Exercises – Long Day

Exercises – Long Day#

Exercise 1: Linearity and the Rank-Nullity Theorem#

The two maps $f: \mathbb{R}^2 \to \mathbb{R}^2$ and $g: \mathbb{R}^2 \to \mathbb{R}^2$ are given by:

\[f((x_1, x_2))=(x_1-x_2,-x_1+x_2) \quad \text{and} \quad g((x_1,x_2))=(-x_2,x_1^{2}).\]

Question a#

Show that only one of the two maps is a linear map between the real vector spaces $\mathbb R^2$ and $\mathbb R^2$. Find out which by investigating whether they fulfill the two linearity requirements in Definition 10.0.1 in the textbook.

Answer

It is the map $f$ that is linear. That $g$ is not linear is easily shown with a counterexample.

Question b#

Provide a basis for the kernel of $f$ as well as the dimension of the kernel.

Hint

Definition 10.1.2 explains what the kernel of a linear map is.

Answer

We find that $\mathrm{ker}(f)=\{t\cdot(1,1) \, \mid \, t \in \mathbb{R} \}.$

A basis for the kernel is hence $\{(1,1)\}.$ Since the basis only contains one vector, we have that $\dim (\mathrm{ker}(f))=1.$

Question c#

State the image space of the found linear map. Also provide a basis for the image space as well as the dimension of the image space.

Hint

The image space is the set of all possible outputs from the map. One must thus find out which vectors $(b_1,b_2) \in \mathbb{R}^2\,$ that can appear as a right-hand side in the equation $(x_1-x_2,-x_1+x_2)=(b_1,b_2)\,$ when $x_1$ and $x_2$ can be chosen freely from $\mathbb{R}$.

Answer

$\mathrm{image}(f)=\{t\cdot(-1,1) \, \mid \, t\in \mathbb{R}\}.$

A basis for $\mathrm{image}(f)\,$ is $\{(-1,1)\}.$ The image space hense has a dimension of $1$ in this case.

Question d#

Check that the rank-nullity theorem (see Corollary 10.4.3 in the textbook) is fulfilled for the found linear map.

Answer

The rank-nullity theorem is fulfilled because $\dim(\mathbb{R}^2)=2$ and $\dim(\mathrm{ker}(f))=\dim(\mathrm{image}(f))=1$.

Question 2: Investigation of a Linear Map#

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 ${\mathbf A} \in \mathbb{R}^{3 \times 4}$ for which it applies that

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

Use Definition 10.1.1 and Lemma 10.1.1 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 10.1.1 from the textbook 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

Lemma 10.3.2 explains how to compute the mapping matrix ${}_\gamma[L]_\beta$.

Question c#

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

Hint

Since $L=L_{\mathbf A}$, it applies that the kernel of $L$ is the same as the kernel of $\mathbf A$. A basis for the kernel of $\mathbf A$ can be found by following the same procedure as in Example 10.1.4 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:

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

Answer

The vectors

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

constitute a basis for $\mathrm{ker} A$ and hence also for $\mathrm{ker} L$. The number of elements in the found 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 $L=L_{\mathbf A}$ to realise that the image space of $L$ is the same as 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, then Theorem 9.3.3 in the textbook implies that

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

is a basis for the column space of ${\mathbf A}$. Since the column space of ${\mathbf A}$ is the same as the image space of $L$, we have now found the wanted basis. The number of elements in the found basis is $2$ and hence $\dim(\mathrm{image}(L))=2.$

Question e#

Check that the rank-nullity theorem (see Corollary 10.4.3 in the textbook) is fulfilled for the given linear map $L$.

Answer

The rank-nullity theorem is fulfilled since $\dim(\mathbb{R}^4)=4$ and $\dim(\mathrm{ker}(L))=\dim(\mathrm{image}(L))=2$.

Question f#

Do the vectors

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

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

Hint

A basis for $\mathrm{image}(L)$ has already been found in Question d. Can the given vectors be written as a linear combination of these basis vectors?

Answer

The vector $\mathbf u$ is not an element in $\mathrm{image}(L)$. The vector $\mathbf w$ is an element in $\mathrm{image}(L)$.

Exercise 3: Investigation of another Linear Map#

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

\[ M(a+bZ+cZ^2+dZ^3)=(a+b+3c+d)+(3a-b+2c+4d)Z+(2a+2b+6c+2d)Z^2 \]

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 10.3.3 explains how to compute 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#

Note that the mapping matrix you just found is exactly the same matrix as the matrix $\mathbf A$ found in Exercise 2b. Use the result from Exercise 2 to state a basis for the kernel and for the image space of $M$.

Hint

No computations are needed if one uses the results from the previous exercise.

Answer

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

Question c#

Use the results from Exercise 2 to determine 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, $\mathrm{image}(M)$.

Answer

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

Exercise 4: Linear Maps and Differentiation#

As in Example 9.3.4 in the textbook the notation $C_\infty(\mathbb R)$ symbolises a real vector space consisting of all functions from $\mathbb{R}$ to $\mathbb{R}$ that can be differentiated any arbitrary 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#

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

Hint

You can use the formula in Lemma 10.3.3.

Hint

Lemma 10.3.3 concerns two vector spaces $V_1$ and $V_2$ as well as two ordered bases, one for each vector space. Note that here $V_1=V_2$, and also note that the ordered basis $\beta$ is used for both $V_1$ and $V_2$.

Answer

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

Question c#

Another ordered basis is now chosen for $V_1$: $\gamma=(\mathrm{sinh(t)},\mathrm{cosh(t)},\sin(t),\cos(t))$. The functions $\mathrm{sinh(t)}$ and $\mathrm{cosh(t)}$ were defined in Exercise 10 on the Long Day of Week 2. Determine the change-of-basis matrices ${}_\gamma [\mathrm{id}_{V_1}] _\beta$ and ${}_\beta [\mathrm{id}_{V_1}] _\gamma$.

Partial answer

\[\begin{split}{}_\gamma [\mathrm{id}_{V_1}] _\beta= \left[\begin{array}{rrrr} 1 & -1 & 0 & 0\\ 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right].\end{split}\]

Question d#

Determine the mapping matrix ${}_\gamma [L] _\gamma$ directly from the formula in Lemma 10.3.3 in the textbook. As before we have $\gamma=(\mathrm{sinh(t)},\mathrm{cosh(t)},\sin(t),\cos(t))$.

Answer

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

Question e#

Use the second part of Theorem 10.3.4 from the textbook to realise that ${}_\gamma [\mathrm{id}_{V_1}] _\beta \cdot {}_\beta [L] _\beta = {}_\gamma [L] _\gamma \cdot {}_\gamma [\mathrm{id}_{V_1}] _\beta.$ Also check that the equation holds using SymPy and the results from the previous questions.

Answer

The equation holds because according to Theorem 10.3.4 it applies that ${}_\gamma [\mathrm{id}_{V_1}] _\beta \cdot {}_\beta [L] _\beta = {}_\gamma [\mathrm{id}_{V_1} \circ L] _\beta = {}_\gamma [L] _\beta$ and ${}_\gamma [L] _\gamma \cdot {}_\gamma [\mathrm{id}_{V_1}] _\beta={}_\gamma [L \circ \mathrm{id}_{V_1}] _\beta={}_\gamma [L] _\beta$. It was here used that $\mathrm{id}_{V_1} \circ L=L \circ \mathrm{id}_{V_1} =L$, which applies since $\mathrm{id}_{V_1}$ is the identity function.

Exercise 5: Linear Maps defined with 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 the linear map $L_{\mathbf A}: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $L_{\mathbf A}\left({\mathbf v}\right)={\mathbf A}\cdot {\mathbf v}$ is investigated.

Question a#

Sketch the set $M=\{(v_1,v_2) \, \mid \, 0 \le v_1 \le 1, \, 0 \le v_2 \le 1\}$ in $\mathbb{R}^2$.

Answer

The sketch is not shown here, but the set $M$ can be described as a square in $\mathbb{R}^2$ with the corner points $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$.

Question b#

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

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

Hint

Each pair ${\mathbf v}=(v_1,v_2) \in M$ fulfills that

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

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

Answer

It applies 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 $L_{\mathbf A}(M)$ is hence a rectangle with the corner points $(0,0),(\lambda_1,0),(0,\lambda_2),(\lambda_1,\lambda_2)$.

Question c#

With $M$ and ${\mathbf A} \in \mathbb{R}^{2 \times 2}$ as before, check that the area of $L_{\mathbf A}(M)$ is equal to $|\det({\mathbf A})|.$

Question d#

Now let

\[\begin{split}{\mathbf A}=\left[\begin{array}{rr} 1 & \lambda\\ 0 & 1\end{array}\right]\quad,\,\lambda \in\mathbb R.\end{split}\]

Check that $L_{\mathbf A}(M)$ has an area of $1$ and that $\det({\mathbf A})=1$. More generally it can be shown that for an arbitrary matrix ${\mathbf A} \in \mathbb{R}^{2 \times 2}$ the area of $L_{\mathbf A}(M)$ always equals $|\det({\mathbf A})|.$

Exercise 6: Change of Basis and Mapping Matrices#

The vectors $\mathbf{v}_1=(1,2)\,$ and $\mathbf{v}_2=(3,7)\,$ in $\mathbb{R}^2\,$ as well as $\mathbf{w}_1=(1,2,2)\,,$ $\mathbf{w}_2=(2,3,1)\,,$ and $\mathbf{w}_3=(1,2,1)\,$ in $\mathbb{R}^3\,$ are given. A linear map $L:\mathbb{R}^2\to \mathbb{R}^3\,$ fulfills that

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

In this exercise you may perform matrix products and computations of inverse matrices using SymPy.

Question a#

Show that $\mathbf{v}_1\,$ and $\mathbf{v}_2\,$ constitute a basis for $\mathbb{R}^2\,$ and that $\mathbf{w}_1\,$, $\mathbf{w}_2\,$, and $\mathbf{w}_3\,$ constitute a basis for $\mathbb{R}^3\,.$

Hint

First show that the vectors $\mathbf{v}_1\,$ and $\mathbf{v}_2\,$ are linearly independent and that the vectors $\mathbf{w}_1\,$, $\mathbf{w}_2\,$, and $\mathbf{w}_3\,$ are linearly independent. This can be done in various ways which you can read more about in Chapters 7 and 8 in our textbook. One method is to follow the same strategy as in Example 7.1.4 in the textbook. Alternatively, one can use Corollary 8.3.5.

Hint

If $n$ vectors in $\mathbb{F}^n$ are linearly independent do they then constitute a basis for $\mathbb{F}^n$? The answer can be derived from Theorem 9.2.7 of the textbook.

Question b#

Provide the mapping matrix ${}_\epsilon[L]_\delta$ for $L$ with respect to the ordered bases $\delta=(\mathbf{v}_1,\mathbf{v}_2)\,$ in $\mathbb{R}^2\,$ and $\epsilon=(\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3)$ in $\mathbb{R}^3$.

Hint

The formula in Lemma 10.3.3 can be of use here.

Answer

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

Question c#

Provide the mapping matrix ${}_\gamma[L]_\delta$ for $L$ where $\delta$ is as before and $\gamma$ is the ordered standard basis for $\mathbb{R}^3\,.$

Hint

Instead of starting over from scratch one can also use the second part of Theorem 10.3.4 from the textbook to realise that ${}_\gamma[\mathrm{id}_{\mathbb{R}^3}]_\epsilon \cdot {}_\epsilon [L] _\delta = {}_\gamma [L] _\delta.$

Hint

First determine the change-of-basis matrix ${}_\gamma[\mathrm{id}_{\mathbb{R}^3}]_\epsilon$. The computation is not complicated since $\gamma$ is the ordered standard basis for $\mathbb{R}^3$.

Answer

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

Question d#

Provide the mapping matrix ${}_\epsilon[L]_\beta$ for $L$ where $\epsilon$ is as before and $\beta$ is the ordered standard basis for $\mathbb{R}^2\,.$

Hint

One can use the second part of Theorem 10.3.4 from the textbook to realise that ${}_\epsilon [L] _\delta \cdot {}_\delta[\mathrm{id}_{\mathbb{R}^2}]_\beta = {}_\epsilon [L] _\beta.$ So, first compute the change-of-basis matrix ${}_\delta[\mathrm{id}_{\mathbb{R}^2}]_\beta.$

Hint

The matrix ${}_\delta[\mathrm{id}_{\mathbb{R}^2}]_\beta$ can be a bit difficult to compute, but $({}_\delta[\mathrm{id}_{\mathbb{R}^2}]_\beta)^{-1}={}_\beta[\mathrm{id}_{\mathbb{R}^2}]_\delta$ according to Lemma 10.3.5 in the textbook.

Answer

\[\begin{split}{}_\epsilon[L]_\beta =\left[\begin{array}{rr} 5 & -2 \\ 9 & -4 \\ -17 & 7 \end{array}\right].\end{split}\]

Question e#

Now provide the mapping matrix ${}_\gamma[L]_\beta$ for $L$ where $\beta$ and $\gamma$ as before are the ordered standard bases for $\mathbb{R}^2$ and $\mathbb{R}^3$.

Hint

It applies that ${}_\gamma[L]_\beta={}_\gamma[\mathrm{id}_{\mathbb{R}^3}]_\epsilon \cdot {}_\epsilon[L]_\delta \cdot {}_\delta[\mathrm{id}_{\mathbb{R}^2}]_\beta$. All matrices needed to reach the answer have thus already been computed in the previous questions.

Answer

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

Exercise 7: 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 10.2.1 in the textbook. Lemma 9.3.2 in the textbook can be of use to determine whether it is a subspace of $V_1$.

Exercise 8: 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 $\beta$ in $\mathbb R^3$ the mapping matrix

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

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

Answer

A possible basis is

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

Exercise 9: Powers of Matrices#

Let ${\mathbf A} \in \mathbb{F}^{n \times n}$ be a matrix and ${\mathbf Q} \in \mathbb{F}^{n \times n}$ an invertible matrix. Let $k$ be a natural number. The $k$^th power of ${\mathbf A}$, with the notation ${\mathbf A}^k$, is defined as the matrix that is achieved by multiplying ${\mathbf A}$ $k$ times by itself. More formally ${\mathbf A}^k$ is defined recursively as follows:

\[\begin{split}{\mathbf A}^k = \left\{ \begin{array}{rl} {\mathbf A} & \text{if $k=1$}\\ {\mathbf A}^{k-1}\cdot {\mathbf A} & \text{if $k \ge 2$.}\end{array} \right.\end{split}\]

Question a#

Assume that ${\mathbf A}$ is a diagonal matrix. Show that for all natural numbers $k$ the matrix ${\mathbf A}^k$ is also a diagonal matrix.

Question b#

Show via induction on $k$ that $({\mathbf Q}^{-1} \cdot {\mathbf A} \cdot {\mathbf Q})^k={\mathbf Q}^{-1} \cdot {\mathbf A}^k \cdot {\mathbf Q}.$

Question c#

Let

\[\begin{split} {\mathbf A}=\left[\begin{array}{rr} 0 & 1\\ -6 & 5\end{array}\right] \quad \text{and} \quad {\mathbf Q}=\left[\begin{array}{rr} 1 & 1\\ 2 & 3\end{array}\right]. \end{split}\]

Check that ${\mathbf Q}^{-1} \cdot {\mathbf A} \cdot {\mathbf Q}$ is a diagonal matrix and use this to find a closed-form expression for ${\mathbf A}^k$.

Answer

\[\begin{split}{\mathbf A}^k=\left[\begin{array}{rr}3\cdot 2^k-2 \cdot 3^k & -2^k+3^k\\3 \cdot 2^{k+1}-2 \cdot 3^{k+1} & -2^{k+1}+3^{k+1}\end{array}\right].\end{split}\]

Exercises – Long Day

Contents

Exercises – Long Day#

Exercise 1: Linearity and the Rank-Nullity Theorem#

Question a#

Question b#

Question c#

Question d#

Question 2: Investigation of a Linear Map#

Question a#

Question b#

Question c#

Question d#

Question e#

Question f#

Exercise 3: Investigation of another Linear Map#

Question a#

Question b#

Question c#

Exercise 4: Linear Maps and Differentiation#

Question a#

Question b#

Question c#

Question d#

Question e#

Exercise 5: Linear Maps defined with Diagonal Matrices#

Question a#

Question b#

Question c#

Question d#

Exercise 6: Change of Basis and Mapping Matrices#

Question a#

Question b#

Question c#

Question d#

Question e#

Exercise 7: Kernel of a Linear Map#

Exercise 8: The Rank-Nullity Theorem in an Example#

Exercise 9: Powers of Matrices#

Question a#

Question b#

Question c#