \(A=\begin{bmatrix}1 & -2\\1 & 3\end{bmatrix}\). We are looking for \(\vec {v}\) such that \(\begin{bmatrix}1 & -2\\1 & 3\end{bmatrix}\vec {v}=\begin{bmatrix}20\\-30\end{bmatrix}\). Multiplying both sides by \(A^{-1}\) gives us
\[\vec {v}=\begin{bmatrix} 0.6 & 0.4\\ -0.2 & 0.2 \end{bmatrix}\begin{bmatrix}20\\-30\end{bmatrix}=\begin{bmatrix}0\\-10\end{bmatrix}\]

Bibliography

Some of the problems come from Chapter 5 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 272–315.