Additional Exercises for Ch 6

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Additional Exercises for Chapter 6: Linear Transformations

Review Exercises

Show the map $T:\mathbb {R}^{n}\rightarrow \mathbb {R}^{m}$ defined by $T\left ( \vec {x} \right ) =A\vec {x}$ where $A$ is an $m\times n$ matrix and $\vec {x}$ is an $m\times 1$ column vector is a linear transformation.

This result follows from the properties of matrix multiplication.

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.

Click on the arrow to see answer.

$\begin{eqnarray*} T_{\vec{u}}\left( a\vec{v}+b\vec{w}\right) &=&a\vec{v}+b\vec{w}-\frac{\left( a\vec{v}+b\vec{w}\right)\dotp \vec{u} }{\norm{ \vec{u} } ^{2}}\vec{u} \\ &=&a\vec{v}-a\frac{\left( \vec{v}\dotp \vec{u}\right) }{\norm{ \vec{u} } ^{2}}\vec{u}+b\vec{w}-b\frac{\left( \vec{w}\dotp \vec{u} \right) }{\norm{ \vec{u}} ^{2}}\vec{u} \\ &=&aT_{\vec{u}}\left( \vec{v}\right) +bT_{\vec{u}}\left( \vec{w} \right) \end{eqnarray*}$

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 this is not a linear transformation. Explain why it is not possible to represent $T_{\vec {u}}$ in $\mathbb {R}^{3}$ by multiplying by a $3\times 3$ matrix.

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 $\pi /3.$

Click on the arrow to see answer.

$\left [ \begin {array}{cc} \cos \left ( \frac {\pi }{3}\right ) & -\sin \left ( \frac {\pi }{3}\right ) \\ \sin \left ( \frac {\pi }{3}\right ) & \cos \left ( \frac {\pi }{3}\right )\end {array} \right ] = \left [ \begin {array}{cc} \frac {1}{2} & -\frac {1}{2}\sqrt {3} \\ \frac {1}{2}\sqrt {3} & \frac {1}{2} \end {array} \right ]$

Find the matrix for the linear transformation which rotates every vector in $\mathbb {R}^{2}$ through an angle of $\pi /4.$

Click on the arrow to see answer.

$\left [ \begin {array}{cc} \cos \left ( \frac {\pi }{4}\right ) & -\sin \left ( \frac {\pi }{4}\right ) \\ \sin \left ( \frac {\pi }{4}\right ) & \cos \left ( \frac {\pi }{4}\right ) \end {array} \right ] = \left [ \begin {array}{cc} \frac {1}{2}\sqrt {2} & -\frac {1}{2}\sqrt {2} \\ \frac {1}{2}\sqrt {2} & \frac {1}{2}\sqrt {2} \end {array} \right ]$

Find the matrix for the linear transformation which rotates every vector in $\mathbb {R}^{2}$ through an angle of $-\pi /3.$

Click on the arrow to see answer.

$\left [ \begin {array}{cc} \cos \left ( -\frac {\pi }{3}\right ) & -\sin \left ( -\frac {\pi }{3}\right ) \\ \sin \left ( -\frac {\pi }{3}\right ) & \cos \left ( -\frac {\pi }{3}\right ) \end {array} \right ] = \left [ \begin {array}{cc} \frac {1}{2} & \frac {1}{2}\sqrt {3} \\ -\frac {1}{2}\sqrt {3} & \frac {1}{2} \end {array} \right ]$

Find the matrix for the linear transformation which rotates every vector in $\mathbb {R}^{2}$ through an angle of $2\pi /3.$

Click on the arrow to see answer.

Find the matrix for the linear transformation which rotates every vector in $\mathbb {R}^{2}$ through an angle of $\pi /12.$

Note that $\pi /12=\pi /3-\pi /4.$

Click on the arrow to see answer.

$\begin{eqnarray*} &&\left[ \begin{array}{cc} \cos \left( \frac{\pi }{3}\right) & -\sin \left( \frac{\pi }{3}\right) \\ \sin \left( \frac{\pi }{3}\right) & \cos \left( \frac{\pi }{3}\right) \end{array} \right] \left[ \begin{array}{cc} \cos \left( -\frac{\pi }{4}\right) & -\sin \left( -\frac{\pi }{4}\right) \\ \sin \left( -\frac{\pi }{4}\right) & \cos \left( -\frac{\pi }{4}\right) \end{array} \right] \\ &=&\left[ \begin{array}{cc} \frac{1}{4}\sqrt{2}\sqrt{3}+\frac{1}{4}\sqrt{2} & \frac{1}{4}\sqrt{2}-\frac{1 }{4}\sqrt{2}\sqrt{3} \\ \frac{1}{4}\sqrt{2}\sqrt{3}-\frac{1}{4}\sqrt{2} & \frac{1}{4}\sqrt{2}\sqrt{3} +\frac{1}{4}\sqrt{2} \end{array} \right] \end{eqnarray*}$

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.

Click on the arrow to see answer.

$\left [ \begin {array}{rr} 1 & 0 \\ 0 & -1 \end {array} \right ] \left [ \begin {array}{cc} \cos \left ( \frac {2\pi }{3}\right ) & -\sin \left ( \frac {2\pi }{3}\right ) \\ \sin \left ( \frac {2\pi }{3}\right ) & \cos \left ( \frac {2\pi }{3}\right ) \end {array} \right ] = \left [ \begin {array}{cc} -\frac {1}{2} & -\frac {1}{2}\sqrt {3} \\ -\frac {1}{2}\sqrt {3} & \frac {1}{2} \end {array} \right ]$

Find the matrix for the linear transformation which rotates every vector in $\mathbb {R}^{2}$ through an angle of $\pi /3$ and then reflects across the $x$ axis.

Click on the arrow to see answer.

$\left [ \begin {array}{rr} 1 & 0 \\ 0 & -1 \end {array} \right ] \left [ \begin {array}{cc} \cos \left ( \frac {\pi }{3}\right ) & -\sin \left ( \frac {\pi }{3}\right ) \\ \sin \left ( \frac {\pi }{3}\right ) & \cos \left ( \frac {\pi }{3}\right ) \end {array} \right ] = \left [ \begin {array}{cc} \frac {1}{2} & -\frac {1}{2}\sqrt {3} \\ -\frac {1}{2}\sqrt {3} & -\frac {1}{2} \end {array} \right ]$

Find the matrix for the linear transformation which rotates every vector in $\mathbb {R}^{2}$ through an angle of $\pi /4$ and then reflects across the $x$ axis.

Click on the arrow to see answer.

$\left [ \begin {array}{rr} 1 & 0 \\ 0 & -1 \end {array} \right ] \left [ \begin {array}{cc} \cos \left ( \frac {\pi }{4}\right ) & -\sin \left ( \frac {\pi }{4}\right ) \\ \sin \left ( \frac {\pi }{4}\right ) & \cos \left ( \frac {\pi }{4}\right ) \end {array} \right ] = \left [ \begin {array}{cc} \frac {1}{2}\sqrt {2} & -\frac {1}{2}\sqrt {2} \\ -\frac {1}{2}\sqrt {2} & -\frac {1}{2}\sqrt {2} \end {array} \right ]$

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.

Click on the arrow to see answer.

$\left [ \begin {array}{rr} -1 & 0 \\ 0 & 1 \end {array} \right ] \left [ \begin {array}{cc} \cos \left ( \frac {\pi }{6}\right ) & -\sin \left ( \frac {\pi }{6}\right ) \\ \sin \left ( \frac {\pi }{6}\right ) & \cos \left ( \frac {\pi }{6}\right ) \end {array} \right ] = \left [ \begin {array}{cc} -\frac {1}{2}\sqrt {3} & \frac {1}{2} \\ \frac {1}{2} & \frac {1}{2}\sqrt {3} \end {array} \right ]$

Find the matrix for the linear transformation which reflects every vector in $\mathbb {R}^{2}$ across the $x$ axis and then rotates every vector through an angle of $\pi /4$ .

Click on the arrow to see answer.

$\left [ \begin {array}{cc} \cos \left ( \frac {\pi }{4}\right ) & -\sin \left ( \frac {\pi }{4}\right ) \\ \sin \left ( \frac {\pi }{4}\right ) & \cos \left ( \frac {\pi }{4}\right ) \end {array} \right ] \left [ \begin {array}{rr} 1 & 0 \\ 0 & -1 \end {array} \right ] = \left [ \begin {array}{cc} \frac {1}{2}\sqrt {2} & \frac {1}{2}\sqrt {2} \\ \frac {1}{2}\sqrt {2} & -\frac {1}{2}\sqrt {2} \end {array} \right ]$

Find the matrix for the linear transformation which reflects every vector in $\mathbb {R}^{2}$ across the $y$ axis and then rotates every vector through an angle of $\pi /4$ .

Click on the arrow to see answer.

$\left [ \begin {array}{cc} \cos \left ( \frac {\pi }{4}\right ) & -\sin \left ( \frac {\pi }{4}\right ) \\ \sin \left ( \frac {\pi }{4}\right ) & \cos \left ( \frac {\pi }{4}\right ) \end {array} \right ] \left [ \begin {array}{rr} -1 & 0 \\ 0 & 1 \end {array} \right ] = \left [ \begin {array}{cc} -\frac {1}{2}\sqrt {2} & -\frac {1}{2}\sqrt {2} \\ -\frac {1}{2}\sqrt {2} & \frac {1}{2}\sqrt {2} \end {array} \right ]$

Find the matrix for the linear transformation which reflects every vector in $\mathbb {R}^{2}$ across the $x$ axis and then rotates every vector through an angle of $\pi /6$ .

Click on the arrow to see answer.

$\left [ \begin {array}{cc} \cos \left ( \frac {\pi }{6}\right ) & -\sin \left ( \frac {\pi }{6}\right ) \\ \sin \left ( \frac {\pi }{6}\right ) & \cos \left ( \frac {\pi }{6}\right ) \end {array} \right ] \left [ \begin {array}{rr} 1 & 0 \\ 0 & -1 \end {array} \right ] = \left [ \begin {array}{cc} \frac {1}{2}\sqrt {3} & \frac {1}{2} \\ \frac {1}{2} & -\frac {1}{2}\sqrt {3} \end {array} \right ]$

Find the matrix for the linear transformation which reflects every vector in $\mathbb {R}^{2}$ across the $y$ axis and then rotates every vector through an angle of $\pi /6$ .

Click on the arrow to see answer.

$\left [ \begin {array}{cc} \cos \left ( \frac {\pi }{6}\right ) & -\sin \left ( \frac {\pi }{6}\right ) \\ \sin \left ( \frac {\pi }{6}\right ) & \cos \left ( \frac {\pi }{6}\right ) \end {array} \right ] \left [ \begin {array}{rr} -1 & 0 \\ 0 & 1 \end {array} \right ] = \left [ \begin {array}{cc} -\frac {1}{2}\sqrt {3} & -\frac {1}{2} \\ -\frac {1}{2} & \frac {1}{2}\sqrt {3} \end {array} \right ]$

Find the matrix for the linear transformation which rotates every vector in $\mathbb {R}^{2}$ through an angle of $5\pi /12.$

Note that $5\pi /12=2\pi /3-\pi /4.$

Click on the arrow to see answer.

$\left [ \begin {array}{cc} \cos \left ( \frac {2\pi }{3}\right ) & -\sin \left ( \frac {2\pi }{3}\right ) \\ \sin \left ( \frac {2\pi }{3}\right ) & \cos \left ( \frac {2\pi }{3}\right ) \end {array} \right ] \left [ \begin {array}{cc} \cos \left ( -\frac {\pi }{4}\right ) & -\sin \left ( -\frac {\pi }{4}\right ) \\ \sin \left ( -\frac {\pi }{4}\right ) & \cos \left ( -\frac {\pi }{4}\right ) \end {array} \right ] =$ $\left [ \begin {array}{cc} \frac {1}{4}\sqrt {2}\sqrt {3}-\frac {1}{4}\sqrt {2} & -\frac {1}{4}\sqrt {2}\sqrt {3 }-\frac {1}{4}\sqrt {2} \\ \frac {1}{4}\sqrt {2}\sqrt {3}+\frac {1}{4}\sqrt {2} & \frac {1}{4}\sqrt {2}\sqrt {3} -\frac {1}{4}\sqrt {2} \end {array} \right ]$ Note that it doesn’t matter about the order in this case.

Let $T$ be a linear transformation induced by the matrix $A = \left [ \begin {array}{rr} 3 & 1 \\ -1 & 2 \end {array}\right ]$ and $S$ 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 $\vec {x} = \left [ \begin {array}{r} 2 \\ -1 \end {array} \right ]$ .

Click on the arrow to see answer.

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 $\vec {x} = \left [ \begin {array}{r} 1 \\ -4 \end {array} \right ]$ .

Click on the arrow to see answer.

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 induced by the matrix $A = \left [ \begin {array}{rr} 2 & 3 \\ 1 & 1 \end {array}\right ]$ and $S$ a linear transformation induced by $B = \left [ \begin {array}{rr} -1 & 3 \\ 1 & -2 \end {array}\right ]$ . Find matrix of $S \circ T$ and find $\left ( S \circ T \right ) \left ( \vec {x} \right )$ for $\vec {x} = \left [ \begin {array}{r} 5 \\ 6 \end {array} \right ]$ .

Let $T$ be a linear transformation induced by the matrix $A = \left [ \begin {array}{rr} 2 & 1 \\ 5 & 2 \end {array} \right ]$ . Find the matrix of $T^{-1}$ .

Click on the arrow to see answer.

The matrix of $T^{-1}$ is $A^{-1}$ . $\left [ \begin {array}{rr} 2 & 1 \\ 5 & 2 \end {array} \right ] ^{-1} = \left [ \begin {array}{rr} -2 & 1 \\ 5 & -2 \end {array} \right ]$

Let $T$ be a linear transformation induced by the matrix $A = \left [ \begin {array}{rr} 4 & -3 \\ 2 & -2 \end {array} \right ]$ . Find the matrix of $T^{-1}$ .

Let $T$ be a linear transformation and suppose $T \left ( \left [ \begin {array}{r} 1 \\ 2 \end {array}\right ] \right ) = \left [ \begin {array}{r} 9 \\ 8 \end {array} \right ]$ , $T \left ( \left [ \begin {array}{r} 0 \\ -1 \end {array}\right ] \right ) = \left [ \begin {array}{r} -4 \\ -3 \end {array}\right ]$ . Find the matrix of $T^{-1}$ .

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 $\mbox {im} \left ( T\right )$ .

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, but in this case $\dim \left ( \ker \left ( T\right ) \right )=1$ 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 $\mbox {im} \left ( T\right )$ .

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 $\mathbb {R}^2$ ).

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 $\dim ( \ker \left ( T \right ) ) = 3 - \dim ( \mbox {im} \left ( T \right ) ) = 3- 1 = 2$ .

Challenge Exercises

Argue geometrically to prove that the following transformations are linear:

(a): Rotation of the plane about the origin through angle $\theta$ .
(b): Reflection of the plane about the line $y=mx$ .

Think in terms of linearity diagrams.

What happens when you rotate two vectors first, then add them, versus adding the two vectors first, then rotating the sum?

Find the matrix of the linear transformation which rotates every vector in $\mathbb {R}^{3}$ counter clockwise about the $z$ axis when viewed from the positive $z$ axis through an angle of 30 $^{\circ }$ and then reflects through the $xy$ plane.

Click on the arrow to see answer.

$\left [ \begin {array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end {array} \right ] \left [ \begin {array}{ccc} \cos \left ( \frac {\pi }{6}\right ) & -\sin \left ( \frac {\pi }{6}\right ) & 0 \\ \sin \left ( \frac {\pi }{6}\right ) & \cos \left ( \frac {\pi }{6}\right ) & 0 \\ 0 & 0 & 1 \end {array} \right ] = \left [ \begin {array}{ccc} \frac {1}{2}\sqrt {3} & -\frac {1}{2} & 0 \\ \frac {1}{2} & \frac {1}{2}\sqrt {3} & 0 \\ 0 & 0 & -1 \end {array} \right ]$

Let $\vec {u}=\left [ \begin {array}{r} a \\ b \end {array} \right ]$ be a unit vector in $\mathbb {R}^{2}.$ Find the matrix which reflects all vectors across this vector, as shown in the following picture.

Notice that $\left [ \begin {array}{r} a \\ b \end {array} \right ] =\left [ \begin {array}{c} \cos \theta \\ \sin \theta \end {array} \right ]$ for some $\theta .$ First rotate through $-\theta .$ Next reflect through the $x$ axis. Finally rotate through $\theta$ .

$\begin{eqnarray*} &&\left[ \begin{array}{cc} \cos \left( \theta \right) & -\sin \left( \theta \right) \\ \sin \left( \theta \right) & \cos \left( \theta \right) \end{array} \right] \left[ \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \right] \left[ \begin{array}{cc} \cos \left( -\theta \right) & -\sin \left( -\theta \right) \\ \sin \left( -\theta \right) & \cos \left( -\theta \right) \end{array} \right] \\ &=& \left[ \begin{array}{cc} \cos ^{2}\theta -\sin ^{2}\theta & 2\cos \theta \sin \theta \\ 2\cos \theta \sin \theta & \sin ^{2}\theta -\cos ^{2}\theta% \end{array} \right] \end{eqnarray*}$

Now to write in terms of $\left ( a,b\right ) ,$ note that $a/\sqrt {a^{2}+b^{2} }=\cos \theta ,b/\sqrt {a^{2}+b^{2}}=\sin \theta .$ Now plug this in to the above. The result is $\left [ \begin {array}{cc} \frac {a^{2}-b^{2}}{a^{2}+b^{2}} & 2\frac {ab}{a^{2}+b^{2}} \\ 2\frac {ab}{a^{2}+b^{2}} & \frac {b^{2}-a^{2}}{a^{2}+b^{2}} \end {array} \right ] =\frac {1}{a^{2}+b^{2}}\left [ \begin {array}{cc} a^{2}-b^{2} & 2ab \\ 2ab & b^{2}-a^{2} \end {array} \right ]$ Since this is a unit vector, $a^{2}+b^{2}=1$ and so you get $\left [ \begin {array}{cc} a^{2}-b^{2} & 2ab \\ 2ab & b^{2}-a^{2} \end {array} \right ]$

Suppose $A$ is an $m\times n$ matrix and $B$ is an $n\times p$ matrix. Let $T_A$ and $T_B$ be linear transformations induced by $A$ and $B$ , respectively. Show that $\begin{equation*} \dim \left( \mbox{null} \left( AB\right) \right) \leq \dim \left( \mbox{null} \left( A\right) \right) +\dim \left( \mbox{null} \left( B\right) \right) . \end{equation*}$ Alternatively, $\begin{equation*} \dim \left( \mbox{ker} \left( T_A\circ T_B\right) \right) \leq \dim \left( \mbox{ker} \left( T_A\right) \right) +\dim \left( \mbox{ker} \left( T_B\right) \right) . \end{equation*}$

Consider the subspace, $T_B\left ( \mathbb {R}^{p}\right ) \cap \mbox {ker} \left ( T_A\right )$ and suppose a basis for this subspace is $\left \{ \vec {w}_{1},\cdots ,\vec {w}_{k}\right \} .$ Now suppose $\left \{ \vec {u}_{1},\cdots ,\vec {u}_{r}\right \}$ is a basis for $\mbox {ker} \left ( T_B\right ) .$ Let $\left \{ \vec {z}_{1},\cdots ,\vec {z}_{k}\right \}$ be such that $B\vec {z}_{i}=\vec {w}_{i}$ and argue that $\begin{equation*} \mbox{ker} \left( T_A\circ T_B\right) \subseteq \mbox{span}\left\{ \vec{u}_{1},\cdots , \vec{u}_{r},\vec{z}_{1},\cdots ,\vec{z}_{k}\right\} . \end{equation*}$

Suppose $AB\vec {x}=\vec {0}.$ Then $B\vec {x}\in \mbox {ker} \left ( T_A\right ) \cap T_B\left ( \mathbb {R}^{p}\right )$ and so $B\vec {x} =\sum _{i=1}^{k}B\vec {z}_{i}$ showing that $\vec {x}-\sum _{i=1}^{k}\vec {z}_{i}\in \mbox {ker} \left ( T_B\right )$ Consider $T_B\left ( \mathbb {R}^{p}\right ) \cap \mbox {ker} \left ( T_A\right )$ and let a basis be $\left \{ \vec {w}_{1},\cdots ,\vec {w}_{k}\right \} .$ Then each $\vec {w}_{i}$ is of the form $B\vec {z}_{i}=\vec {w}_{i}$ . Therefore, $\left \{ \vec {z}_{1},\cdots ,\vec {z}_{k}\right \}$ is linearly independent and $AB\vec {z}_{i}=0.$ Now let $\left \{ \vec {u} _{1},\cdots ,\vec {u}_{r}\right \}$ be a basis for $\mbox {ker} \left ( T_B\right ) .$ If $AB\vec {x}=\vec {0}$ , then $B\vec {x} \in \mbox {ker} \left ( T_A\right ) \cap T_B\left ( \mathbb {R}^{p}\right )$ and so $B\vec {x}=\sum _{i=1}^{k}c_{i}B\vec {z} _{i}$ which implies $\vec {x}-\sum _{i=1}^{k}c_{i}\vec {z}_{i}\in \mbox {ker} \left ( T_B\right )$ and so it is of the form $\vec {x}-\sum _{i=1}^{k}c_{i}\vec {z}_{i}=\sum _{j=1}^{r}d_{j}\vec {u} _{j}$ It follows that if $AB\vec {x}=\vec {0}$ so that $\vec {x}\in \mbox {ker} \left ( T_A\circ T_B\right ) ,$ then $\vec {x}\in \mbox {span}\left ( \vec {z}_{1},\cdots ,\vec {z}_{k}, \vec {u}_{1},\cdots ,\vec {u}_{r}\right ) .$ Therefore,

$\begin{eqnarray*} \dim \left( \mbox{ker} \left( T_A\circ T_B\right) \right) &\leq &k+r=\dim \left( T_B\left( \mathbb{R}^{p}\right) \cap \mbox{ker} \left( T_A\right) \right) +\dim \left( \mbox{ker} \left( T_B\right) \right) \\ &\leq &\dim \left( \mbox{ker} \left( T_A\right) \right) +\dim \left( \mbox{ker} \left( T_B\right) \right) \end{eqnarray*}$

Practice Problem Source

These 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.