Exercises – Long Day

Exercises – Long Day#

Exercise 1: Linearity#

Two maps $f: \mathbb{R}^2 \to \mathbb{R}^2$ and $g: \mathbb{R}^2 \to \mathbb{R}^2$, both between real vector spaces, are given by:

\[\begin{split}f\left(\left[\begin{array}{r} x_1\\ x_2 \end{array}\right]\right)=\left[\begin{array}{r} x_1-x_2\\-x_1+x_2\end{array}\right] \quad \text{and} \quad g\left(\left[\begin{array}{r}x_1\\x_2\end{array}\right]\right)=\left[\begin{array}{r}-x_2\\x_1^{2}\end{array}\right]. \end{split}\]

Show that only one of the two maps is linear.

Hint

Check whether they satisfy the two linearity requirements in Definition 11.0.1 in the textbook.

Answer

It is the map $f$ that is linear. That $g$ is not linear is easily shown with a counterexample. For example, we see that $g\left(\left[\begin{array}{r}1\\1\end{array}\right]\right)+g\left(\left[\begin{array}{r}1\\1\end{array}\right]\right)$ is not the same as $g\left(\left[\begin{array}{r}1\\1\end{array}\right]+\left[\begin{array}{r}1\\1\end{array}\right]\right)$.

Exercise 2: Linear maps defined by diagonal matrices#

Let $\lambda_1,\lambda_2 \in \mathbb{R}$ and define

\[\begin{split}{\mathbf A}=\left[\begin{array}{rr} \lambda_1 & 0\\ 0 & \lambda_2\end{array}\right].\end{split}\]

In this exercise we wish to investigate what effect the linear map $L_{\mathbf A}: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $L_{\mathbf A}\left({\mathbf v}\right)={\mathbf A}\cdot {\mathbf v}$ has on the set $\mathcal{K}=\{(v_1,v_2) \, \mid \, 0 \le v_1 \le 1, \, 0 \le v_2 \le 1\}$.

Question a#

Sketch the set $\mathcal{K}$ in $\mathbb{R}^2$.

Answer

The set $\mathcal{K}$ is a square in $\mathbb{R}^2$ with corners $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$.

Question b#

Sketch the set $L_{\mathbf A}(\mathcal{K})$, meaning the set $\{L_{\mathbf A}({\mathbf v}) \, \mid \, {\mathbf v} \in \mathcal{K} \}$, in the following cases:

For $\lambda_1=1$ and $\lambda_2=1$.
For $\lambda_1=2$ and $\lambda_2=3$.
For $\lambda_1=-1$ and $\lambda_2=2$.
For $\lambda_1=-4$ and $\lambda_2=-3$.

Hint

The vectors ${\mathbf v}=(v_1,v_2) \in \mathcal{K}$ satisfy

\[\begin{split} \left[\begin{array}{r}v_1\\v_2\end{array}\right] = v_1\cdot \left[\begin{array}{r}1\\0\end{array}\right]+v_2\cdot \left[\begin{array}{r}0\\1\end{array}\right]\, , \quad 0 \le v_1 \le 1 \ \text{and} \ 0 \le v_2 \le 1. \end{split}\]

Now use the fact that $L_{\mathbf A}$ is a linear map.

Answer

It holds that

\[\begin{split} L_{\mathbf A}\left(\left[\begin{array}{r}v_1\\v_2\end{array}\right]\right) = v_1\cdot \left[\begin{array}{r}\lambda_1\\0\end{array}\right]+v_2\cdot \left[\begin{array}{r}0\\\lambda_2\end{array}\right]\, , \quad 0 \le v_1 \le 1 \ \text{and} \ 0 \le v_2 \le 1. \end{split}\]

In all cases $L_{\mathbf A}(\mathcal{K})$ is thus a rectangle with corners $(0,0),(\lambda_1,0),(0,\lambda_2),(\lambda_1,\lambda_2)$.

Question c#

With $\mathcal{K}$ and the four matrices ${\mathbf A} \in \mathbb{R}^{2 \times 2}$ from Question b, check that the area of $L_{\mathbf A}(\mathcal{K})$ equals $|\det({\mathbf A})|$.

Note: In fact, it can be shown that the area of $L_{\mathbf A}(\mathcal{K})$ equals $|\det({\mathbf A})|$ for any matrix ${\mathbf A} \in \mathbb{R}^{2 \times 2}$.

Exercise 3: Linear maps and differentiation#

As in Example 10.4.5 in the textbook, the notation $C_\infty(\mathbb R)$ denotes a real vector space consisting of all functions from $\mathbb{R}$ to $\mathbb{R}$ that can be differentiated any number of times. In this exercise we consider the subspace $V_1$ of $C_\infty(\mathbb{R})$ that is spanned by the functions $\mathrm e^t, \mathrm e^{-t}, \cos(t)$ and $\sin(t)$. You may use the fact that these four functions are linearly independent over $\mathbb{R}$ and hence that they constitute an ordered basis for $V_1$.

We now define a function $L: V_1 \to V_1$ by the expression $L(f)=f'+f$, where $f \in V_1$ and where $f'$ denotes the derivative of the function $f$.

Question a#

Show that the function $L$ is a linear map between real vector spaces.

Question b#

Determine the mapping matrix ${}_\beta [L] _\beta$, given the ordered basis $\beta=(\mathrm e^t,\mathrm e^{-t},\cos(t),\sin(t))$.

Hint

The formula for a mapping matrix is found in Definition 11.3.1 in the textbook.

Hint

Definition 11.3.1 concerns two vector spaces $V_1$ and $V_2$ as well as two ordered bases, one for each vector space. Note that we here have $V_1=V_2$, and also note that the ordered basis $\beta$ is used for both $V_1$ and $V_2$.

Answer

\[\begin{split}{}_\beta [L] _\beta= \left[\begin{array}{rrrr} 2 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & -1 & 1 \end{array}\right].\end{split}\]

Exercise 4: Change of basis and mapping matrices#

We are given the vectors

\[\begin{split}\mathbf v_1 = \left[\begin{array}{r} 1\\ 2\end{array}\right] \text{ and } \mathbf v_2=\left[\begin{array}{r} 3\\ 7\end{array}\right] \,\text{ from $\mathbb{R}^2$},\end{split}\]

as well as

\[\begin{split}\mathbf w_1 = \left[\begin{array}{r} 1\\ 2\\ 2\end{array}\right], \mathbf w_2 = \left[\begin{array}{r} 2\\ 3\\1\end{array}\right] \text{ and } \mathbf w_3 = \left[\begin{array}{r} 1\\ 2\\1\end{array}\right] \,\text{ from $\mathbb{R}^3$}.\end{split}\]

We now define the ordered bases

\[\begin{split}\beta=\left(\mathbf v_1,\mathbf v_2\right) \quad \text{and} \quad \epsilon=\left( \left[\begin{array}{r} 1\\ 0\end{array}\right],\left[\begin{array}{r} 0\\ 1\end{array}\right]\right) \text{ for $\mathbb{R}^2$}\end{split}\]

as well as

\[\begin{split}\gamma=\left(\mathbf w_1,\mathbf w_2,\mathbf w_3\right) \quad \text{and} \quad \eta=\left( \left[\begin{array}{r} 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{r} 0\\ 1\\0\end{array}\right],\left[\begin{array}{r} 0\\ 0\\1\end{array}\right]\right) \text{ for $\mathbb{R}^3$}.\end{split}\]

A linear map $L:\mathbb{R}^2\to \mathbb{R}^3\,$ satisfies

\[L(\mathbf{v}_1)=\mathbf{w}_1+\mathbf{w}_2-3\mathbf{w}_3 \quad \mathrm{and} \quad L(\mathbf{v}_2)=\mathbf{w}_1-\mathbf{w}_2-2\mathbf{w}_3\,.\]

We wish to find the mapping matrix ${}_\eta[L]_\epsilon$, meaning the mapping matrix of $L$ with respect to the ordered standard bases.

Question a#

Start by finding the mapping matrix ${}_\gamma[L]_\beta$ of $L$.

Hint

The formula in Definition 11.3.1 can be of use here. Also remember to use the given information about $L$ so that you can avoid doing too many calculations.

Answer

\[\begin{split}{}_\gamma[L]_\beta =\left[\begin{array}{rr} 1 & 1 \\ 1 & -1 \\ -3 & -2 \end{array}\right].\end{split}\]

Question b#

Use your answer above to find the mapping matrix ${}_\eta[L]_\beta$ of $L$.

Hint

Instead of starting from the definition of a mapping matrix, you can use the second part of Theorem 11.3.3 in the textbook to realise that ${}_\eta[\mathrm{id}_{\mathbb{R}^3}]_\gamma \cdot {}_\gamma [L] _\beta = {}_\eta [L] _\beta$.

Hint

Finding the change-of-basis matrix ${}_\eta[\mathrm{id}_{\mathbb{R}^3}]_\gamma$ is not that difficult as $\eta$ is the ordered standard basis for $\mathbb{R}^3$.

Answer

\[\begin{split}{}_\eta[L]_\beta =\left[\begin{array}{rr} 0 & -3 \\ -1 & -5 \\ 0 & -1 \end{array}\right].\end{split}\]

Question c#

And now for the final step, use your above answer to find the mapping matrix ${}_\eta[L]_\epsilon$ of $L$.

Hint

A consequence of Theorem 11.3.3 in the textbook is that ${}_\eta[L]_\epsilon={}_\eta[L]_\beta \cdot {}_\beta[\mathrm{id}_{\mathbb{R}^2}]_\epsilon$.

The change-of-basis matrix ${}_\beta[\mathrm{id}_{\mathbb{R}^2}]_\epsilon$ can be found using Lemma 11.3.5, since it implies that ${}_\beta[\mathrm{id}_{\mathbb{R}^2}]_\epsilon=({}_\epsilon[\mathrm{id}_{\mathbb{R}^2}]_\beta)^{-1}$. Note that the matrix ${}_\epsilon[\mathrm{id}_{\mathbb{R}^2}]_\beta$ is easy to construct without calculations.

Answer

\[\begin{split}{}_\eta[L]_\epsilon =\left[\begin{array}{rr} 6 & -3 \\ 3 & -2 \\ 2 & -1 \end{array}\right].\end{split}\]

Exercise 5: Kernel of a linear map#

Let $V_1$ and $V_2$ be two vector spaces over a field $\mathbb{F}$. Show that the kernel of a linear map $L: V_1 \to V_2$ is a subspace of $V_1$.

Hint

The kernel of a linear map is defined in Definition 11.2.1 in the textbook. Lemma 10.4.2 can be of use to check whether it is a subspace of $V_1$.

Exercise 6: Linear map from a vector space of polynomials#

Let $V_1=\{aZ^2+bZ+c \, \mid \, a,b,c \in \mathbb{R}\}$ be a subspace of the real vector space $\mathbb{R}[Z]$ consisting of all polynomials of degree no higher than two. We define a function $L: V_1 \to \mathbb{R}$ by the expression $p(Z) \mapsto p'(1)$. The expression $p'(1)$ is the number you obtain from substituting $1$ into $p'(Z)$, where $p'(Z)$ denotes the derivative of $p(Z)$.

Question a#

Calculate $L(Z^2+Z)$ and $L(5Z^2-10Z+4)$.

Answer

Since $(Z^2+Z)'=2Z+1$, it holds that $L(Z^2+Z)=2\cdot 1 +1=3$.

Likewise it holds that $(5Z^2-10Z+4)'=10Z-10$ and hence that $L(5Z^2-10Z+4)=10\cdot 1-10=0$.

Question b#

Show that $L$ is a linear map between real vector spaces.

Hint

The conditions for being a linear map can be found in Definition 11.0.1 in the textbook.

Question c#

Provide a basis for $\mathrm{ker}(L)$.

Hint

If $L(aZ^2+bZ+c)=0$, which condition must the coefficients $a$, $b$ and $c$ then satisfy?

Answer

A possible basis is $\{1,Z^2-2Z\}$.

Question d#

Show that $\mathrm{image}(L)=\mathbb{R}$. Conclude that the map $L$ is surjective.

Hint

First try finding a polynomial $p(Z)$ in $V_1$ that satisfies $L(p(Z))=1$. Then reflect on what $L(ap(Z))$ can then be, where $a \in \mathbb{R}$ is chosen freely.

Hint

Saying that a linear map $M:V_1 \to V_2$ is surjective is precisely the same as saying that the image space of the map is identical to the entire $V_2$. Have a look at the text before Example 2.2.4 in the textbook where it is explained what it means for a function to be surjective.

If we thus have shown that $\mathrm{image}(L)=\mathbb{R}$, then it follows directly that $L$ is surjective.

Answer

$L(aZ)=a$ for an arbitrary $a \in \mathbb{R}$. Hence $\mathrm{image}(L)=\mathbb{R}$.

In other words, the image space of $L$ equals its codomain. This means that $L$ is surjective.

Exercise 7: Injective linear maps#

A function $f: A \to B$ is called injective if and only if, for all $a_1,a_2 \in A$, $f(a_1)=f(a_2)$ implies $a_1=a_2$. Reread the text before Example 2.2.4 in the textbook for a review of this concept.

In this exercise we will investigate what it means for a linear map $L: V_1 \to V_2$ between two vector spaces $V_1$ and $V_2$ over $\mathbb{F}$ to be injective.

Question a#

Show that if $L$ is injective, then $\mathrm{ker}(L)=\{{\mathbf 0}\}$.

Hint

By the contrapositive, this proposition is logically equivalent to the proposition:

If $\mathrm{ker}(L) \neq \{{\mathbf 0}\}$, then $L$ is not injective.

Hint

We know that $L({\mathbf 0})={\mathbf 0}$ for all linear maps $L$ (see Equation (11.1) in the textbook). If $\mathrm{ker}(L) \neq \{{\mathbf 0}\}$, then a non-zero ${\mathbf v} \in V_1$ exists such that $L({\mathbf v})={\mathbf 0}$.

Partial Answer

If $\mathrm{ker}(L) \neq \{{\mathbf 0}\}$, then non-zero ${\mathbf v} \in V_1$ exist such that $L({\mathbf v})={\mathbf 0}$. Hence $L$ maps both ${\mathbf v}$ and ${\mathbf 0}$ to the zero vector in $V_2$. This means that $L$ is not injective.

Question b#

Show that if $\mathrm{ker}(L)=\{{\mathbf 0}\}$, then $L$ is injective.

Hint

This proposition is logically equivalent to the proposition:

If $L$ is not injective, then $\mathrm{ker}(L) \neq \{{\mathbf 0}\}$.

Hint

Assume that two different vectors ${\mathbf v}_1,{\mathbf v}_2 \in V_1$ exist such that $L({\mathbf v}_1)=L({\mathbf v}_2)$. Use the linearity of $L$ to show that then $L({\mathbf v}_2-{\mathbf v}_1)={\mathbf 0}$.

Answer

If $L$ is not injective then two different vectors ${\mathbf v}_1,{\mathbf v}_2 \in V_1$ exist such that $L({\mathbf v}_1)=L({\mathbf v}_2)$. This implies that $L({\mathbf v}_2-{\mathbf v}_1)={\mathbf 0}$ and thus that ${\mathbf v}_2-{\mathbf v}_1 \in \ker(L)$. Hence the kernel contains a non-zero vector, implying that $\mathrm{ker}(L) \neq \{{\mathbf 0}\}$.

Question c#

Conclude based on the two previous questions that a linear map $L: V_1 \to V_2$ between two vector spaces $V_1$ and $V_2$ over $\Bbb F$ is injective if and only if $\mathrm{ker}(L) = \{{\mathbf 0}\}$.

Hint

Use Equation (1.22) from Chapter 1 in the textbook.

Exercise 8: Investigation of a linear map from $\Bbb R^4$ to $\Bbb R^3$#

Let $L:\mathbb{R}^4 \to \mathbb{R}^3$ be given by the expression

\[\begin{split} L\left(\left[\begin{array}{r} v_1\\ v_2\\ v_3\\ v_4 \end{array}\right] \right)= \left[\begin{array}{r} v_1+v_2+3v_3+v_4\\ 3v_1-v_2+2v_3+4v_4\\ 2v_1+2v_2+6v_3+2v_4 \end{array}\right]. \end{split}\]

Question a#

Find a matrix ${\mathbf A} \in \mathbb{R}^{3 \times 4}$ that satisfies

\[\begin{split} L\left(\left[\begin{array}{r} v_1\\ v_2\\ v_3\\ v_4 \end{array}\right] \right)={\mathbf A}\cdot \left[\begin{array}{r} v_1\\ v_2\\ v_3\\ v_4 \end{array}\right]. \end{split}\]

Use Lemma 11.1.1 from the textbook to conclude that $L$ is a linear map.

Answer

\[\begin{split} {\mathbf A}= \left[\begin{array}{rrrr} 1 & 1 & 3 & 1\\ 3 & -1 & 2 & 4\\ 2 & 2 & 6 & 2 \end{array}\right]. \end{split}\]

Since $L=L_{\mathbf A}$ where $\mathbf A$ is the above matrix, Lemma 11.1.1 implies that $L$ is a linear map.

Question b#

Let $\beta$ be the ordered standard basis for $\mathbb{R}^4$ and $\gamma$ the one for $\mathbb{R}^3$. Check that ${\mathbf A} = {}_\gamma[L]_\beta$.

Hint

Definition 11.3.1 describes how the mapping matrix ${}_\gamma[L]_\beta$ looks.

Hint

It is useful to keep in mind that $[\mathbf w]_\gamma=\mathbf w$ for all $\mathbf w \in \mathbb{R}^3$, since $\gamma$ is the ordered standard basis for $\mathbb{R}^3$ in this exercise.

Question c#

Provide an ordered basis for the kernel of $L$ as well as the dimension of the kernel.

Hint

Since $L=L_{\mathbf A}$, the kernel of $L$ is the same as the kernel of $\mathbf A$ (also known as the nullspace of $\mathbf A$). An ordered basis for the kernel of $\mathbf A$ can be found by following the procedure used in Example 11.1.4 in the textbook.

Hint

First, check that the matrix

\[\begin{split}{\mathbf A}=\left[\begin{array}{rrrr} 1 & 1 & 3 & 1\\ 3 & -1 & 2 & 4\\ 2 & 2 & 6 & 2 \end{array}\right]\end{split}\]

has the reduced row-echelon form:

\[\begin{split}\mathrm{rref}({\mathbf A})=\left[\begin{array}{rrrr} 1 & 0 & 5/4 & 5/4\\ 0 & 1 & 7/4 & -1/4\\ 0 & 0 & 0 & 0 \end{array}\right].\end{split}\]

Answer

The list of vectors

\[\begin{split}\left(\left[\begin{array}{r} -5/4 \\ -7/4\\ 1\\ 0 \end{array}\right],\left[\begin{array}{r} -5/4\\ 1/4\\ 0\\ 1 \end{array}\right]\right)\end{split}\]

constitutes an ordered basis for $\mathrm{ker} \mathbf A$ and hence also for $\mathrm{ker} L$. The number of elements in the found ordered basis is $2$, hence $\mathrm{dim}(\mathrm{ker}(L))=2$.

Question d#

Provide a basis for the image space of $L$ as well as the dimension of the image space.

Hint

Use the fact that $L=L_{\mathbf A}$ along with Lemma 11.1.2 to show that the image space of $L$ is the same as the column space of $\mathbf A$. The question can thus be answered by finding an ordered basis for the column space of $\mathbf A$.

Answer

Since only the first two columns in the reduced row-echelon form of ${\mathbf A}$ contain pivot elements, Theorem 9.2.1 in the textbook implies that a possible ordered basis for the column space of ${\mathbf A}$ is

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

Since $L=L_{\mathbf A}$, the column space of ${\mathbf A}$ is identical to the image space of $L$, so the required ordered basis has been found. The number of elements in the found ordered basis is $2$, hence $\dim(\mathrm{image}(L))=2$.

Python test 2#

The Python test on the Jupyter Notebook from Python module 2 released last week will open in Möbius at 15:30 and close at 17:00.

Have you finished all the exercises?#

Are you done with the exercises of the day and just waiting for the Python test to start, then click here for some challenging optional extra exercises.

Exercises – Long Day

Contents

Exercises – Long Day#

Exercise 1: Linearity#

Exercise 2: Linear maps defined by diagonal matrices#

Question a#

Question b#

Question c#

Exercise 3: Linear maps and differentiation#

Question a#

Question b#

Exercise 4: Change of basis and mapping matrices#

Question a#

Question b#

Question c#

Exercise 5: Kernel of a linear map#

Exercise 6: Linear map from a vector space of polynomials#

Question a#

Question b#

Question c#

Question d#

Exercise 7: Injective linear maps#

Question a#

Question b#

Question c#

Exercise 8: Investigation of a linear map from \(\Bbb R^4\) to \(\Bbb R^3\)#

Question a#

Question b#

Question c#

Question d#

Python test 2#

Have you finished all the exercises?#