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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.
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.
Click the arrow to see answer.
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 \(\pi /3.\)
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.
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.
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 \(\vec {x} = \left [ \begin{array}{r} 2 \\ -1 \end{array} \right ]\).
\[ T \left [ \begin{array}{r} x\\ y \end{array}\right ] = \left [ \begin{array}{rrr} 1 &1 \\ 1 & 1 \end{array}\right ] \left [ \begin{array}{r} x\\ y \end{array}\right ] \]
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
\(T\).
Click on the arrow to see answer.
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 \(\mbox {im} \left ( T\right ) \).
Click on the arrow to see answer.
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 \(\dim ( \ker \left ( T \right ) )\)?
Click on the arrow to see answer.
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 \(T\).
Click the arrow to see answer.
First, observe that \(\begin{bmatrix}-2\\-2\end{bmatrix}=\begin{bmatrix}2\\-1\end{bmatrix}-\begin{bmatrix}4\\1\end{bmatrix}\). Therefore,
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 \(T(\vec {v})=\begin{bmatrix}20\\-30\end{bmatrix}\).
Click the arrow to see answer.
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
