Exercises – Short Day

Exercises – Short Day#

Exercise 1: In the Image Space or Not?#

We are being informed that the image space $\mathrm{image}(L)$ of a linear map $L:\mathbb{R}^4 \to \mathbb{R}^3$ has an ordered basis $\beta$ given by

\[\begin{split}\beta=\left( \left[\begin{array}{r} 1 \\ 3\\ 2 \end{array}\right], \left[\begin{array}{r} 1 \\ -1\\ 2 \end{array}\right]\right). \end{split}\]

Question a#

Do the vectors

\[\begin{split}{\mathbf v}=\left[\begin{array}{r} 1 \\ 2\\ 3 \end{array}\right] \quad \text{and} \quad {\mathbf w}=\left[\begin{array}{r} 2 \\ 2\\ 4 \end{array}\right]\end{split}\]

belong to the image space $\mathrm{image}(L)$?

Hint

Can ${\mathbf v}$ be written as a linear combination of the vectors that appear in the given ordered basis? What about ${\mathbf w}$?

Hint

Does the equation

\[\begin{split}c_1\left[\begin{array}{r} 1 \\ 3\\ 2 \end{array}\right]+c_2\left[\begin{array}{r} 1 \\ -1\\ 2 \end{array}\right]={\mathbf v}\end{split}\]

have a solution? What about the

\[\begin{split}c_1\left[\begin{array}{r} 1 \\ 3\\ 2 \end{array}\right]+c_2\left[\begin{array}{r} 1 \\ -1\\ 2 \end{array}\right]={\mathbf w}? \end{split}\]

Answer

The vector $\mathbf v$ is not an element within $\mathrm{image}(L)$.

The vector $\mathbf w$ is an element within $\mathrm{image}(L)$.

Exercise 2: The Rank-Nullity Theorem in an Example#

A linear map $L:\mathbb{R}^3\rightarrow \mathbb{R}^3$ has with respect to the ordered standard basis $\epsilon$ for $\mathbb R^3$ the mapping matrix

\[\begin{split}{}_\epsilon[L]_\epsilon =\left[\begin{array}{rrr}1&2&1\\ 2&4&0\\ 3&6&0\end{array}\right]\,.\end{split}\]

We are being informed that $\dim (\mathrm{ker}(L))=1$. In other words, the kernel of $L$ has a dimension of $1$. Find, purely via mental math, an ordered basis for $\mathrm{image}(L)\,.$

Answer

Since the second column of the mapping matrix is twice the first, the remaining two columns span the column space of the matrix. Since we were informed that the kernel of $L$ has a dimension of $1$, the rank-nullity theorem implies that the image space of $L$ has a dimension of $2$. A possible ordered basis of the image space is thus

\[\begin{split}\left(\left[\begin{array}{r}1\\2\\3\end{array}\right],\left[\begin{array}{r}1\\0\\0\end{array}\right]\right).\end{split}\]

Exercise 3: Vector Equations and Image Space#

We define the following function:

\[\begin{split}M: \mathbb{R}^{2 \times 3} \to \mathbb{R}^{2 \times 2}\, , \quad {\mathbf A} \mapsto {\mathbf A} \cdot \left[\begin{array}{rr}1 & 2\\ 0 & 0\\ 2 & 1\end{array}\right].\end{split}\]

Question a#

Show that $M$ is a linear map between real vector spaces, and provide an ordered basis for $\mathrm{ker}(M).$

Hint

To show linearity, use Definition 11.0.1 and maybe first read Theorem 7.2.1, in particular part (v).

To find an ordered basis for $\mathrm{ker}(M)$, remember what a matrix ${\mathbf A} \in \mathbb{R}^{2 \times 3}$ must fulfill in order to be within $\mathrm{ker}(M)\,$.

Hint

\[\begin{split}\left[\begin{array}{rr}a & b & c\\ d & e & f\end{array}\right] \cdot \left[\begin{array}{rr}1 & 2\\ 0 & 0\\ 2 & 1\end{array}\right]=\left[\begin{array}{rr}a+2c & 2a+c\\ d+2f & 2d+f\end{array}\right].\end{split}\]

Answer

A possible basis for $\mathrm{ker}(M)$ is

\[\begin{split}\left( \left[\begin{array}{rr}0 & 1 & 0\\ 0 & 0 & 0\end{array}\right]\, , \left[\begin{array}{rr}0 & 0 & 0\\ 0 & 1 & 0\end{array}\right] \right).\end{split}\]

Question b#

Use the rank-nullity theorem for linear maps (Corollary 11.4.3 in the textbook) to show that $\dim(\mathrm{image}(M))=4$. Conclude that $\mathrm{image}(M)=\mathbb{R}^{2 \times 2}$.

Answer

If $V$ is a vector space with a finite dimension $n$ and $W$ is a subspace of $V$ of the same dimension $n$, then $W=V$. Hence, $\mathrm{image}(M)=\mathbb{R}^{2 \times 2}$.

Question c#

Justify that the equation $M({\mathbf A})=\left[\begin{array}{rr}3 & 3\\ 3 & 3\end{array}\right]$ has a solution based on your result from Question b.

Now, find the solution set.

Hint

Theorem 11.4.1 explains when an equation of the given type has a solution. The structure of the solution set is also described in Theorem 11.4.1.

Answer

The solution set is

\[\begin{split}\left\{ \left[\begin{array}{rr}1 & 0 & 1\\ 1 & 0 & 1\end{array}\right]+t_1 \cdot \left[\begin{array}{rr}0 & 1 & 0\\ 0 & 0 & 0\end{array}\right]+t_2 \cdot \left[\begin{array}{rr}0 & 0 & 0\\ 0 & 1 & 0\end{array}\right] \, \mid \, t_1,t_2 \in \mathbb{R}\right\}.\end{split}\]

Note that the expression

\[\begin{split}\left[\begin{array}{rr}1 & 0 & 1\\ 1 & 0 & 1\end{array}\right]+t_1 \cdot \left[\begin{array}{rr}0 & 1 & 0\\ 0 & 0 & 0\end{array}\right]+t_2 \cdot \left[\begin{array}{rr}0 & 0 & 0\\ 0 & 1 & 0\end{array}\right], \quad t_1,t_2 \in \mathbb{R}\end{split}\]

is termed the general solution of the equation.

Exercise 4: Investigation of a Linear Map#

Define $V_1$ and $V_2$ to be the sets of polynomials in $\mathbb{R}[Z]$ of degree no higher than three, respectively two. Note that both $V_1$ and $V_2$ are real vector spaces. Let the map $M:V_1 \to V_2$ be given by the expression

\[ M(a+bZ+cZ^2+dZ^3)=(a+b+3c+d)+(3a-b+2c+4d)Z+(2a+2b+6c+2d)Z^2\, , \text{where $a,b,c,d \in \mathbb{R}$}. \]

Question a#

We choose the ordered basis $\beta=(1,Z,Z^2,Z^3)$ for $V_1$ and the ordered basis $\gamma=(1,Z,Z^2)$ for $V_2$. Determine the mapping matrix ${}_\gamma[M]_\beta$.

Hint

Lemma 11.3.3 explains how to find the mapping matrix ${}_\gamma[M]_\beta$.

Answer

\[\begin{split} {}_\gamma[M]_\beta = \left[\begin{array}{rrrr} 1 & 1 & 3 & 1\\ 3 & -1 & 2 & 4\\ 2 & 2 & 6 & 2 \end{array}\right]. \end{split}\]

Question b#

Provide an ordered basis for the kernel of $M$.

Hint

First, find an ordered basis for the kernel of the mapping matrix ${}_\gamma[M]_\beta$. Then use the last part of Theorem 11.4.2.

Answer

The polynomials $-5/4-(7/4)Z+Z^2$ and $-5/4+(1/4)Z+Z^3$ constitute an ordered basis for $\mathrm{ker}(M)$.

Question c#

Provide an ordered basis for the image space of $M$.

Hint

First, find an ordered basis for the column space of the mapping matrix ${}_\gamma[M]_\beta$. Consider how this ordered basis can be used to find an ordered basis for $\mathrm{image}(M).$

Answer

The list $(1+3Z+2Z^2, 1-Z+2Z^2)$ is an ordered basis for $\mathrm{image}(M).$

Question d#

Investigate whether the polynomials

\[p_1(Z)=1+2Z+3Z^2 \quad \text{and} \quad p_2(Z)=2+2Z+4Z^2\]

belong to the image space $\mathrm{image}(M)$.

Hint

Use the ordered basis for the column space of $M$ found in Question c.

Hint

Does the equation $c_1\cdot(1+3Z+2Z^2)+c_2\cdot(1-Z+2Z^2) =p_1(Z)$ have a solution $c_1,c_2 \in \mathbb{R}$? What about the equation $c_1\cdot(1+3Z+2Z^2)+c_2\cdot(1-Z+2Z^2) =p_1(Z)$?

Answer

The polynomial $p_1(Z)$ is not an element within $\mathrm{image}(M)$. The polynomial $p_2(Z)$ is an element within $\mathrm{image}(M)$.

Exercise 5: Linear Maps and Differentiation#

As in Example 10.4.5 in the textbook we denote by $C_\infty (\mathbb{R})$ the real vector space that consists of all functions from $\mathbb{R}$ to $\mathbb{R}$ that can be differentiated an arbitrary number of times. In this exercise we consider the linear map $M: C_\infty (\mathbb{R}) \to C_\infty (\mathbb{R})$ defined by the expression $M(f)=f'+f$, where $f \in C_\infty (\mathbb{R})$ and where $f'$ denotes the derivative of the function $f$.

Question a#

We are being informed that $\dim \mathrm{ker}(M)=1$. Provide a basis for $\mathrm{ker}(M)$.

Hint

Since it is given that the kernel of $M$ is a $1$-dimensional subspace of $C_\infty$, then any element within non-zero $\mathrm{ker}(M)$ spans the entire kernel.

Hint

Try finding a non-zero element $f(t)$ in $\mathrm{ker}(M)$ of the form $f(t)=\mathrm e^{at}$ for some $a \in \mathbb{R}$.

Answer

Since $(\mathrm e^{at})'=a\mathrm e^{at}$, then $M(\mathrm e^{-t})=0$. From the first hint we have that $\{\mathrm e^{-t}\}$ is a basis for $\mathrm{ker}(M)$.

Question b#

Find the solution set to the following equations

$M(f)=\mathrm e^{t}$,
$M(f)=t$,
$M(f)=\mathrm e^{-t}$.

Hint

Theorem 11.4.1 explains the structure of the solution set: each solution is in the form ${\mathbf v}_p+{\mathbf v}$, where ${\mathbf v}_p$ is a particular solution and where ${\mathbf v} \in \mathrm{ker}(M)$.

Hint

In order to find a particular solution to the first equation, find an $a \in \mathbb{R}$ such that $f(t)=a\mathrm e^{t}$ fulfills the equation.

In order to find a particular solution to the second equation, try a function of the form$f(t)=t+a$, where $a \in \mathbb{R}$.

In order to find a particular solution to the third equation, try a function of the form $f(t)=at\mathrm e^{-t}$, where $a \in \mathbb{R}$.

Answer

The solution set is $\{(1/2)\mathrm e^{t}+c_1\cdot \mathrm e^{-t} \, \mid \, c_1 \in \mathbb{R}\}.$ It is also said that $(1/2)\mathrm e^{t}+c_1\cdot \mathrm e^{-t}, \ c_1 \in \mathbb{R}$ is the general solutoin to the equation $f'+f=\mathrm e^t$.
The solution set is $\{t-1+c_1\cdot \mathrm e^{-t} \, \mid \, c_1 \in \mathbb{R}\}.$ It is also said that $t-1+c_1\cdot \mathrm e^{-t}, \ c_1 \in \mathbb{R}$ is the general solution to the equation $f'+f=t$.
The solution set is $\{te^{-t}+c_1\cdot \mathrm e^{-t} \, \mid \, c_1 \in \mathbb{R}\}.$ It is also said that $t\mathrm e^{-t}+c_1\cdot \mathrm e^{-t}, \ c_1 \in \mathbb{R}$ is the general solution to the equation $f'+f=\mathrm e^{-t}$.

Exercise 6: Examples of Linear Maps#

Let $n$ be a natural number and $d$ an integer such that $0 \le d \le n$. Find an example of a linear map $L: \mathbb{C}^n \to \mathbb{C}^n$ between complex vector spaces such that $\dim(\mathrm{ker}(L))=d$ and $\dim(\mathrm{image}(L))=n-d$.

Exercises – Short Day

Contents

Exercises – Short Day#

Exercise 1: In the Image Space or Not?#

Question a#

Exercise 2: The Rank-Nullity Theorem in an Example#

Exercise 3: Vector Equations and Image Space#

Question a#

Question b#

Question c#

Exercise 4: Investigation of a Linear Map#

Question a#

Question b#

Question c#

Question d#

Exercise 5: Linear Maps and Differentiation#

Question a#

Question b#

Exercise 6: Examples of Linear Maps#