Computing Linear Maps

The prior section shows that a linear map is determined by its action on a basis. The equation

$h (\vec{v}) = h (c_{1} \cdot {\vec{β}}_{1} + \dots + c_{n} \cdot {\vec{β}}_{n}) = c_{1} \cdot h ({\vec{β}}_{1}) + \dots + c_{n} \cdot h ({\vec{β}}_{n})$

describes how we get the value of the map on any vector $\vec{v}$ by starting from the value of the map on the vectors ${\vec{β}}_{i}$ in a basis and extending linearly.

This section gives a convenient scheme based on matrices to use the representations of $h ({\vec{β}}_{1})$ , …, $h ({\vec{β}}_{n})$ to compute, from the representation of a vector in the domain ${Rep}_{B} (\vec{v})$ , the representation of that vector’s image in the codomain ${Rep}_{D} (h (\vec{v}))$ .

Representing Linear Maps with Matrices

Example 1.1 For the spaces $ℝ^{2}$ and $ℝ^{3}$ fix these bases.

$B = ⟨ (\begin{matrix} 2 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 4 \end{matrix}) ⟩ D = ⟨ (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) ⟩$

Consider the map $h : ℝ^{2} \to ℝ^{3}$ that is determined by this association.

$(\begin{matrix} 2 \\ 0 \end{matrix}) \overset{h}{⟼} (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) (\begin{matrix} 1 \\ 4 \end{matrix}) \overset{h}{⟼} (\begin{matrix} 1 \\ 2 \\ 0 \end{matrix})$

To compute the action of this map on any vector at all from the domain we first represent the vector $h ({\vec{β}}_{1})$

$(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) = 0 (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) - \frac{1}{2} (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}) + 1 (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) {Rep}_{D} (h ({\vec{β}}_{1})) = {(\begin{matrix} 0 \\ - 1 / 2 \\ 1 \end{matrix})}_{D}$

and $h ({\vec{β}}_{2})$ .

$(\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}) = 1 (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) - 1 (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}) + 0 (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) {Rep}_{D} (h ({\vec{β}}_{2})) = {(\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix})}_{D}$

With these, for any member $\vec{v}$ of the domain we can compute $h (\vec{v})$ .

$\begin{aligned} h (\vec{v}) & = h (c_{1} \cdot (\begin{array}{c} 2 \\ 0 \end{array}) + c_{2} \cdot (\begin{array}{c} 1 \\ 4 \end{array})) \\ = c_{1} \cdot h ((\begin{array}{c} 2 \\ 0 \end{array})) + c_{2} \cdot h ((\begin{array}{c} 1 \\ 4 \end{array})) \\ = c_{1} \cdot (0 (\begin{array}{c} 1 \\ 0 \\ 0 \end{array}) - \frac{1}{2} (\begin{array}{c} 0 \\ - 2 \\ 0 \end{array}) + 1 (\begin{array}{c} 1 \\ 0 \\ 1 \end{array})) + c_{2} \cdot (1 (\begin{array}{c} 1 \\ 0 \\ 0 \end{array}) - 1 (\begin{array}{c} 0 \\ - 2 \\ 0 \end{array}) + 0 (\begin{array}{c} 1 \\ 0 \\ 1 \end{array})) \\ = (0 c_{1} + 1 c_{2}) \cdot (\begin{array}{c} 1 \\ 0 \\ 0 \end{array}) + (- \frac{1}{2} c_{1} - 1 c_{2}) \cdot (\begin{array}{c} 0 \\ - 2 \\ 0 \end{array}) + (1 c_{1} + 0 c_{2}) \cdot (\begin{array}{c} 1 \\ 0 \\ 1 \end{array}) \end{aligned}$

Thus,

if ${Rep}_{B} (\vec{v}) = (\begin{matrix} c_{1} \\ c_{2} \end{matrix})$ then ${Rep}_{D} (h (\vec{v})) = (\begin{matrix} 0 c_{1} + 1 c_{2} \\ - (1 / 2) c_{1} - 1 c_{2} \\ 1 c_{1} + 0 c_{2} \end{matrix})$ .

For instance,

since ${Rep}_{B} ((\begin{matrix} 4 \\ 8 \end{matrix})) = {(\begin{matrix} 1 \\ 2 \end{matrix})}_{B}$ we have ${Rep}_{D} (h ((\begin{matrix} 4 \\ 8 \end{matrix}))) = (\begin{matrix} 2 \\ - 5 / 2 \\ 1 \end{matrix})$ .

We express computations like the one above with a matrix notation.

${(\begin{matrix} 0 & 1 \\ - 1 / 2 & - 1 \\ 1 & 0 \end{matrix})}_{B, D} {(\begin{matrix} c_{1} \\ c_{2} \end{matrix})}_{B} = {(\begin{matrix} 0 c_{1} + 1 c_{2} \\ (- 1 / 2) c_{1} - 1 c_{2} \\ 1 c_{1} + 0 c_{2} \end{matrix})}_{D}$

In the middle is the argument $\vec{v}$ to the map, represented with respect to the domain’s basis $B$ by the column vector with components $c_{1}$ and $c_{2}$ . On the right is the value of the map on that argument $h (\vec{v})$ , represented with respect to the codomain’s basis $D$ . The matrix on the left is the new thing. We will use it to represent the map and we will think of the above equation as representing an application of the map to the matrix.

That matrix consists of the coefficients from the vector on the right, $0$ and $1$ from the first row, $- 1 / 2$ and $- 1$ from the second row, and $1$ and $0$ from the third row. That is, we make it by adjoining the vectors representing the $h ({\vec{β}}_{i})$ ’s.

$(\begin{array}{cc} ⋮ & ⋮ \\ {Rep}_{D} (h ({\vec{β}}_{1})) & {Rep}_{D} (h ({\vec{β}}_{2})) \\ ⋮ & ⋮ \end{array})$

Definition 1.2 Suppose that $V$ and $W$ are vector spaces of dimensions $n$ and $m$ with bases $B$ and $D$ , and that $h : V \to W$ is a linear map. If

${Rep}_{D} (h ({\vec{β}}_{1})) = {(\begin{matrix} h_{1, 1} \\ h_{2, 1} \\ ⋮ \\ h_{m, 1} \end{matrix})}_{D} \dots {Rep}_{D} (h ({\vec{β}}_{n})) = {(\begin{matrix} h_{1, n} \\ h_{2, n} \\ ⋮ \\ h_{m, n} \end{matrix})}_{D}$

then

${Rep}_{B, D} (h) = {(\begin{array}{cccc} h_{1, 1} & h_{1, 2} & \dots & h_{1, n} \\ h_{2, 1} & h_{2, 2} & \dots & h_{2, n} \\ ⋮ \\ h_{m, 1} & h_{m, 2} & \dots & h_{m, n} \end{array})}_{B, D}$

is the matrix representation of $h$ with respect to $B, D$ .

In that matrix the number of columns $n$ is the dimension of the map’s domain while the number of rows $m$ is the dimension of the codomain.

Remark 1.3 As with the notation for represenation of a vector, the ${Rep}_{B, D}$ notation here is not standard. The most common alternative is $[h]_{B, D}$ .

We use lower case letters for a map, upper case for the matrix, and lower case again for the entries of the matrix. Thus for the map $h$ , the matrix representing it is $H$ , with entries $h_{i, j}$ .

Example 1.4 If $h : ℝ^{3} \to 𝒫_{1}$ is

$(\begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix}) \overset{h}{⟼} (2 a_{1} + a_{2}) + (- a_{3}) x$

then where

$B = ⟨ (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}), (\begin{matrix} 2 \\ 0 \\ 0 \end{matrix}) ⟩ D = ⟨ 1 + x, - 1 + x ⟩$

the action of $h$ on $B$ is this.

$(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) \overset{h}{⟼} - x (\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}) \overset{h}{⟼} 2 (\begin{matrix} 2 \\ 0 \\ 0 \end{matrix}) \overset{h}{⟼} 4$

A simple calculation

${Rep}_{D} (- x) = {(\begin{matrix} - 1 / 2 \\ - 1 / 2 \end{matrix})}_{D} {Rep}_{D} (2) = {(\begin{matrix} 1 \\ - 1 \end{matrix})}_{D} {Rep}_{D} (4) = {(\begin{matrix} 2 \\ - 2 \end{matrix})}_{D}$

shows that this is the matrix representing $h$ with respect to the bases.

${Rep}_{B, D} (h) = {(\begin{matrix} - 1 / 2 & 1 & 2 \\ - 1 / 2 & - 1 & - 2 \end{matrix})}_{B, D}$

Theorem 1.5 Assume that $V$ and $W$ are vector spaces of dimensions $n$ and $m$ with bases $B$ and $D$ , and that $h : V \to W$ is a linear map. If $h$ is represented by

${Rep}_{B, D} (h) = {(\begin{array}{cccc} h_{1, 1} & h_{1, 2} & \dots & h_{1, n} \\ h_{2, 1} & h_{2, 2} & \dots & h_{2, n} \\ ⋮ \\ h_{m, 1} & h_{m, 2} & \dots & h_{m, n} \end{array})}_{B, D}$

and $\vec{v} \in V$ is represented by

${Rep}_{B} (\vec{v}) = {(\begin{matrix} c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{matrix})}_{B}$

then the representation of the image of $\vec{v}$ is this.

${Rep}_{D} (h (\vec{v})) = {(\begin{matrix} h_{1, 1} c_{1} + h_{1, 2} c_{2} + \dots + h_{1, n} c_{n} \\ h_{2, 1} c_{1} + h_{2, 2} c_{2} + \dots + h_{2, n} c_{n} \\ ⋮ \\ h_{m, 1} c_{1} + h_{m, 2} c_{2} + \dots + h_{m, n} c_{n} \end{matrix})}_{D}$

Proof This formalizes Example 1.1. See Exercise 1.33.

QED

Definition 1.6 The matrix-vector product of a $m \times n$ matrix and a $n \times 1$ vector is this.

$(\begin{array}{cccc} a_{1, 1} & a_{1, 2} & \dots & a_{1, n} \\ a_{2, 1} & a_{2, 2} & \dots & a_{2, n} \\ ⋮ \\ a_{m, 1} & a_{m, 2} & \dots & a_{m, n} \end{array}) (\begin{matrix} c_{1} \\ ⋮ \\ c_{n} \end{matrix}) = (\begin{matrix} a_{1, 1} c_{1} + \dots + a_{1, n} c_{n} \\ a_{2, 1} c_{1} + \dots + a_{2, n} c_{n} \\ ⋮ \\ a_{m, 1} c_{1} + \dots + a_{m, n} c_{n} \end{matrix})$

Briefly, application of a linear map is represented by the matrix-vector product of the map’s representative and the vector’s representative.

Remark 1.7 Theorem 1.5 is not surprising, because we chose the matrix representative in Definition 1.2 precisely to make the theorem true— if the theorem were not true then we would adjust the definition to make it so. Nonetheless, we need the verification.

Example 1.8 For the matrix from Example 1.4 we can calculate where that map sends this vector.

$\vec{v} = (\begin{matrix} 4 \\ 1 \\ 0 \end{matrix})$

With respect to the domain basis $B$ the representation of this vector is

${Rep}_{B} (\vec{v}) = {(\begin{matrix} 0 \\ 1 / 2 \\ 2 \end{matrix})}_{B}$

and so the matrix-vector product gives the representation of the value $h (\vec{v})$ with respect to the codomain basis $D$ .

$\begin{aligned} {Rep}_{D} (h (\vec{v})) & = {(\begin{array}{c} - 1 / 2 & 1 & 2 \\ - 1 / 2 & - 1 & - 2 \end{array})}_{B, D} {(\begin{array}{c} 0 \\ 1 / 2 \\ 2 \end{array})}_{B} \\ = {(\begin{array}{c} (- 1 / 2) \cdot 0 + 1 \cdot (1 / 2) + 2 \cdot 2 \\ (- 1 / 2) \cdot 0 - 1 \cdot (1 / 2) - 2 \cdot 2 \end{array})}_{D} = {(\begin{array}{c} 9 / 2 \\ - 9 / 2 \end{array})}_{D} \end{aligned}$

To find $h (\vec{v})$ itself, not its representation, take $(9 / 2) (1 + x) - (9 / 2) (- 1 + x) = 9$ .

Example 1.9 Let $π : ℝ^{3} \to ℝ^{2}$ be projection onto the $x y$ -plane. To give a matrix representing this map, we first fix some bases.

$B = ⟨ (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}) ⟩ D = ⟨ (\begin{matrix} 2 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 1 \end{matrix}) ⟩$

For each vector in the domain’s basis, find its image under the map.

$(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \overset{π}{⟼} (\begin{matrix} 1 \\ 0 \end{matrix}) (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) \overset{π}{⟼} (\begin{matrix} 1 \\ 1 \end{matrix}) (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}) \overset{π}{⟼} (\begin{matrix} - 1 \\ 0 \end{matrix})$

Then find the representation of each image with respect to the codomain’s basis.

${Rep}_{D} ((\begin{matrix} 1 \\ 0 \end{matrix})) = (\begin{matrix} 1 \\ - 1 \end{matrix}) {Rep}_{D} ((\begin{matrix} 1 \\ 1 \end{matrix})) = (\begin{matrix} 0 \\ 1 \end{matrix}) {Rep}_{D} ((\begin{matrix} - 1 \\ 0 \end{matrix})) = (\begin{matrix} - 1 \\ 1 \end{matrix})$

Finally, adjoining these representations gives the matrix representing $π$ with respect to $B, D$ .

${Rep}_{B, D} (π) = {(\begin{matrix} 1 & 0 & - 1 \\ - 1 & 1 & 1 \end{matrix})}_{B, D}$

We can illustrate Theorem 1.5 by computing the matrix-vector product representing this action by the projection map.

$π ((\begin{matrix} 2 \\ 2 \\ 1 \end{matrix})) = (\begin{matrix} 2 \\ 2 \end{matrix})$

Represent the domain vector with respect to the domain’s basis

${Rep}_{B} ((\begin{matrix} 2 \\ 2 \\ 1 \end{matrix})) = {(\begin{matrix} 1 \\ 2 \\ 1 \end{matrix})}_{B}$

to get this matrix-vector product.

${Rep}_{D} (π ((\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}))) = {(\begin{matrix} 1 & 0 & - 1 \\ - 1 & 1 & 1 \end{matrix})}_{B, D} {(\begin{matrix} 1 \\ 2 \\ 1 \end{matrix})}_{B} = {(\begin{matrix} 0 \\ 2 \end{matrix})}_{D}$

Expanding this into a linear combination of vectors from $D$

$0 \cdot (\begin{matrix} 2 \\ 1 \end{matrix}) + 2 \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) = (\begin{matrix} 2 \\ 2 \end{matrix})$

checks that the map’s action is indeed reflected in the operation of the matrix. We will sometimes compress these three displayed equations into one.

$(\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}) = {(\begin{matrix} 1 \\ 2 \\ 1 \end{matrix})}_{B} \overset{h}{\underset{H}{⟼}} {(\begin{matrix} 0 \\ 2 \end{matrix})}_{D} = (\begin{matrix} 2 \\ 2 \end{matrix})$

We now have two ways to compute the effect of projection, the straightforward formula that drops each three-tall vector’s third component to make a two-tall vector, and the above formula that uses representations and matrix-vector multiplication. The second way may seem complicated compared to the first, but it has advantages. The next example shows that for some maps this new scheme simplifies the formula.

Example 1.10 To represent a rotation map $t_{θ} : ℝ^{2} \to ℝ^{2}$ that turns all vectors in the plane counterclockwise through an angle $θ$

Counterclockwise rotation through pi/6: the vector u=(3,1) on the left becomes ((3 square root of 3 minus 1)/2, (3 plus square root of 3)/2) on the right. Both the map arrow and the image vector are labelled t subscript pi/6.

we start by fixing the standard bases $ℰ_{2}$ for both the domain and codomain basis, Now find the image under the map of each vector in the domain’s basis.

$\begin{matrix} (\begin{matrix} 1 \\ 0 \end{matrix}) \overset{t_{θ}}{⟼} (\begin{matrix} \cos θ \\ \sin θ \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) \overset{t_{θ}}{⟼} (\begin{matrix} - \sin θ \\ \cos θ \end{matrix}) & (*) \end{matrix}$

Represent these images with respect to the codomain’s basis. Because this basis is $ℰ_{2}$ , vectors represent themselves. Adjoin the representations to get the matrix representing the map.

${Rep}_{ℰ_{2}, ℰ_{2}} (t_{θ}) = (\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix})$

The advantage of this scheme is that we get a formula for the image of any vector at all just by knowing in ( $*$ ) how to represent the image of the two basis vectors. For instance, here we rotate a vector by $θ = π / 6$ .

$(\begin{matrix} 3 \\ - 2 \end{matrix}) = {(\begin{matrix} 3 \\ - 2 \end{matrix})}_{ℰ_{2}} \overset{t_{π / 6}}{⟼} (\begin{matrix} \sqrt{3} / 2 & - 1 / 2 \\ 1 / 2 & \sqrt{3} / 2 \end{matrix}) (\begin{matrix} 3 \\ - 2 \end{matrix}) \approx {(\begin{matrix} 3.598 \\ - 0.232 \end{matrix})}_{ℰ_{2}} = (\begin{matrix} 3.598 \\ - 0.232 \end{matrix})$

More generally, we have a formula for rotation by $θ = π / 6$ .

$(\begin{matrix} x \\ y \end{matrix}) \overset{t_{π / 6}}{⟼} (\begin{matrix} \sqrt{3} / 2 & - 1 / 2 \\ 1 / 2 & \sqrt{3} / 2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} (\sqrt{3} / 2) x - (1 / 2) y \\ (1 / 2) x + (\sqrt{3} / 2) y \end{matrix})$

Example 1.11 In the definition of matrix-vector product the width of the matrix equals the height of the vector. Hence, this product is not defined.

$(\begin{matrix} 1 & 0 & 0 \\ 4 & 3 & 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix})$

It is undefined for a reason: the three-wide matrix represents a map with a three-dimensional domain while the two-tall vector represents a member of a two-dimensional space. So the vector cannot be in the domain of the map.

Nothing in Definition 1.6 forces us to view matrix-vector product in terms of representations. We can get some insights by focusing on how the entries combine.

A good way to view matrix-vector product is that it is formed from the dot products of the rows of the matrix with the column vector.

$(\begin{matrix} ⋮ \\ a_{i, 1} & a_{i, 2} & \dots & a_{i, n} \\ ⋮ \end{matrix}) (\begin{matrix} c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{matrix}) = (\begin{matrix} ⋮ \\ a_{i, 1} c_{1} + a_{i, 2} c_{2} + \dots + a_{i, n} c_{n} \\ ⋮ \end{matrix})$

Looked at in this row-by-row way, this new operation generalizes dot product.

We can also view the operation column-by-column.

$\begin{aligned} (\begin{array}{cccc} h_{1, 1} & h_{1, 2} & \dots & h_{1, n} \\ h_{2, 1} & h_{2, 2} & \dots & h_{2, n} \\ ⋮ \\ h_{m, 1} & h_{m, 2} & \dots & h_{m, n} \end{array}) (\begin{array}{c} c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{array}) & = (\begin{array}{c} h_{1, 1} c_{1} + h_{1, 2} c_{2} + \dots + h_{1, n} c_{n} \\ h_{2, 1} c_{1} + h_{2, 2} c_{2} + \dots + h_{2, n} c_{n} \\ ⋮ \\ h_{m, 1} c_{1} + h_{m, 2} c_{2} + \dots + h_{m, n} c_{n} \end{array}) \\ = c_{1} (\begin{array}{c} h_{1, 1} \\ h_{2, 1} \\ ⋮ \\ h_{m, 1} \end{array}) + \dots + c_{n} (\begin{array}{c} h_{1, n} \\ h_{2, n} \\ ⋮ \\ h_{m, n} \end{array}) \end{aligned}$

The result is the columns of the matrix weighted by the entries of the vector.

Example 1.12 $(\begin{matrix} 1 & 0 & - 1 \\ 2 & 0 & 3 \end{matrix}) (\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}) = 2 (\begin{matrix} 1 \\ 2 \end{matrix}) - 1 (\begin{matrix} 0 \\ 0 \end{matrix}) + 1 (\begin{matrix} - 1 \\ 3 \end{matrix}) = (\begin{matrix} 1 \\ 7 \end{matrix})$

This way of looking at matrix-vector product brings us back to the objective stated at the start of this section, to compute $h (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n})$ as $c_{1} h ({\vec{β}}_{1}) + \dots + c_{n} h ({\vec{β}}_{n})$ .

We began this section by noting that the equality of these two enables us to compute the action of $h$ on any argument knowing only $h ({\vec{β}}_{1})$ , …, $h ({\vec{β}}_{n})$ . We have developed this into a scheme to compute the action of the map by taking the matrix-vector product of the matrix representing the map with the vector representing the argument. In this way, with respect to any bases, for any linear map there is a matrix representation. The next subsection will show the converse, that if we fix bases then for any matrix there is an associated linear map.

Exercises

Exercise 1.13 Worked answer

Recommended. Multiply the matrix

$(\begin{matrix} 1 & 3 & 1 \\ 0 & - 1 & 2 \\ 1 & 1 & 0 \end{matrix})$

by each vector, or state “not defined.”
1. $(\begin{matrix} 2 \\ 1 \\ 0 \end{matrix})$
2. $(\begin{matrix} - 2 \\ - 2 \end{matrix})$
3. $(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$
Back to Exercise 1.13

Answer.
1. $(\begin{matrix} 1 \cdot 2 + 3 \cdot 1 + 1 \cdot 0 \\ 0 \cdot 2 + (- 1) \cdot 1 + 2 \cdot 0 \\ 1 \cdot 2 + 1 \cdot 1 + 0 \cdot 0 \end{matrix}) = (\begin{matrix} 5 \\ - 1 \\ 3 \end{matrix})$
2. Not defined.
3. $(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$
Exercise 1.14 Worked answer

Multiply this matrix by each vector or state “not defined.”

$(\begin{matrix} 3 & 1 \\ 2 & 4 \end{matrix})$
1. $(\begin{matrix} 1 \\ 2 \\ 1 \end{matrix})$
2. $(\begin{matrix} 0 \\ - 1 \end{matrix})$
3. $(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$
Back to Exercise 1.14

Answer.
1. This is not defined.
2. This is defined.
  
  $(\begin{matrix} 3 & 1 \\ 2 & 4 \end{matrix}) (\begin{matrix} 0 \\ - 1 \end{matrix}) = (\begin{matrix} - 1 \\ - 4 \end{matrix})$
3. Not defined.
Exercise 1.15 Worked answer

Perform, if possible, each matrix-vector multiplication.
1. $(\begin{matrix} 2 & 1 \\ 3 & - 1 / 2 \end{matrix}) (\begin{matrix} 4 \\ 2 \end{matrix})$
2. $(\begin{matrix} 1 & 1 & 0 \\ - 2 & 1 & 0 \end{matrix}) (\begin{matrix} 1 \\ 3 \\ 1 \end{matrix})$
3. $(\begin{matrix} 1 & 1 \\ - 2 & 1 \end{matrix}) (\begin{matrix} 1 \\ 3 \\ 1 \end{matrix})$
Back to Exercise 1.15

Answer.
1. $(\begin{matrix} 2 \cdot 4 + 1 \cdot 2 \\ 3 \cdot 4 - (1 / 2) \cdot 2 \end{matrix}) = (\begin{matrix} 10 \\ 11 \end{matrix})$
2. $(\begin{matrix} 4 \\ 1 \end{matrix})$
3. Not defined.
Exercise 1.16 Worked answer

This matrix equation expresses a linear system. Solve it.

$(\begin{matrix} 2 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & - 1 & 2 \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 8 \\ 4 \\ 4 \end{matrix})$

Back to Exercise 1.16

Answer. Matrix-vector multiplication gives rise to a linear system.

$\begin{array}{rcrcrcr} 2 x & + & y & + & z & = & 8 \\ y & + & 3 z & = & 4 \\ x & - & y & + & 2 z & = & 4 \end{array}$

Gaussian reduction shows that $z = 1$ , $y = 1$ , and $x = 3$ .
Exercise 1.17 Worked answer

Recommended. For a homomorphism from $𝒫_{2}$ to $𝒫_{3}$ that sends

$1 \mapsto 1 + x, x \mapsto 1 + 2 x, and x^{2} \mapsto x - x^{3}$

where does $1 - 3 x + 2 x^{2}$ go?

Back to Exercise 1.17

Answer. Here are two ways to get the answer.

First, obviously $1 - 3 x + 2 x^{2} = 1 \cdot 1 - 3 \cdot x + 2 \cdot x^{2}$ , and so we can apply the general property of preservation of combinations to get $h (1 - 3 x + 2 x^{2}) = h (1 \cdot 1 - 3 \cdot x + 2 \cdot x^{2}) = 1 \cdot h (1) - 3 \cdot h (x) + 2 \cdot h (x^{2}) = 1 \cdot (1 + x) - 3 \cdot (1 + 2 x) + 2 \cdot (x - x^{3}) = - 2 - 3 x - 2 x^{3}$ .

The other way uses the computation scheme developed in this subsection. Because we know where these elements of the space go, we consider this basis $B = ⟨ 1, x, x^{2} ⟩$ for the domain. Arbitrarily, we can take $D = ⟨ 1, x, x^{2}, x^{3} ⟩$ as a basis for the codomain. With those choices, we have that

${Rep}_{B, D} (h) = {(\begin{matrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & - 1 \end{matrix})}_{B, D}$

and, as

${Rep}_{B} (1 - 3 x + 2 x^{2}) = {(\begin{matrix} 1 \\ - 3 \\ 2 \end{matrix})}_{B}$

the matrix-vector multiplication calculation gives this.

${Rep}_{D} (h (1 - 3 x + 2 x^{2})) = {(\begin{matrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & - 1 \end{matrix})}_{B, D} {(\begin{matrix} 1 \\ - 3 \\ 2 \end{matrix})}_{B} = {(\begin{matrix} - 2 \\ - 3 \\ 0 \\ - 2 \end{matrix})}_{D}$

Thus, $h (1 - 3 x + 2 x^{2}) = - 2 \cdot 1 - 3 \cdot x + 0 \cdot x^{2} - 2 \cdot x^{3} = - 2 - 3 x - 2 x^{3}$ , as above.
Exercise 1.18 Worked answer

Let $h : ℝ^{2} \to ℳ_{2 \times 2}$ be the linear transformation with this action.

$(\begin{matrix} 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) \mapsto (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$

What is its effect on the general vector with entries $x$ and $y$ ?

Back to Exercise 1.18

Answer. Fix this natural basis for $ℳ_{2 \times 2}$ .

$D = ⟨ (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) ⟩$

The representation of the map $h$ with respect to $ℰ_{2}, D$ is this.

${Rep}_{ℰ_{2}, D} (h) = (\begin{matrix} 1 & 0 \\ 2 & - 1 \\ 0 & 1 \\ 1 & 0 \end{matrix})$

Since the general vector is represented with respect to $ℰ_{2}$ by itself, and similarly every matrix is represented with respect to $D$ by itself, we have this for the effect of the map.

${Rep}_{D} (h ((\begin{matrix} x \\ y \end{matrix}))) = (\begin{matrix} 1 & 0 \\ 2 & - 1 \\ 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x \\ 2 x - y \\ y \\ x \end{matrix})$
Exercise 1.19 Worked answer

Recommended. Assume that $h : ℝ^{2} \to ℝ^{3}$ is determined by this action.

$(\begin{matrix} 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 2 \\ 2 \\ 0 \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) \mapsto (\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix})$

Using the standard bases, find
1. the matrix representing this map;
2. a general formula for $h (\vec{v})$ .
Back to Exercise 1.19

Answer. Again, as recalled in the subsection, with respect to $ℰ_{i}$ , a column vector represents itself.
1. To represent $h$ with respect to $ℰ_{2}, ℰ_{3}$ take the images of the basis vectors from the domain, and represent them with respect to the basis for the codomain. The first is this
  
  ${Rep}_{ℰ_{3}} (h ({\vec{e}}_{1})) = {Rep}_{ℰ_{3}} ((\begin{matrix} 2 \\ 2 \\ 0 \end{matrix})) = (\begin{matrix} 2 \\ 2 \\ 0 \end{matrix})$
  
  while the second is this.
  
  ${Rep}_{ℰ_{3}} (h ({\vec{e}}_{2})) = {Rep}_{ℰ_{3}} ((\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix})) = (\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix})$
  
  Adjoin these to make the matrix.
  
  ${Rep}_{ℰ_{2}, ℰ_{3}} (h) = (\begin{matrix} 2 & 0 \\ 2 & 1 \\ 0 & - 1 \end{matrix})$
2. For any $\vec{v}$ in the domain $ℝ^{2}$ ,
  
  ${Rep}_{ℰ_{2}} (\vec{v}) = {Rep}_{ℰ_{2}} ((\begin{matrix} v_{1} \\ v_{2} \end{matrix})) = (\begin{matrix} v_{1} \\ v_{2} \end{matrix})$
  
  and so
  
  ${Rep}_{ℰ_{3}} (h (\vec{v})) = (\begin{matrix} 2 & 0 \\ 2 & 1 \\ 0 & - 1 \end{matrix}) (\begin{matrix} v_{1} \\ v_{2} \end{matrix}) = (\begin{matrix} 2 v_{1} \\ 2 v_{1} + v_{2} \\ - v_{2} \end{matrix})$
  
  is the desired representation.
Exercise 1.20 Worked answer

Represent the homomorphism $h : ℝ^{3} \to ℝ^{2}$ given by this formula and with respect to these bases.

$(\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} x + y \\ x + z \end{matrix}) B = ⟨ (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) ⟩ D = ⟨ (\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 2 \end{matrix}) ⟩$

Back to Exercise 1.20

Answer. The action of the map on the domain’s basis vectors is this.

$(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) \mapsto (\begin{matrix} 2 \\ 2 \end{matrix}) (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 2 \\ 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 1 \\ 1 \end{matrix})$

Represent those with respect to the codomain’s basis.

${Rep}_{D} ((\begin{matrix} 2 \\ 2 \end{matrix})) = {(\begin{matrix} 2 \\ 1 \end{matrix})}_{D} {Rep}_{D} ((\begin{matrix} 2 \\ 1 \end{matrix})) = {(\begin{matrix} 2 \\ 1 / 2 \end{matrix})}_{D} {Rep}_{D} ((\begin{matrix} 1 \\ 1 \end{matrix})) = {(\begin{matrix} 1 \\ 1 / 2 \end{matrix})}_{D}$

Concatenate them together into a matrix.

${Rep}_{B, D} (h) = (\begin{matrix} 2 & 2 & 1 \\ 1 & 1 / 2 & 1 / 2 \end{matrix})$
Exercise 1.21 Worked answer

Recommended. Let $d / d x : 𝒫_{3} \to 𝒫_{3}$ be the derivative transformation.
1. Represent $d / d x$ with respect to $B, B$ where $B = ⟨ 1, x, x^{2}, x^{3} ⟩$ .
2. Represent $d / d x$ with respect to $B, D$ where $D = ⟨ 1, 2 x, 3 x^{2}, 4 x^{3} ⟩$ .
Back to Exercise 1.21

Answer.
1. We must first find the image of each vector from the domain’s basis, and then represent that image with respect to the codomain’s basis.
  
  $\begin{matrix} {Rep}_{B} (\frac{d 1}{d x}) = (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}) {Rep}_{B} (\frac{d x}{d x}) = (\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}) {Rep}_{B} (\frac{d x^{2}}{d x}) = (\begin{matrix} 0 \\ 2 \\ 0 \\ 0 \end{matrix}) \\ {Rep}_{B} (\frac{d x^{3}}{d x}) = (\begin{matrix} 0 \\ 0 \\ 3 \\ 0 \end{matrix}) \end{matrix}$
  
  Those representations are then adjoined to make the matrix representing the map.
  
  ${Rep}_{B, B} (\frac{d}{d x}) = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{matrix})$
2. Proceeding as in the prior item, we represent the images of the domain’s basis vectors
  
  $\begin{matrix} {Rep}_{D} (\frac{d 1}{d x}) = (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}) {Rep}_{D} (\frac{d x}{d x}) = (\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}) {Rep}_{D} (\frac{d x^{2}}{d x}) = (\begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix}) \\ {Rep}_{D} (\frac{d x^{3}}{d x}) = (\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix}) \end{matrix}$
  
  and adjoin to make the matrix.
  
  ${Rep}_{B, D} (\frac{d}{d x}) = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix})$
Exercise 1.22 Worked answer

Recommended. Represent each linear map with respect to each pair of bases.
1. $d / d x : 𝒫_{n} \to 𝒫_{n}$ with respect to $B, B$ where $B = ⟨ 1, x, \dots, x^{n} ⟩$ , given by
  
  $a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} \mapsto a_{1} + 2 a_{2} x + \dots + n a_{n} x^{n - 1}$
2. $\int : 𝒫_{n} \to 𝒫_{n + 1}$ with respect to $B_{n}, B_{n + 1}$ where $B_{i} = ⟨ 1, x, \dots, x^{i} ⟩$ , given by
  
  $a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} \mapsto a_{0} x + \frac{a_{1}}{2} x^{2} + \dots + \frac{a_{n}}{n + 1} x^{n + 1}$
3. $\int_{0}^{1} : 𝒫_{n} \to ℝ$ with respect to $B, ℰ_{1}$ where $B = ⟨ 1, x, \dots, x^{n} ⟩$ and $ℰ_{1} = ⟨ 1 ⟩$ , given by
  
  $a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} \mapsto a_{0} + \frac{a_{1}}{2} + \dots + \frac{a_{n}}{n + 1}$
4. ${eval}_{3} : 𝒫_{n} \to ℝ$ with respect to $B, ℰ_{1}$ where $B = ⟨ 1, x, \dots, x^{n} ⟩$ and $ℰ_{1} = ⟨ 1 ⟩$ , given by
  
  $a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} \mapsto a_{0} + a_{1} \cdot 3 + a_{2} \cdot 3^{2} + \dots + a_{n} \cdot 3^{n}$
5. ${slide}_{- 1} : 𝒫_{n} \to 𝒫_{n}$ with respect to $B, B$ where $B = ⟨ 1, x, \dots, x^{n} ⟩$ , given by
  
  $a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} \mapsto a_{0} + a_{1} \cdot (x + 1) + \dots + a_{n} \cdot (x + 1)^{n}$
Back to Exercise 1.22

Answer. For each, we must find the image of each of the domain’s basis vectors, represent each image with respect to the codomain’s basis, and then adjoin those representations to get the matrix.
1. The basis vectors from the domain have these images
  
  $1 \mapsto 0 x \mapsto 1 x^{2} \mapsto 2 x \dots$
  
  and these images are represented with respect to the codomain’s basis in this way.
  
  $\begin{matrix} {Rep}_{B} (0) = (\begin{matrix} 0 \\ 0 \\ 0 \\ ⋮ \end{matrix}) {Rep}_{B} (1) = (\begin{matrix} 1 \\ 0 \\ 0 \\ ⋮ \end{matrix}) {Rep}_{B} (2 x) = (\begin{matrix} 0 \\ 2 \\ 0 \\ ⋮ \end{matrix}) \\ \dots {Rep}_{B} (n x^{n - 1}) = (\begin{matrix} 0 \\ 0 \\ 0 \\ ⋮ \\ n \\ 0 \end{matrix}) \end{matrix}$
  
  The matrix
  
  ${Rep}_{B, B} (\frac{d}{d x}) = (\begin{matrix} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 2 & \dots & 0 \\ ⋮ \\ 0 & 0 & 0 & \dots & n \\ 0 & 0 & 0 & \dots & 0 \end{matrix})$
  
  has $n + 1$ rows and columns.
2. Once the images under this map of the domain’s basis vectors are determined
  
  $1 \mapsto x x \mapsto x^{2} / 2 x^{2} \mapsto x^{3} / 3 \dots$
  
  then they can be represented with respect to the codomain’s basis
  
  $\begin{matrix} {Rep}_{B_{n + 1}} (x) = (\begin{matrix} 0 \\ 1 \\ 0 \\ ⋮ \end{matrix}) {Rep}_{B_{n + 1}} (x^{2} / 2) = (\begin{matrix} 0 \\ 0 \\ 1 / 2 \\ ⋮ \end{matrix}) \\ \dots {Rep}_{B_{n + 1}} (x^{n + 1} / (n + 1)) = (\begin{matrix} 0 \\ 0 \\ 0 \\ ⋮ \\ 1 / (n + 1) \end{matrix}) \end{matrix}$
  
  and put together to make the matrix.
  
  ${Rep}_{B_{n}, B_{n + 1}} (\int) = (\begin{matrix} 0 & 0 & \dots & 0 & 0 \\ 1 & 0 & \dots & 0 & 0 \\ 0 & 1 / 2 & \dots & 0 & 0 \\ ⋮ \\ 0 & 0 & \dots & 0 & 1 / (n + 1) \end{matrix})$
3. The images of the basis vectors of the domain are
  
  $1 \mapsto 1 x \mapsto 1 / 2 x^{2} \mapsto 1 / 3 \dots$
  
  and they are represented with respect to the codomain’s basis as
  
  ${Rep}_{ℰ_{1}} (1) = 1 {Rep}_{ℰ_{1}} (1 / 2) = 1 / 2 \dots$
  
  so the matrix is
  
  ${Rep}_{B, ℰ_{1}} (\int) = (\begin{matrix} 1 & 1 / 2 & \dots & 1 / n & 1 / (n + 1) \end{matrix})$
  
  (this is an $1 \times (n + 1)$ matrix).
4. The images of the domain’s basis vectors are
  
  $1 \mapsto 1 x \mapsto 3 x^{2} \mapsto 9 \dots$
  
  and they are represented in the codomain as
  
  ${Rep}_{ℰ_{1}} (1) = 1 {Rep}_{ℰ_{1}} (3) = 3 {Rep}_{ℰ_{1}} (9) = 9 \dots$
  
  and so the matrix is this.
  
  ${Rep}_{B, ℰ_{1}} ({eval}_{3}) = (\begin{matrix} 1 & 3 & 9 & \dots & 3^{n} \end{matrix})$
5. The images of the basis vectors from the domain are $1 \mapsto 1$ , and $x \mapsto x + 1 = 1 + x$ , and $x^{2} \mapsto (x + 1)^{2} = 1 + 2 x + x^{2}$ , and $x^{3} \mapsto (x + 1)^{3} = 1 + 3 x + 3 x^{2} + x^{3}$ , etc. The representations are here.
  
  ${Rep}_{B} (1) = (\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \\ ⋮ \\ 0 \end{matrix}) {Rep}_{B} (1 + x) = (\begin{matrix} 1 \\ 1 \\ 0 \\ 0 \\ ⋮ \\ 0 \end{matrix}) {Rep}_{B} (1 + 2 x + x^{2}) = (\begin{matrix} 1 \\ 2 \\ 1 \\ 0 \\ ⋮ \\ 0 \end{matrix}) \dots$
  
  The resulting matrix
  
  ${Rep}_{B, B} ({slide}_{- 1}) = (\begin{matrix} 1 & 1 & 1 & 1 & \dots & 1 \\ 0 & 1 & 2 & 3 & \dots & (\binom{n}{1}) \\ 0 & 0 & 1 & 3 & \dots & (\binom{n}{2}) \\ ⋮ \\ 0 & 0 & 0 & \dots & 1 \end{matrix})$
  
  is Pascal’s triangle (recall that $(\binom{n}{r})$ is the number of ways to choose $r$ things, without order and without repetition, from a set of size $n$ ).
Exercise 1.23 Worked answer

Represent the identity map on any nontrivial space with respect to $B, B$ , where $B$ is any basis.

Back to Exercise 1.23

Answer. Where the space is $n$ -dimensional,

${Rep}_{B, B} (id) = {(\begin{matrix} 1 & 0 \dots & 0 \\ 0 & 1 \dots & 0 \\ ⋮ \\ 0 & 0 \dots & 1 \end{matrix})}_{B, B}$

is the $n \times n$ identity matrix.
Exercise 1.24 Worked answer

Represent, with respect to the natural basis, the transpose transformation on the space $ℳ_{2 \times 2}$ of $2 \times 2$ matrices.

Back to Exercise 1.24

Answer. Taking this as the natural basis

$B = ⟨ {\vec{β}}_{1}, {\vec{β}}_{2}, {\vec{β}}_{3}, {\vec{β}}_{4} ⟩ = ⟨ (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) ⟩$

the transpose map acts in this way

${\vec{β}}_{1} \mapsto {\vec{β}}_{1} {\vec{β}}_{2} \mapsto {\vec{β}}_{3} {\vec{β}}_{3} \mapsto {\vec{β}}_{2} {\vec{β}}_{4} \mapsto {\vec{β}}_{4}$

so that representing the images with respect to the codomain’s basis and adjoining those column vectors together gives this.

${Rep}_{B, B} (trans) = {(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix})}_{B, B}$
Exercise 1.25 Worked answer

Assume that $B = ⟨ {\vec{β}}_{1}, {\vec{β}}_{2}, {\vec{β}}_{3}, {\vec{β}}_{4} ⟩$ is a basis for a vector space. Represent with respect to $B, B$ the transformation that is determined by each.
1. ${\vec{β}}_{1} \mapsto {\vec{β}}_{2}$ , ${\vec{β}}_{2} \mapsto {\vec{β}}_{3}$ , ${\vec{β}}_{3} \mapsto {\vec{β}}_{4}$ , ${\vec{β}}_{4} \mapsto \vec{0}$
2. ${\vec{β}}_{1} \mapsto {\vec{β}}_{2}$ , ${\vec{β}}_{2} \mapsto \vec{0}$ , ${\vec{β}}_{3} \mapsto {\vec{β}}_{4}$ , ${\vec{β}}_{4} \mapsto \vec{0}$
3. ${\vec{β}}_{1} \mapsto {\vec{β}}_{2}$ , ${\vec{β}}_{2} \mapsto {\vec{β}}_{3}$ , ${\vec{β}}_{3} \mapsto \vec{0}$ , ${\vec{β}}_{4} \mapsto \vec{0}$
Back to Exercise 1.25

Answer.
1. With respect to the basis of the codomain, the images of the members of the basis of the domain are represented as
  
  ${Rep}_{B} ({\vec{β}}_{2}) = (\begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix}) {Rep}_{B} ({\vec{β}}_{3}) = (\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix}) {Rep}_{B} ({\vec{β}}_{4}) = (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}) {Rep}_{B} (\vec{0}) = (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix})$
  
  and consequently, the matrix representing the transformation is this.
  
  $(\begin{matrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix})$
2. $(\begin{matrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix})$
3. $(\begin{matrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})$
Exercise 1.26 Worked answer

Example 1.10 shows how to represent the rotation transformation of the plane with respect to the standard basis. Express these other transformations also with respect to the standard basis.
1. the dilation map $d_{s}$ , which multiplies all vectors by the same scalar $s$
2. the reflection map $f_{ℓ}$ , which reflects all all vectors across a line $ℓ$ through the origin
Back to Exercise 1.26

Answer.
1. The picture of $d_{s} : ℝ^{2} \to ℝ^{2}$ is this.
  
  This map’s effect on the vectors in the standard basis for the domain is
  
  $(\begin{matrix} 1 \\ 0 \end{matrix}) \overset{d_{s}}{⟼} (\begin{matrix} s \\ 0 \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) \overset{d_{s}}{⟼} (\begin{matrix} 0 \\ s \end{matrix})$
  
  and those images are represented with respect to the codomain’s basis (again, the standard basis) by themselves.
  
  ${Rep}_{ℰ_{2}} ((\begin{matrix} s \\ 0 \end{matrix})) = (\begin{matrix} s \\ 0 \end{matrix}) {Rep}_{ℰ_{2}} ((\begin{matrix} 0 \\ s \end{matrix})) = (\begin{matrix} 0 \\ s \end{matrix})$
  
  Thus the representation of the dilation map is this.
  
  ${Rep}_{ℰ_{2}, ℰ_{2}} (d_{s}) = (\begin{matrix} s & 0 \\ 0 & s \end{matrix})$
2. The picture of $f_{ℓ} : ℝ^{2} \to ℝ^{2}$ is this.
  
  Some calculation (see Exercise I.1.33) shows that when the line has slope $k$
  
  $(\begin{matrix} 1 \\ 0 \end{matrix}) \overset{f_{ℓ}}{⟼} (\begin{matrix} (1 - k^{2}) / (1 + k^{2}) \\ 2 k / (1 + k^{2}) \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) \overset{f_{ℓ}}{⟼} (\begin{matrix} 2 k / (1 + k^{2}) \\ - (1 - k^{2}) / (1 + k^{2}) \end{matrix})$
  
  (the case of a line with undefined slope is separate but easy) and so the matrix representing reflection is this.
  
  ${Rep}_{ℰ_{2}, ℰ_{2}} (f_{ℓ}) = \frac{1}{1 + k^{2}} \cdot (\begin{matrix} 1 - k^{2} & 2 k \\ 2 k & - (1 - k^{2}) \end{matrix})$
Exercise 1.27 Worked answer

Recommended. Consider a linear transformation of $ℝ^{2}$ determined by these two.

$(\begin{matrix} 1 \\ 1 \end{matrix}) \mapsto (\begin{matrix} 2 \\ 0 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} - 1 \\ 0 \end{matrix})$
1. Represent this transformation with respect to the standard bases.
2. Where does the transformation send this vector?
  
  $(\begin{matrix} 0 \\ 5 \end{matrix})$
3. Represent this transformation with respect to these bases.
  
  $B = ⟨ (\begin{matrix} 1 \\ - 1 \end{matrix}), (\begin{matrix} 1 \\ 1 \end{matrix}) ⟩ D = ⟨ (\begin{matrix} 2 \\ 2 \end{matrix}), (\begin{matrix} - 1 \\ 1 \end{matrix}) ⟩$
4. Using $B$ from the prior item, represent the transformation with respect to $B, B$ .
Back to Exercise 1.27

Answer. Call the map $t : ℝ^{2} \to ℝ^{2}$ .
1. To represent this map with respect to the standard bases, we must find, and then represent, the images of the vectors ${\vec{e}}_{1}$ and ${\vec{e}}_{2}$ from the domain’s basis. The image of ${\vec{e}}_{1}$ is given.
  
  One way to find the image of ${\vec{e}}_{2}$ is by eye—we can see this.
  
  $(\begin{matrix} 1 \\ 1 \end{matrix}) - (\begin{matrix} 1 \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 1 \end{matrix}) \overset{t}{⟼} (\begin{matrix} 2 \\ 0 \end{matrix}) - (\begin{matrix} - 1 \\ 0 \end{matrix}) = (\begin{matrix} 3 \\ 0 \end{matrix})$
  
  A more systematic way to find the image of ${\vec{e}}_{2}$ is to use the given information to represent the transformation, and then use that representation to determine the image. Taking this for a basis,
  
  $C = ⟨ (\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix}) ⟩$
  
  the given information says this.
  
  ${Rep}_{C, ℰ_{2}} (t) (\begin{matrix} 2 & - 1 \\ 0 & 0 \end{matrix})$
  
  As
  
  ${Rep}_{C} ({\vec{e}}_{2}) = {(\begin{matrix} 1 \\ - 1 \end{matrix})}_{C}$
  
  we have that
  
  ${Rep}_{ℰ_{2}} (t ({\vec{e}}_{2})) = {(\begin{matrix} 2 & - 1 \\ 0 & 0 \end{matrix})}_{C, ℰ_{2}} {(\begin{matrix} 1 \\ - 1 \end{matrix})}_{C} = {(\begin{matrix} 3 \\ 0 \end{matrix})}_{ℰ_{2}}$
  
  and consequently we know that $t ({\vec{e}}_{2}) = 3 \cdot {\vec{e}}_{1}$ (since, with respect to the standard basis, this vector is represented by itself). Therefore, this is the representation of $t$ with respect to $ℰ_{2}, ℰ_{2}$ .
  
  ${Rep}_{ℰ_{2}, ℰ_{2}} (t) = {(\begin{matrix} - 1 & 3 \\ 0 & 0 \end{matrix})}_{ℰ_{2}, ℰ_{2}}$
2. To use the matrix developed in the prior item, note that
  
  ${Rep}_{ℰ_{2}} ((\begin{matrix} 0 \\ 5 \end{matrix})) = {(\begin{matrix} 0 \\ 5 \end{matrix})}_{ℰ_{2}}$
  
  and so we have this is the representation, with respect to the codomain’s basis, of the image of the given vector.
  
  ${Rep}_{ℰ_{2}} (t ((\begin{matrix} 0 \\ 5 \end{matrix}))) = {(\begin{matrix} - 1 & 3 \\ 0 & 0 \end{matrix})}_{ℰ_{2}, ℰ_{2}} {(\begin{matrix} 0 \\ 5 \end{matrix})}_{ℰ_{2}} = {(\begin{matrix} 15 \\ 0 \end{matrix})}_{ℰ_{2}}$
  
  Because the codomain’s basis is the standard one, and so vectors in the codomain are represented by themselves, we have this.
  
  $t ((\begin{matrix} 0 \\ 5 \end{matrix})) = (\begin{matrix} 15 \\ 0 \end{matrix})$
3. We first find the image of each member of $B$ , and then represent those images with respect to $D$ . For the first step, we can use the matrix developed earlier.
  
  ${Rep}_{ℰ_{2}} ((\begin{matrix} 1 \\ - 1 \end{matrix})) = {(\begin{matrix} - 1 & 3 \\ 0 & 0 \end{matrix})}_{ℰ_{2}, ℰ_{2}} {(\begin{matrix} 1 \\ - 1 \end{matrix})}_{ℰ_{2}} = {(\begin{matrix} - 4 \\ 0 \end{matrix})}_{ℰ_{2}} so t ((\begin{matrix} 1 \\ - 1 \end{matrix})) = (\begin{matrix} - 4 \\ 0 \end{matrix})$
  
  Actually, for the second member of $B$ there is no need to apply the matrix because the problem statement gives its image.
  
  $t ((\begin{matrix} 1 \\ 1 \end{matrix})) = (\begin{matrix} 2 \\ 0 \end{matrix})$
  
  Now representing those images with respect to $D$ is routine.
  
  ${Rep}_{D} ((\begin{matrix} - 4 \\ 0 \end{matrix})) = {(\begin{matrix} - 1 \\ 2 \end{matrix})}_{D} and {Rep}_{D} ((\begin{matrix} 2 \\ 0 \end{matrix})) = {(\begin{matrix} 1 / 2 \\ - 1 \end{matrix})}_{D}$
  
  Thus, the matrix is this.
  
  ${Rep}_{B, D} (t) = {(\begin{matrix} - 1 & 1 / 2 \\ 2 & - 1 \end{matrix})}_{B, D}$
4. We know the images of the members of the domain’s basis from the prior item.
  
  $t ((\begin{matrix} 1 \\ - 1 \end{matrix})) = (\begin{matrix} - 4 \\ 0 \end{matrix}) t ((\begin{matrix} 1 \\ 1 \end{matrix})) = (\begin{matrix} 2 \\ 0 \end{matrix})$
  
  We can compute the representation of those images with respect to the codomain’s basis.
  
  ${Rep}_{B} ((\begin{matrix} - 4 \\ 0 \end{matrix})) = {(\begin{matrix} - 2 \\ - 2 \end{matrix})}_{B} and {Rep}_{B} ((\begin{matrix} 2 \\ 0 \end{matrix})) = {(\begin{matrix} 1 \\ 1 \end{matrix})}_{B}$
  
  Thus this is the matrix.
  
  ${Rep}_{B, B} (t) = {(\begin{matrix} - 2 & 1 \\ - 2 & 1 \end{matrix})}_{B, B}$
Exercise 1.28 Worked answer

Suppose that $h : V \to W$ is one-to-one so that by Theorem 2.20, for any basis $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩ \subset V$ the image $h (B) = ⟨ h ({\vec{β}}_{1}), \dots, h ({\vec{β}}_{n}) ⟩$ is a basis for $h (V)$ . (If $h$ is onto then $h (V) = W$ .)
1. Represent the map $h$ with respect to $⟨ B, h (B) ⟩$ .
2. For a member $\vec{v}$ of the domain, where the representation of $\vec{v}$ has components $c_{1}$ , …, $c_{n}$ , represent the image vector $h (\vec{v})$ with respect to the image basis $h (B)$ .
Back to Exercise 1.28

Answer.
1. The images of the members of the domain’s basis are
  
  ${\vec{β}}_{1} \mapsto h ({\vec{β}}_{1}) {\vec{β}}_{2} \mapsto h ({\vec{β}}_{2}) \dots {\vec{β}}_{n} \mapsto h ({\vec{β}}_{n})$
  
  and those images are represented with respect to the codomain’s basis in this way.
  
  $\begin{matrix} {Rep}_{h (B)} (h ({\vec{β}}_{1})) = (\begin{matrix} 1 \\ 0 \\ ⋮ \\ 0 \end{matrix}) {Rep}_{h (B)} (h ({\vec{β}}_{2})) = (\begin{matrix} 0 \\ 1 \\ ⋮ \\ 0 \end{matrix}) \\ \dots {Rep}_{h (B)} (h ({\vec{β}}_{n})) = (\begin{matrix} 0 \\ 0 \\ ⋮ \\ 1 \end{matrix}) \end{matrix}$
  
  Hence, the matrix is the identity.
  
  ${Rep}_{B, h (B)} (h) = (\begin{matrix} 1 & 0 & \dots & 0 \\ 0 & 1 & 0 \\ ⋱ \\ 0 & 0 & 1 \end{matrix})$
2. Using the matrix in the prior item, the representation is this.
  
  ${Rep}_{h (B)} (h (\vec{v})) = {(\begin{matrix} c_{1} \\ ⋮ \\ c_{n} \end{matrix})}_{h (B)}$
Exercise 1.29 Worked answer

Give a formula for the product of a matrix and ${\vec{e}}_{i}$ , the column vector that is all zeroes except for a single one in the $i$ -th position.

Back to Exercise 1.29

Answer. The product

$(\begin{matrix} h_{1, 1} & \dots & h_{1, i} & \dots & h_{1, n} \\ h_{2, 1} & \dots & h_{2, i} & \dots & h_{2, n} \\ ⋮ \\ h_{m, 1} & \dots & h_{m, i} & \dots & h_{1, n} \end{matrix}) (\begin{matrix} 0 \\ ⋮ \\ 1 \\ ⋮ \\ 0 \end{matrix}) = (\begin{matrix} h_{1, i} \\ h_{2, i} \\ ⋮ \\ h_{m, i} \end{matrix})$

gives the $i$ -th column of the matrix.
Exercise 1.30 Worked answer

Recommended. For each vector space of functions of one real variable, represent the derivative transformation with respect to $B, B$ .
1. ${a \cos x + b \sin x ∣ a, b \in ℝ}$ , $B = ⟨ \cos x, \sin x ⟩$
2. ${a e^{x} + b e^{2 x} ∣ a, b \in ℝ}$ , $B = ⟨ e^{x}, e^{2 x} ⟩$
3. ${a + b x + c e^{x} + d x e^{x} ∣ a, b, c, d \in ℝ}$ , $B = ⟨ 1, x, e^{x}, x e^{x} ⟩$
Back to Exercise 1.30

Answer.
1. The images of the basis vectors for the domain are $\cos x \overset{d / d x}{⟼} - \sin x$ and $\sin x \overset{d / d x}{⟼} \cos x$ . Representing those with respect to the codomain’s basis (again, $B$ ) and adjoining the representations gives this matrix.
  
  ${Rep}_{B, B} (\frac{d}{d x}) = {(\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})}_{B, B}$
2. The images of the vectors in the domain’s basis are $e^{x} \overset{d / d x}{⟼} e^{x}$ and $e^{2 x} \overset{d / d x}{⟼} 2 e^{2 x}$ . Representing with respect to the codomain’s basis and adjoining gives this matrix.
  
  ${Rep}_{B, B} (\frac{d}{d x}) = {(\begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix})}_{B, B}$
3. The images of the members of the domain’s basis are $1 \overset{d / d x}{⟼} 0$ , $x \overset{d / d x}{⟼} 1$ , $e^{x} \overset{d / d x}{⟼} e^{x}$ , and $x e^{x} \overset{d / d x}{⟼} e^{x} + x e^{x}$ . Representing these images with respect to $B$ and adjoining gives this matrix.
  
  ${Rep}_{B, B} (\frac{d}{d x}) = {(\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{matrix})}_{B, B}$
Exercise 1.31 Worked answer

Find the range of the linear transformation of $ℝ^{2}$ represented with respect to the standard bases by each matrix.
1. $(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$
2. $(\begin{matrix} 0 & 0 \\ 3 & 2 \end{matrix})$
3. a matrix of the form $(\begin{matrix} a & b \\ 2 a & 2 b \end{matrix})$
Back to Exercise 1.31

Answer.
1. It is the set of vectors of the codomain represented with respect to the codomain’s basis in this way.
  
  ${(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) ∣ x, y \in ℝ} = {(\begin{matrix} x \\ 0 \end{matrix}) ∣ x, y \in ℝ}$
  
  As the codomain’s basis is $ℰ_{2}$ , and so each vector is represented by itself, the range of this transformation is the $x$ -axis.
2. It is the set of vectors of the codomain represented in this way.
  
  ${(\begin{matrix} 0 & 0 \\ 3 & 2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) ∣ x, y \in ℝ} = {(\begin{matrix} 0 \\ 3 x + 2 y \end{matrix}) ∣ x, y \in ℝ}$
  
  With respect to $ℰ_{2}$ vectors represent themselves so this range is the $y$ axis.
3. The set of vectors represented with respect to $ℰ_{2}$ as
  
  $\begin{aligned} {(\begin{array}{c} a & b \\ 2 a & 2 b \end{array}) (\begin{array}{c} x \\ y \end{array}) ∣ x, y \in ℝ} & = {(\begin{array}{c} a x + b y \\ 2 a x + 2 b y \end{array}) ∣ x, y \in ℝ} \\ = {(a x + b y) \cdot (\begin{array}{c} 1 \\ 2 \end{array}) ∣ x, y \in ℝ} \end{aligned}$
  
  is the line $y = 2 x$ , provided either $a$ or $b$ is not zero, and is the set consisting of just the origin if both are zero.
Exercise 1.32 Worked answer

Recommended. Can one matrix represent two different linear maps? That is, can ${Rep}_{B, D} (h) = {Rep}_{\hat{B}, \hat{D}} (\hat{h})$ ?

Back to Exercise 1.32

Answer. Yes, for two reasons.

First, the two maps $h$ and $\hat{h}$ need not have the same domain and codomain. For instance,

$(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})$

represents a map $h : ℝ^{2} \to ℝ^{2}$ with respect to the standard bases that sends

$(\begin{matrix} 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 1 \\ 3 \end{matrix}) and (\begin{matrix} 0 \\ 1 \end{matrix}) \mapsto (\begin{matrix} 2 \\ 4 \end{matrix})$

and also represents a $\hat{h} : 𝒫_{1} \to ℝ^{2}$ with respect to $⟨ 1, x ⟩$ and $ℰ_{2}$ that acts in this way.

$1 \mapsto (\begin{matrix} 1 \\ 3 \end{matrix}) and x \mapsto (\begin{matrix} 2 \\ 4 \end{matrix})$

The second reason is that, even if the domain and codomain of $h$ and $\hat{h}$ coincide, different bases produce different maps. An example is the $2 \times 2$ identity matrix

$I = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$

which represents the identity map on $ℝ^{2}$ with respect to $ℰ_{2}, ℰ_{2}$ . However, with respect to $ℰ_{2}$ for the domain but the basis $D = ⟨ {\vec{e}}_{2}, {\vec{e}}_{1} ⟩$ for the codomain, the same matrix $I$ represents the map that swaps the first and second components

$(\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} y \\ x \end{matrix})$

(that is, reflection about the line $y = x$ ).
Exercise 1.33 Worked answer

Prove Theorem 1.5.

Back to Exercise 1.33

Answer. We mimic Example 1.1, just replacing the numbers with letters.

Write $B$ as $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ and $D$ as $⟨ {\vec{δ}}_{1}, \dots, {\vec{δ}}_{m} ⟩$ . By definition of representation of a map with respect to bases, the assumption that

${Rep}_{B, D} (h) = (\begin{matrix} h_{1, 1} & \dots & h_{1, n} \\ ⋮ & ⋮ \\ h_{m, 1} & \dots & h_{m, n} \end{matrix})$

means that $h ({\vec{β}}_{i}) = h_{i, 1} {\vec{δ}}_{1} + \dots + h_{i, n} {\vec{δ}}_{n}$ . And, by the definition of the representation of a vector with respect to a basis, the assumption that

${Rep}_{B} (\vec{v}) = (\begin{matrix} c_{1} \\ ⋮ \\ c_{n} \end{matrix})$

means that $\vec{v} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ . Substituting gives

$\begin{aligned} h (\vec{v}) & = h (c_{1} \cdot {\vec{β}}_{1} + \dots + c_{n} \cdot {\vec{β}}_{n}) \\ = c_{1} \cdot h ({\vec{β}}_{1}) + \dots + c_{n} \cdot {\vec{β}}_{n} \\ = c_{1} \cdot (h_{1, 1} {\vec{δ}}_{1} + \dots + h_{m, 1} {\vec{δ}}_{m}) + \dots + c_{n} \cdot (h_{1, n} {\vec{δ}}_{1} + \dots + h_{m, n} {\vec{δ}}_{m}) \\ = (h_{1, 1} c_{1} + \dots + h_{1, n} c_{n}) \cdot {\vec{δ}}_{1} + \dots + (h_{m, 1} c_{1} + \dots + h_{m, n} c_{n}) \cdot {\vec{δ}}_{m} \end{aligned}$

and so $h (\vec{v})$ is represented as required.
Exercise 1.34 Worked answer

Recommended. Example 1.10 shows how to represent rotation of all vectors in the plane through an angle $θ$ about the origin, with respect to the standard bases.
1. Rotation of all vectors in three-space through an angle $θ$ about the $x$ -axis is a transformation of $ℝ^{3}$ . Represent it with respect to the standard bases. Arrange the rotation so that to someone whose feet are at the origin and whose head is at $(1, 0, 0)$ , the movement appears clockwise.
2. Repeat the prior item, only rotate about the $y$ -axis instead. (Put the person’s head at ${\vec{e}}_{2}$ .)
3. Repeat, about the $z$ -axis.
4. Extend the prior item to $ℝ^{4}$ . (Hint: we can restate ‘rotate about the $z$ -axis’ as ‘rotate parallel to the $x y$ -plane’.)
Back to Exercise 1.34

Answer.
1. The picture is this.
  
  The images of the vectors from the domain’s basis
  
  $(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 0 \\ \cos θ \\ - \sin θ \end{matrix}) (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) \mapsto (\begin{matrix} 0 \\ \sin θ \\ \cos θ \end{matrix})$
  
  are represented with respect to the codomain’s basis (again, $ℰ_{3}$ ) by themselves, so adjoining the representations to make the matrix gives this.
  
  ${Rep}_{ℰ_{3}, ℰ_{3}} (r_{θ}) = (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos θ & \sin θ \\ 0 & - \sin θ & \cos θ \end{matrix})$
2. The picture is similar to the one in the prior answer. The images of the vectors from the domain’s basis
  
  $(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \mapsto (\begin{matrix} \cos θ \\ 0 \\ \sin θ \end{matrix}) (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) \mapsto (\begin{matrix} - \sin θ \\ 0 \\ \cos θ \end{matrix})$
  
  are represented with respect to the codomain’s basis $ℰ_{3}$ by themselves, so this is the matrix.
  
  $(\begin{matrix} \cos θ & 0 & - \sin θ \\ 0 & 1 & 0 \\ \sin θ & 0 & \cos θ \end{matrix})$
3. To a person standing up, with the vertical $z$ -axis, a rotation of the $x y$ -plane that is clockwise proceeds from the positive $y$ -axis to the positive $x$ -axis. That is, it rotates opposite to the direction in Example 1.10. The images of the vectors from the domain’s basis
  
  $(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \mapsto (\begin{matrix} \cos θ \\ - \sin θ \\ 0 \end{matrix}) (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} \sin θ \\ \cos θ \\ 0 \end{matrix}) (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) \mapsto (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})$
  
  are represented with respect to $ℰ_{3}$ by themselves, so the matrix is this.
  
  $(\begin{matrix} \cos θ & \sin θ & 0 \\ - \sin θ & \cos θ & 0 \\ 0 & 0 & 1 \end{matrix})$
4. $(\begin{matrix} \cos θ & \sin θ & 0 & 0 \\ - \sin θ & \cos θ & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix})$
Exercise 1.35 Worked answer

(Schur’s Triangularization Lemma)
1. Let $U$ be a subspace of $V$ and fix bases $B_{U} \subseteq B_{V}$ . What is the relationship between the representation of a vector from $U$ with respect to $B_{U}$ and the representation of that vector (viewed as a member of $V$ ) with respect to $B_{V}$ ?
2. What about maps?
3. Fix a basis $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ for $V$ and observe that the spans
  
  $[\emptyset] = {\vec{0}} \subset [{{\vec{β}}_{1}}] \subset [{{\vec{β}}_{1}, {\vec{β}}_{2}}] \subset \dots \subset [B] = V$
  
  form a strictly increasing chain of subspaces. Show that for any linear map $h : V \to W$ there is a chain $W_{0} = {\vec{0}} \subseteq W_{1} \subseteq \dots \subseteq W_{m} = W$ of subspaces of $W$ such that
  
  $h ([{{\vec{β}}_{1}, \dots, {\vec{β}}_{i}}]) \subseteq W_{i}$
  
  for each $i$ .
4. Conclude that for every linear map $h : V \to W$ there are bases $B, D$ so the matrix representing $h$ with respect to $B, D$ is upper-triangular (that is, each entry $h_{i, j}$ with $i > j$ is zero).
5. Is an upper-triangular representation unique?
Back to Exercise 1.35

Answer.
1. Write the basis $B_{U}$ as $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{k} ⟩$ and then write $B_{V}$ as the extension $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{k}, {\vec{β}}_{k + 1}, \dots, {\vec{β}}_{n} ⟩$ . If
  
  ${Rep}_{B_{U}} (\vec{v}) = (\begin{matrix} c_{1} \\ ⋮ \\ c_{k} \end{matrix})$
  
  so that $\vec{v} = c_{1} \cdot {\vec{β}}_{1} + \dots + c_{k} \cdot {\vec{β}}_{k}$ then
  
  ${Rep}_{B_{V}} (\vec{v}) = (\begin{matrix} c_{1} \\ ⋮ \\ c_{k} \\ 0 \\ ⋮ \\ 0 \end{matrix})$
  
  because $\vec{v} = c_{1} \cdot {\vec{β}}_{1} + \dots + c_{k} \cdot {\vec{β}}_{k} + 0 \cdot {\vec{β}}_{k + 1} + \dots + 0 \cdot {\vec{β}}_{n}$ .
2. We must first decide what the question means. Compare $h : V \to W$ with its restriction to the subspace $h ↾_{U} : U \to W$ . The range space of the restriction is a subspace of $W$ , so fix a basis $D_{h (U)}$ for this range space and extend it to a basis $D_{V}$ for $W$ . We want the relationship between these two.
  
  ${Rep}_{B_{V}, D_{V}} (h) and {Rep}_{B_{U}, D_{h (U)}} (h ↾_{U})$
  
  The answer falls right out of the prior item: if
  
  ${Rep}_{B_{U}, D_{h (U)}} (h ↾_{U}) = (\begin{matrix} h_{1, 1} & \dots & h_{1, k} \\ ⋮ & ⋮ \\ h_{p, 1} & \dots & h_{p, k} \end{matrix})$
  
  then the extension is represented in this way.
  
  ${Rep}_{B_{V}, D_{V}} (h) = (\begin{matrix} h_{1, 1} & \dots & h_{1, k} & h_{1, k + 1} & \dots & h_{1, n} \\ ⋮ & ⋮ \\ h_{p, 1} & \dots & h_{p, k} & h_{p, k + 1} & \dots & h_{p, n} \\ 0 & \dots & 0 & h_{p + 1, k + 1} & \dots & h_{p + 1, n} \\ ⋮ & ⋮ \\ 0 & \dots & 0 & h_{m, k + 1} & \dots & h_{m, n} \end{matrix})$
3. Take $W_{i}$ to be the span of ${h ({\vec{β}}_{1}), \dots, h ({\vec{β}}_{i})}$ .
4. Apply the answer from the second item to the third item.
5. No. For instance $π_{x} : ℝ^{2} \to ℝ^{2}$ , projection onto the $x$ axis, is represented by these two upper-triangular matrices
  
  ${Rep}_{ℰ_{2}, ℰ_{2}} (π_{x}) = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) and {Rep}_{C, ℰ_{2}} (π_{x}) = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$
  
  where $C = ⟨ {\vec{e}}_{2}, {\vec{e}}_{1} ⟩$ .

Any Matrix Represents a Linear Map

The prior subsection shows that the action of a linear map $h$ is described by a matrix $H$ , with respect to appropriate bases, in this way.

$\begin{matrix} \vec{v} = {(\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix})}_{B} \overset{h}{\underset{H}{⟼}} h (\vec{v}) = {(\begin{matrix} h_{1, 1} v_{1} + \dots + h_{1, n} v_{n} \\ ⋮ \\ h_{m, 1} v_{1} + \dots + h_{m, n} v_{n} \end{matrix})}_{D} & (*) \end{matrix}$

Here we will show the converse, that each matrix represents a linear map.

So we start with a matrix

$H = (\begin{array}{cccc} h_{1, 1} & h_{1, 2} & \dots & h_{1, n} \\ h_{2, 1} & h_{2, 2} & \dots & h_{2, n} \\ ⋮ \\ h_{m, 1} & h_{m, 2} & \dots & h_{m, n} \end{array})$

and we will describe how it defines a map $h$ . We require that the map be represented by the matrix so first note that in ( $*$ ) the dimension of the map’s domain is the number of columns $n$ of the matrix and the dimension of the codomain is the number of rows $m$ . Thus, for $h$ ’s domain fix an $n$ -dimensional vector space $V$ and for the codomain fix an $m$ -dimensional space $W$ . Also fix bases $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ and $D = ⟨ {\vec{δ}}_{1}, \dots, {\vec{δ}}_{m} ⟩$ for those spaces.

Now let $h : V \to W$ be: where $\vec{v}$ in the domain has the representation

${Rep}_{B} (\vec{v}) = {(\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix})}_{B}$

then its image $h (\vec{v})$ is the member of the codomain with this representation.

${Rep}_{D} (h (\vec{v})) = {(\begin{matrix} h_{1, 1} v_{1} + \dots + h_{1, n} v_{n} \\ ⋮ \\ h_{m, 1} v_{1} + \dots + h_{m, n} v_{n} \end{matrix})}_{D}$

That is, to compute the action of $h$ on any $\vec{v} \in V$ , first express $\vec{v}$ with respect to the basis $\vec{v} = v_{1} {\vec{β}}_{1} + \dots + v_{n} {\vec{β}}_{n}$ and then $h (\vec{v}) = (h_{1, 1} v_{1} + \dots + h_{1, n} v_{n}) \cdot {\vec{δ}}_{1} + \dots + (h_{m, 1} v_{1} + \dots + h_{m, n} v_{n}) \cdot {\vec{δ}}_{m}$ .

Above we have made some choices; for instance $V$ can be any $n$ -dimensional space and $B$ could be any basis for $V$ , so $H$ does not define a unique function. However, note once we have fixed $V$ , $B$ , $W$ , and $D$ then $h$ is well-defined since $\vec{v}$ has a unique representation with respect to the basis $B$ and the calculation of $\vec{w}$ from its representation is also uniquely determined.

Example 2.1 Consider this matrix.

$H = (\begin{matrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{matrix})$

It is $3 \times 2$ so any map that it defines must carry a dimension $2$ domain to a dimension $3$ codomain. We can choose the domain and codomain to be $ℝ^{2}$ and $𝒫_{2}$ , with these bases.

$B = ⟨ (\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ - 1 \end{matrix}) ⟩ D = ⟨ x^{2}, x^{2} + x, x^{2} + x + 1 ⟩$

Then let $h : ℝ^{2} \to 𝒫_{2}$ be the function defined by $H$ . We will compute the image under $h$ of this member of the domain.

$\vec{v} = (\begin{matrix} - 3 \\ 2 \end{matrix})$

The computation is straightforward.

${Rep}_{D} (h (\vec{v})) = H \cdot {Rep}_{B} (\vec{v}) = (\begin{matrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{matrix}) (\begin{matrix} - 1 / 2 \\ - 5 / 2 \end{matrix}) = (\begin{matrix} - 11 / 2 \\ - 23 / 2 \\ - 35 / 2 \end{matrix})$

From its representation, computation of $h (\vec{v})$ is routine $(- 11 / 2) (x^{2}) - (23 / 2) (x^{2} + x) - (35 / 2) (x^{2} + x + 1) = (- 69 / 2) x^{2} - (58 / 2) x - (35 / 2)$ .

Theorem 2.2 Any matrix represents a homomorphism between vector spaces of appropriate dimensions, with respect to any pair of bases.

Proof We must check that for any matrix $H$ and any domain and codomain bases $B, D$ , the defined map $h$ is linear. If $\vec{v}, \vec{u} \in V$ are such that

${Rep}_{B} (\vec{v}) = (\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix}) {Rep}_{B} (\vec{u}) = (\begin{matrix} u_{1} \\ ⋮ \\ u_{n} \end{matrix})$

and $c, d \in ℝ$ then the calculation

$\begin{aligned} h (c \vec{v} + d \vec{u}) & = (h_{1, 1} (c v_{1} + d u_{1}) + \dots + h_{1, n} (c v_{n} + d u_{n})) \cdot {\vec{δ}}_{1} + \\ \dots + (h_{m, 1} (c v_{1} + d u_{1}) + \dots + h_{m, n} (c v_{n} + d u_{n})) \cdot {\vec{δ}}_{m} \\ = c \cdot h (\vec{v}) + d \cdot h (\vec{u}) \end{aligned}$

supplies that check.

QED

Example 2.3 Even if the domain and codomain are the same, the map that the matrix represents depends on the bases that we choose. If

$H = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), B_{1} = D_{1} = ⟨ (\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix}) ⟩, and B_{2} = D_{2} = ⟨ (\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix}) ⟩,$

then $h_{1} : ℝ^{2} \to ℝ^{2}$ represented by $H$ with respect to $B_{1}, D_{1}$ maps

$(\begin{matrix} c_{1} \\ c_{2} \end{matrix}) = {(\begin{matrix} c_{1} \\ c_{2} \end{matrix})}_{B_{1}} \mapsto {(\begin{matrix} c_{1} \\ 0 \end{matrix})}_{D_{1}} = (\begin{matrix} c_{1} \\ 0 \end{matrix})$

while $h_{2} : ℝ^{2} \to ℝ^{2}$ represented by $H$ with respect to $B_{2}, D_{2}$ is this map.

$(\begin{matrix} c_{1} \\ c_{2} \end{matrix}) = {(\begin{matrix} c_{2} \\ c_{1} \end{matrix})}_{B_{2}} \mapsto {(\begin{matrix} c_{2} \\ 0 \end{matrix})}_{D_{2}} = (\begin{matrix} 0 \\ c_{2} \end{matrix})$

These are different functions. The first is projection onto the $x$ -axis while the second is projection onto the $y$ -axis.

This result means that when convenient we can work solely with matrices, just doing the computations without having to worry whether a matrix of interest represents a linear map on some pair of spaces.

When we are working with a matrix but we do not have particular spaces or bases in mind then we can take the domain and codomain to be $ℝ^{n}$ and $ℝ^{m}$ , with the standard bases. This is convenient because with the standard bases vector representation is transparent— the representation of $\vec{v}$ is $\vec{v}$ . (In this case the column space of the matrix equals the range of the map and consequently the column space of $H$ is often denoted by $ℛ (H)$ .)

Given a matrix, to come up with an associated map we can choose among many domain and codomain spaces, and many bases for those. So a matrix can represent many maps. We finish this section by illustrating how the matrix can give us information about the associated maps.

Theorem 2.4 The rank of a matrix equals the rank of any map that it represents.

Proof Suppose that the matrix $H$ is $m \times n$ . Fix domain and codomain spaces $V$ and $W$ of dimension $n$ and $m$ with bases $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ and $D$ . Then $H$ represents some linear map $h$ between those spaces with respect to these bases whose range space

$\begin{aligned} {h (\vec{v}) ∣ \vec{v} \in V} & = {h (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}) ∣ c_{1}, \dots, c_{n} \in ℝ} \\ = {c_{1} h ({\vec{β}}_{1}) + \dots + c_{n} h ({\vec{β}}_{n}) ∣ c_{1}, \dots, c_{n} \in ℝ} \end{aligned}$

is the span $[{h ({\vec{β}}_{1}), \dots, h ({\vec{β}}_{n})}]$ . The rank of the map $h$ is the dimension of this range space.

The rank of the matrix is the dimension of its column space, the span of the set of its columns $[{{Rep}_{D} (h ({\vec{β}}_{1})), \dots, {Rep}_{D} (h ({\vec{β}}_{n}))}]$ .

To see that the two spans have the same dimension, recall from the proof of Lemma I.2.5 that if we fix a basis then representation with respect to that basis gives an isomorphism ${Rep}_{D} : W \to ℝ^{m}$ . Under this isomorphism there is a linear relationship among members of the range space if and only if the same relationship holds in the column space, e.g, $\vec{0} = c_{1} \cdot h ({\vec{β}}_{1}) + \dots + c_{n} \cdot h ({\vec{β}}_{n})$ if and only if $\vec{0} = c_{1} \cdot {Rep}_{D} (h ({\vec{β}}_{1})) + \dots + c_{n} \cdot {Rep}_{D} (h ({\vec{β}}_{n}))$ . Hence, a subset of the range space is linearly independent if and only if the corresponding subset of the column space is linearly independent. Therefore the size of the largest linearly independent subset of the range space equals the size of the largest linearly independent subset of the column space, and so the two spaces have the same dimension.

QED

That settles the apparent ambiguity in our use of the same word ‘rank’ to apply both to matrices and to maps.

Example 2.5 Any map represented by

$(\begin{matrix} 1 & 2 & 2 \\ 1 & 2 & 1 \\ 0 & 0 & 3 \\ 0 & 0 & 2 \end{matrix})$

must have three-dimensional domain and a four-dimensional codomain. In addition, because the rank of this matrix is two (we can spot this by eye or get it with Gauss’s Method), any map represented by this matrix has a two-dimensional range space.

Corollary 2.6 Let $h$ be a linear map represented by a matrix $H$ . Then $h$ is onto if and only if the rank of $H$ equals the number of its rows, and $h$ is one-to-one if and only if the rank of $H$ equals the number of its columns.

Proof For the onto half, the dimension of the range space of $h$ is the rank of $h$ , which equals the rank of $H$ by the theorem. Since the dimension of the codomain of $h$ equals the number of rows in $H$ , if the rank of $H$ equals the number of rows then the dimension of the range space equals the dimension of the codomain. But a subspace with the same dimension as its superspace must equal that superspace (because any basis for the range space is a linearly independent subset of the codomain whose size is equal to the dimension of the codomain, and thus this basis for the range space must also be a basis for the codomain).

For the other half, a linear map is one-to-one if and only if it is an isomorphism between its domain and its range, that is, if and only if its domain has the same dimension as its range. The number of columns in $H$ is the dimension of $h$ ’s domain and by the theorem the rank of $H$ equals the dimension of $h$ ’s range.

QED

Definition 2.7 A linear map that is one-to-one and onto is nonsingular , otherwise it is singular . That is, a linear map is nonsingular if and only if it is an isomorphism.

Remark 2.8 Some authors use ‘nonsingular’ as a synonym for one-to-one while others use it the way that we have here. The difference is slight because any map is onto its range space, so a one-to-one map is an isomorphism with its range.

In the first chapter we defined a matrix to be nonsingular if it is square and is the matrix of coefficients of a linear system with a unique solution. The next result justifies our dual use of the term.

Lemma 2.9 A nonsingular linear map is represented by a square matrix. A square matrix represents nonsingular maps if and only if it is a nonsingular matrix. Thus, a matrix represents isomorphisms if and only if it is square and nonsingular.

Proof Assume that the map $h : V \to W$ is nonsingular. Corollary 2.6 says that for any matrix $H$ representing that map, because $h$ is onto the number of rows of $H$ equals the rank of $H$ , and because $h$ is one-to-one the number of columns of $H$ is also equal to the rank of $H$ . Hence $H$ is square.

Next assume that $H$ is square, $n \times n$ . The matrix $H$ is nonsingular if and only if its row rank is $n$ , which is true if and only if $H$ ’s rank is $n$ by Theorem Two.III.3.11, which is true if and only if $h$ ’s rank is $n$ by Theorem 2.4, which is true if and only if $h$ is an isomorphism by Theorem I.2.3. (This last holds because the domain of $h$ is $n$ -dimensional as it is the number of columns in $H$ .)

QED

Example 2.10 Any map from $ℝ^{2}$ to $𝒫_{1}$ represented with respect to any pair of bases by

$(\begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix})$

is nonsingular because this matrix has rank two.

Example 2.11 Any map $g : V \to W$ represented by

$(\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix})$

is singular because this matrix is singular.

We’ve now seen that the relationship between maps and matrices goes both ways: for a particular pair of bases, any linear map is represented by a matrix and any matrix describes a linear map. That is, by fixing spaces and bases we get a correspondence between maps and matrices. In the rest of this chapter we will explore this correspondence. For instance, we’ve defined for linear maps the operations of addition and scalar multiplication and we shall see what the corresponding matrix operations are. We shall also see the matrix operation that represent the map operation of composition. And, we shall see how to find the matrix that represents a map’s inverse.

Exercises

Exercise 2.12 Worked answer

For each matrix, state the dimension of the domain and codomain of any map that the matrix represents.
1. $(\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix})$
2. $(\begin{matrix} 1 & 1 & - 3 \\ 2 & 5 & 0 \end{matrix})$
3. $(\begin{matrix} 1 & 3 \\ 1 & 4 \\ 1 & - 1 \end{matrix})$
4. $(\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})$
5. $(\begin{matrix} 1 & - 1 & 4 & 5 \\ 0 & 0 & 0 & 0 \end{matrix})$
Back to Exercise 2.12

Answer.
1. domain: $2$ , codomain: $2$
2. domain: $3$ , codomain: $2$
3. domain: $2$ , codomain: $3$
4. domain: $3$ , codomain: $2$
5. domain: $4$ , codomain: $2$
Exercise 2.13 Worked answer

Consider a linear map $f : V \to W$ represented with respect to some bases $B, D$ by the matrix. Decide if that map is nonsingular.
1. $(\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix})$
2. $(\begin{matrix} 1 & 1 \\ - 3 & - 3 \end{matrix})$
3. $(\begin{matrix} 3 & 0 & 0 \\ 2 & 1 & 0 \\ 4 & 4 & 4 \end{matrix})$
4. $(\begin{matrix} 2 & 0 & - 2 \\ 1 & 1 & 0 \\ 4 & 1 & - 4 \end{matrix})$
Back to Exercise 2.13

Answer. For each we just have to decide if the matrix is nonsingular, perhaps by doing Gauss’s Method. (In truth, we can do each of these by eye.)
1. This matrix is nonsingular, since the second row is not a multiple of the first, so the map is nonsingular.
2. The matrix is singular so the map is singular.
3. Nonsingular.
4. Nonsingular.
Exercise 2.14 Worked answer

Recommended. Let $h$ be the linear map defined by this matrix on the domain $𝒫_{1}$ and codomain $ℝ^{2}$ with respect to the given bases.

$H = (\begin{matrix} 2 & 1 \\ 4 & 2 \end{matrix}) B = ⟨ 1 + x, x ⟩, D = ⟨ (\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix}) ⟩$

What is the image under $h$ of the vector $\vec{v} = 2 x - 1$ ?

Back to Exercise 2.14

Answer. With respect to $B$ the vector’s representation is this.

${Rep}_{B} (2 x - 1) = (\begin{matrix} - 1 \\ 3 \end{matrix})$

Using the matrix-vector product we can compute ${Rep}_{D} (h (\vec{v}))$

${Rep}_{D} (h (2 x - 1)) = (\begin{matrix} 2 & 1 \\ 4 & 2 \end{matrix}) {(\begin{matrix} - 1 \\ 3 \end{matrix})}_{B} = {(\begin{matrix} 1 \\ 2 \end{matrix})}_{D}$

From that representation we can compute $h (\vec{v})$ .

$h (2 x - 1) = 1 \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) + 2 \cdot (\begin{matrix} 1 \\ 0 \end{matrix}) = (\begin{matrix} 3 \\ 1 \end{matrix})$
Exercise 2.15 Worked answer

Recommended. Decide if each vector lies in the range of the map from $ℝ^{3}$ to $ℝ^{2}$ represented with respect to the standard bases by the matrix.
1. $(\begin{matrix} 1 & 1 & 3 \\ 0 & 1 & 4 \end{matrix})$ , $(\begin{matrix} 1 \\ 3 \end{matrix})$
2. $(\begin{matrix} 2 & 0 & 3 \\ 4 & 0 & 6 \end{matrix})$ , $(\begin{matrix} 1 \\ 1 \end{matrix})$
Back to Exercise 2.15

Answer. As described in the subsection, with respect to the standard bases, representations are transparent, and so, for instance, the first matrix describes this map.

$(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})}_{ℰ_{3}} \mapsto {(\begin{matrix} 1 \\ 0 \end{matrix})}_{ℰ_{2}} = (\begin{matrix} 1 \\ 0 \end{matrix}) (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 1 \\ 1 \end{matrix}) (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) \mapsto (\begin{matrix} 3 \\ 4 \end{matrix})$

So, for this first one, we are asking whether there are scalars such that

$c_{1} (\begin{matrix} 1 \\ 0 \end{matrix}) + c_{2} (\begin{matrix} 1 \\ 1 \end{matrix}) + c_{3} (\begin{matrix} 3 \\ 4 \end{matrix}) = (\begin{matrix} 1 \\ 3 \end{matrix})$

that is, whether the vector is in the column space of the matrix.
1. Yes. We can get this conclusion by setting up the resulting linear system and applying Gauss’s Method, as usual. Another way to get it is to note by inspection of the equation of columns that taking $c_{3} = 3 / 4$ , and $c_{1} = - 5 / 4$ , and $c_{2} = 0$ will do. Still a third way to get this conclusion is to note that the rank of the matrix is two, which equals the dimension of the codomain, and so the map is onto—the range is all of $ℝ^{2}$ and in particular includes the given vector.
2. No; note that all of the columns in the matrix have a second component that is twice the first, while the vector does not. Alternatively, the column space of the matrix is
  
  ${c_{1} (\begin{matrix} 2 \\ 4 \end{matrix}) + c_{2} (\begin{matrix} 0 \\ 0 \end{matrix}) + c_{3} (\begin{matrix} 3 \\ 6 \end{matrix}) ∣ c_{1}, c_{2}, c_{3} \in ℝ} = {c (\begin{matrix} 1 \\ 2 \end{matrix}) ∣ c \in ℝ}$
  
  (which is the fact already noted, but we got it by calculation rather than inspiration), and the given vector is not in this set.
Exercise 2.16 Worked answer

Recommended. Consider this matrix, representing a transformation of $ℝ^{2}$ with respect to the bases.

$\frac{1}{2} \cdot (\begin{matrix} 1 & 1 \\ - 1 & 1 \end{matrix}) B = ⟨ (\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix}) ⟩ D = ⟨ (\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ - 1 \end{matrix}) ⟩$
1. To what vector in the codomain is the first member of $B$ mapped?
2. The second member?
3. Where is a general vector from the domain (a vector with components $x$ and $y$ ) mapped? That is, what transformation of $ℝ^{2}$ is represented with respect to $B, D$ by this matrix?
Back to Exercise 2.16

Answer.
1. The first member of the basis
  
  $(\begin{matrix} 0 \\ 1 \end{matrix}) = {(\begin{matrix} 1 \\ 0 \end{matrix})}_{B}$
  
  maps to
  
  ${(\begin{matrix} 1 / 2 \\ - 1 / 2 \end{matrix})}_{D}$
  
  which is this member of the codomain.
  
  $\frac{1}{2} \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) - \frac{1}{2} \cdot (\begin{matrix} 1 \\ - 1 \end{matrix}) = (\begin{matrix} 0 \\ 1 \end{matrix})$
2. The second member of the basis maps
  
  $(\begin{matrix} 1 \\ 0 \end{matrix}) = {(\begin{matrix} 0 \\ 1 \end{matrix})}_{B} \mapsto {(\begin{matrix} (1 / 2 \\ 1 / 2 \end{matrix})}_{D}$
  
  to this member of the codomain.
  
  $\frac{1}{2} \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) + \frac{1}{2} \cdot (\begin{matrix} 1 \\ - 1 \end{matrix}) = (\begin{matrix} 1 \\ 0 \end{matrix})$
3. Because the map that the matrix represents is the identity map on the basis, it must be the identity on all members of the domain. We can come to the same conclusion in another way by considering
  
  $(\begin{matrix} x \\ y \end{matrix}) = {(\begin{matrix} y \\ x \end{matrix})}_{B}$
  
  which maps to
  
  ${(\begin{matrix} (x + y) / 2 \\ (x - y) / 2 \end{matrix})}_{D}$
  
  which represents this member of $ℝ^{2}$ .
  
  $\frac{x + y}{2} \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) + \frac{x - y}{2} \cdot (\begin{matrix} 1 \\ - 1 \end{matrix}) = (\begin{matrix} x \\ y \end{matrix})$
Exercise 2.17 Worked answer

Consider a homomorphism $h : ℝ^{2} \to ℝ^{2}$ represented with respect to the standard bases $ℰ_{2}, ℰ_{2}$ by this matrix.

$(\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix})$

Find the image under $h$ of each vector.
1. $(\begin{matrix} 2 \\ 3 \end{matrix})$
2. $(\begin{matrix} 0 \\ 1 \end{matrix})$
3. $(\begin{matrix} - 1 \\ 1 \end{matrix})$
Back to Exercise 2.17

Answer. Since, with respect to the standard basis $ℰ_{2}$ , a matrix is represented by itself, we just need to do the matrix-vector multiplication.
1. $\begin{aligned} {Rep}_{ℰ_{2}} (h ((\begin{array}{c} 2 \\ 3 \end{array}))) & = {Rep}_{ℰ_{2}, ℰ_{2}} (h) {Rep}_{ℰ_{2}} ((\begin{array}{c} 2 \\ 3 \end{array})) \\ = {(\begin{array}{c} 1 & 3 \\ 2 & 4 \end{array})}_{ℰ_{2}, ℰ_{2}} {(\begin{array}{c} 2 \\ 3 \end{array})}_{ℰ_{2}} \\ = {(\begin{array}{c} 11 \\ 16 \end{array})}_{ℰ_{2}} = (\begin{array}{c} 11 \\ 16 \end{array}) \end{aligned}$
2. $(\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) = (\begin{matrix} 3 \\ 4 \end{matrix})$
3. $(\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix}) (\begin{matrix} - 1 \\ 1 \end{matrix}) = (\begin{matrix} 2 \\ 2 \end{matrix})$
Exercise 2.18 Worked answer

What transformation of $F = {a \cos θ + b \sin θ ∣ a, b \in ℝ}$ is represented with respect to $B = ⟨ \cos θ - \sin θ, \sin θ ⟩$ and $D = ⟨ \cos θ + \sin θ, \cos θ ⟩$ by this matrix?

$(\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$

Back to Exercise 2.18

Answer. A general member of the domain, represented with respect to the domain’s basis as

$a \cos θ + b \sin θ = {(\begin{matrix} a \\ a + b \end{matrix})}_{B}$

maps to

${(\begin{matrix} 0 \\ a \end{matrix})}_{D} representing 0 \cdot (\cos θ + \sin θ) + a \cdot (\cos θ)$

and so the linear map represented by the matrix with respect to these bases

$a \cos θ + b \sin θ \mapsto a \cos θ$

is projection onto the first component.
Exercise 2.19 Worked answer

Recommended. Decide whether $1 + 2 x$ is in the range of the map from $ℝ^{3}$ to $𝒫_{2}$ represented with respect to $ℰ_{3}$ and $⟨ 1, 1 + x^{2}, x ⟩$ by this matrix.

$(\begin{matrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix})$

Back to Exercise 2.19

Answer. Denote the given basis of $𝒫_{2}$ by $B$ . Application of the linear map is represented by matrix-vector multiplication. Thus the first vector in $ℰ_{3}$ maps to the element of $𝒫_{2}$ represented with respect to $B$ by

$(\begin{matrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) = (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix})$

and that element is $1 + x$ . Calculate the other two images of basis vectors in the same way.

$(\begin{matrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}) (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) = (\begin{matrix} 3 \\ 1 \\ 0 \end{matrix}) = {Rep}_{B} (4 + x^{2}) (\begin{matrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}) (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) = {Rep}_{B} (x)$

So the range of $h$ is the span of three polynomials $1 + x$ , $4 + x^{2}$ , and $x$ . We can thus decide if $1 + 2 x$ is in the range of the map by looking for scalars $c_{1}$ , $c_{2}$ , and $c_{3}$ such that

$c_{1} \cdot (1 + x) + c_{2} \cdot (4 + x^{2}) + c_{3} \cdot (x) = 1 + 2 x$

and obviously $c_{1} = 1$ , $c_{2} = 0$ , and $c_{3} = 1$ suffice. Thus $1 + 2 x$ is in the range, since it is the image of this vector.

$1 \cdot (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + 0 \cdot (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) + 1 \cdot (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})$

Comment. A slicker argument is to note that the matrix is nonsingular, so it has rank $3$ , so the range has dimension $3$ , and since the codomain has dimension $3$ the map is onto. Thus every polynomial is the image of some vector and in particular $1 + 2 x$ is the image of a vector in the domain.
Exercise 2.20 Worked answer

Find the map that this matrix represents with respect to $B, B$ .

$(\begin{matrix} 2 & 1 \\ - 1 & 0 \end{matrix}) B = ⟨ (\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \end{matrix}) ⟩$

Back to Exercise 2.20

Answer. Where $B = ⟨ {\vec{β}}_{1}, {\vec{β}}_{2} ⟩$ we can find ${Rep}_{B} (\vec{v})$ by eye, where $\vec{v}$ is the general vector, with entries $x$ and $y$ .

$(\begin{matrix} x \\ y \end{matrix}) = a \cdot (\begin{matrix} 1 \\ 0 \end{matrix}) + b \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) gives b = y, a = x - y$

Thus the representation of general vector with respect to $B$ is this.

${Rep}_{B} ((\begin{matrix} x \\ y \end{matrix})) = (\begin{matrix} x - y \\ y \end{matrix})$

Compute the effect of the map with matrix-vector multiplication.

${(\begin{matrix} 2 & 1 \\ - 1 & 0 \end{matrix})}_{B} {(\begin{matrix} x - y \\ y \end{matrix})}_{B} = {(\begin{matrix} 2 (x - y) + y \\ - (x - y) \end{matrix})}_{B} = {(\begin{matrix} 2 x - y \\ - x + y \end{matrix})}_{B}$

Finish by converting back to the standard vector representation.

${(\begin{matrix} 2 x - y \\ - x + y \end{matrix})}_{B} = (2 x - y) \cdot (\begin{matrix} 1 \\ 0 \end{matrix}) + (- x + y) \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) = (\begin{matrix} x \\ - x + y \end{matrix})$
Exercise 2.21 Worked answer

Example 2.11 gives a matrix that is singular and is therefore associated with maps that are singular. We cannot state the action of the associated map $g$ on domain elements $\vec{v} \in V$ , because do not know the domain $V$ or codomain $W$ or the starting and ending bases $B$ and $D$ . But we can compute what happens to the representations ${Rep}_{B, D} (\vec{v})$ .
1. Find the set of column vectors representing the members of the null space of any map $g$ represented by this matrix.
2. Find the nullity of any such map $g$ .
3. Find the set of column vectors representing the members of the range space of any map $g$ represented by the matrix.
4. Find the rank of any such map $g$ .
5. Check that rank plus nullity equals the dimension of the domain.
Back to Exercise 2.21

Answer. Let the matrix be $G$ , and suppose that it represents $g : V \to W$ with respect to bases $B$ and $D$ . Because $G$ has two columns, the domain $V$ is two-dimensional. Because $G$ has two rows, the codomain $W$ is two-dimensional. The action of $g$ on a representation ${Rep}_{B, D} (\vec{v})$ of a general member of the domain is this.

${(\begin{matrix} x \\ y \end{matrix})}_{B} \mapsto {(\begin{matrix} x + 2 y \\ 3 x + 6 y \end{matrix})}_{D}$
1. No matter what is the codomain’s basis $D$ , the only representation of the zero vector ${\vec{0}}_{W}$ is
  
  ${Rep}_{D} (\vec{0}) = {(\begin{matrix} 0 \\ 0 \end{matrix})}_{D}$
  
  and so the set of representations of members of the null space is this.
  
  ${{(\begin{matrix} x \\ y \end{matrix})}_{B} ∣ x + 2 y = 0 and 3 x + 6 y = 0} = {y \cdot {(\begin{matrix} - 1 / 2 \\ 1 \end{matrix})}_{D} ∣ y \in ℝ}$
2. The nullity is $1$ . The representation map ${Rep}_{D} : W \to ℝ^{2}$ and its inverse are isomorphisms and so preserve the dimension of subspaces. The subspace of $ℝ^{2}$ that is in the prior item is one-dimensional. Therefore, the image of that subspace under the inverse of the representation map—the null space of $G$ , is also one-dimensional.
3. The set of representations of members of the range space is this.
  
  $\begin{aligned} {{(\begin{array}{c} x + 2 y \\ 3 x + 6 y \end{array})}_{D} ∣ x, y \in ℝ} & = {x \cdot {(\begin{array}{c} 1 \\ 3 \end{array})}_{D} + y \cdot {(\begin{array}{c} 2 \\ 6 \end{array})}_{D} ∣ x, y \in ℝ} \\ = {k \cdot {(\begin{array}{c} 1 \\ 3 \end{array})}_{D} ∣ k \in ℝ} \end{aligned}$
4. Of course, Theorem 2.4 gives that the rank of the map equals the rank of the matrix, which is one. Alternatively, the same argument that we used above for the null space gives here that the dimension of the range space is one.
5. One plus one equals two.
Exercise 2.22 Worked answer

Recommended. Take each matrix to represent $h : ℝ^{m} \to ℝ^{n}$ with respect to the standard bases. For each (i) state $m$ and $n$ . Then set up an augmented matrix with the given matrix on the left and a vector representing a range space element on the right (e.g., if the codomain is $ℝ^{3}$ then in the right-hand column put the three entries $a$ , $b$ , and $c$ ). Perform Gauss-Jordan reduction. Use that to (ii) find $ℛ (h)$ and $rank (h)$ (and state whether the underlying map is onto), and (iii) find $𝒩 (h)$ and $nullity (h)$ (and state whether the underlying map is one-to-one).
1. $(\begin{matrix} 2 & 1 \\ - 1 & 3 \end{matrix})$
2. $(\begin{matrix} 0 & 1 & 3 \\ 2 & 3 & 4 \\ - 2 & - 1 & 2 \end{matrix})$
3. $(\begin{matrix} 1 & 1 \\ 2 & 1 \\ 3 & 1 \end{matrix})$
Back to Exercise 2.22

Answer.
1. (i) The dimension of the domain space is the number of columns $m = 2$ . The dimension of the codomain space is the number of rows $n = 2$ .
  
  For the rest, we consider this matrix-vector equation.
  
  $\begin{matrix} (\begin{matrix} 2 & 1 \\ - 1 & 3 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} a \\ b \end{matrix}) & (*) \end{matrix}$
  
  We solve for $x$ and $y$ .
  
  $(\begin{array}{ccc} 2 & 1 & a \\ - 1 & 3 & b \end{array}) ⟶_{\begin{array}{l} \end{array}}^{(1 / 2) ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} (2 / 7) ρ_{2} \end{array}}^{(1 / 2) ρ_{1}} ⟶_{\begin{array}{l} \end{array}}^{- (1 / 2) ρ_{2} + ρ_{1}} (\begin{array}{ccc} 1 & 0 & (3 / 7) a - (1 / 7) b \\ 0 & 1 & (1 / 7) a + (2 / 7) b \end{array})$
  
  (ii) For all
  
  $(\begin{matrix} a \\ b \end{matrix}) \in ℝ^{2}$
  
  in equation ( $*$ ) the system has a solution, by the calculation. So the range space is all of the codomain $ℛ (h) = ℝ^{2}$ . The map’s rank is the dimension of the range, $2$ . The map is onto because the range space is all of the codomain.
  
  (iii) Again by the calculation, to find the nullspace, setting $a = b = 0$ in equation ( $*$ ) gives that $x = y = 0$ . The null space is the trivial subspace of the domain.
  
  $𝒩 (h) = {(\begin{matrix} 0 \\ 0 \end{matrix})}$
  
  The nullity is the dimension of that null space, $0$ . The map is one-to-one because the null space is trivial.
2. (i) The dimension of the domain space is the number of matrix columns, $m = 3$ , and the dimension of the codomain space is the number of rows, $n = 3$ .
  
  The calculation is this.
  
  $\begin{matrix} (\begin{array}{cccc} 0 & 1 & 3 & a \\ 2 & 3 & 4 & b \\ - 2 & - 1 & 2 & c \end{array}) ⟶_{\begin{array}{l} \end{array}}^{ρ_{1} \leftrightarrow ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{ρ_{1} + ρ_{3}} ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{2} + ρ_{3}} \\ ⟶_{\begin{array}{l} \end{array}}^{(1 / 2) ρ_{1}} ⟶_{\begin{array}{l} \end{array}}^{- (3 / 2) ρ_{2} + ρ_{1}} (\begin{array}{cccc} 1 & 0 & - 5 / 2 & - (3 / 2) a + (1 / 2) b \\ 0 & 1 & 3 & a \\ 0 & 0 & 0 & - 2 a + b + c \end{array}) \end{matrix}$
  
  (ii) There are codomain triples
  
  $(\begin{matrix} a \\ b \\ c \end{matrix}) \in ℝ^{3}$
  
  for which the system does not have a solution, specifically the system only has a solution if $- 2 a + b + c = 0$ .
  
  $ℛ (h) = {(\begin{matrix} a \\ b \\ c \end{matrix}) ∣ a = (b + c) / 2} = {(\begin{matrix} 1 / 2 \\ 1 \\ 0 \end{matrix}) b + (\begin{matrix} 1 / 2 \\ 0 \\ 1 \end{matrix}) c ∣ b, c \in ℝ}$
  
  The map’s rank is the range’s dimension, $2$ . The map is not onto because the range space is not all of the codomain.
  
  (iii) Setting $a = b = c = 0$ in the calculation gives infinitely many solutions. Parametrizing using the free variable $z$ leads to this description of the nullspace.
  
  $𝒩 (h) = {(\begin{matrix} x \\ y \\ z \end{matrix}) ∣ y = - 3 z and x = (5 / 2) z} = {(\begin{matrix} 5 / 2 \\ - 3 \\ 1 \end{matrix}) z ∣ z \in ℝ}$
  
  The nullity is the dimension of that null space, $1$ . The map is not one-to-one because the null space is not trivial.
3. (i) The domain has dimension $m = 2$ while the codomain has dimension $n = 3$ . Here is the calculation.
  
  $(\begin{array}{ccc} 1 & 1 & a \\ 2 & 1 & b \\ 3 & 1 & c \end{array}) ⟶_{\begin{array}{l} - 3 ρ_{1} + ρ_{3} \end{array}}^{- 2 ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{2} + ρ_{3}} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{2} + ρ_{1}} (\begin{array}{ccc} 1 & 0 & - a + b \\ 0 & 1 & 2 a - b \\ 0 & 0 & a - 2 b + c \end{array})$
  
  (ii) The range is this subspace of the codomain.
  
  $ℛ (h) = {(\begin{matrix} 2 b - c \\ b \\ c \end{matrix}) ∣ b, c \in ℝ} = {(\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}) b + (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}) c ∣ b, c \in ℝ}$
  
  The rank is $2$ . The map is not onto.
  
  (iii) The null space is the trivial subspace of the domain.
  
  $𝒩 (h) = {(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})}$
  
  The nullity is $0$ . The map is one-to-one.
Exercise 2.23 Worked answer

Use the method from the prior exercise on this matrix.

$(\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 0 \\ 2 & 2 & 2 \end{matrix})$

Back to Exercise 2.23

Answer. Here is the Gauss-Jordan reduction.

$\begin{aligned} (\begin{array}{cccc} 1 & 0 & - 1 & a \\ 2 & 1 & 0 & b \\ 2 & 2 & 2 & c \end{array}) & ⟶_{\begin{matrix} - 2 ρ_{1} + ρ_{3} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} (\begin{array}{cccc} 1 & 0 & - 1 & a \\ 0 & 1 & 2 & - 2 a + b \\ 0 & 2 & 4 & - 2 a + c \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{- 2 ρ_{2} + ρ_{3}} (\begin{array}{cccc} 1 & 0 & - 1 & a \\ 0 & 1 & 2 & - 2 a + b \\ 0 & 0 & 0 & 2 a - 2 b + c \end{array}) \end{aligned}$

(i) The dimensions are $m = n = 3$ . (ii) The range space is the set containing all of the members of the codomain for which this system has a solution.

$ℛ (h) = {(\begin{matrix} b - (1 / 2) c \\ b \\ c \end{matrix}) ∣ b, c \in ℝ}$

The rank is 2. Because the rank is less than the dimension $n = 3$ of the codomain, the map is not onto.

(iii) The null space is the set of members of the domain that map to $a = 0$ , $b = 0$ , and $c = 0$ .

$𝒩 (h) = {(\begin{matrix} z \\ - 2 z \\ z \end{matrix}) ∣ z \in ℝ}$

The nullity is $1$ . Because the nullity is not $0$ the map is not one-to-one.
Exercise 2.24 Worked answer

Verify that the map represented by this matrix is an isomorphism.

$(\begin{matrix} 2 & 1 & 0 \\ 3 & 1 & 1 \\ 7 & 2 & 1 \end{matrix})$

Back to Exercise 2.24

Answer. For any map represented by this matrix, the domain and codomain are each of dimension $3$ . To show that the map is an isomorphism, we must show it is both onto and one-to-one. For that we don’t need to augment the matrix with $a$ , $b$ , and $c$ ; this calculation

$\begin{matrix} (\begin{matrix} 2 & 1 & 0 \\ 3 & 1 & 1 \\ 7 & 2 & 1 \end{matrix}) ⟶_{\begin{array}{l} - (7 / 2) ρ_{1} + ρ_{3} \end{array}}^{- (3 / 2) ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{- 3 ρ_{2} + ρ_{3}} ⟶_{\begin{array}{l} - 2 ρ_{2} \\ - (1 / 2) ρ_{3} \end{array}}^{(1 / 2) ρ_{1}} ⟶_{\begin{array}{l} \end{array}}^{2 ρ_{3} + ρ_{2}} \\ ⟶_{\begin{array}{l} \end{array}}^{- (1 / 2) ρ_{2} + ρ_{1}} (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) \end{matrix}$

gives that for each codomain vector there is one and only one associated domain vector.
Exercise 2.25 Worked answer

This is an alternative proof of Lemma 2.9. Given an $n \times n$ matrix $H$ , fix a domain $V$ and codomain $W$ of appropriate dimension $n$ , and bases $B, D$ for those spaces, and consider the map $h$ represented by the matrix.
1. Show that $h$ is onto if and only if there is at least one ${Rep}_{B} (\vec{v})$ associated by $H$ with each ${Rep}_{D} (\vec{w})$ .
2. Show that $h$ is one-to-one if and only if there is at most one ${Rep}_{B} (\vec{v})$ associated by $H$ with each ${Rep}_{D} (\vec{w})$ .
3. Consider the linear system $H \cdot {Rep}_{B} (\vec{v}) = {Rep}_{D} (\vec{w})$ . Show that $H$ is nonsingular if and only if there is exactly one solution ${Rep}_{B} (\vec{v})$ for each ${Rep}_{D} (\vec{w})$ .
Back to Exercise 2.25

Answer.
1. The defined map $h$ is onto if and only if for every $\vec{w} \in W$ there is a $\vec{v} \in V$ such that $h (\vec{v}) = \vec{w}$ . Since for every vector there is exactly one representation, converting to representations gives that $h$ is onto if and only if for every representation ${Rep}_{D} (\vec{w})$ there is a representation ${Rep}_{B} (\vec{v})$ such that $H \cdot {Rep}_{B} (\vec{v}) = {Rep}_{D} (\vec{w})$ .
2. This is just like the prior part.
3. As described at the start of this subsection, by definition the map $h$ defined by the matrix $H$ associates this domain vector $\vec{v}$ with this codomain vector $\vec{w}$ .
  
  ${Rep}_{B} (\vec{v}) = (\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix}) {Rep}_{D} (\vec{w}) = H \cdot {Rep}_{B} (\vec{v}) = (\begin{matrix} h_{1, 1} v_{1} + \dots + h_{1, n} v_{n} \\ ⋮ \\ h_{m, 1} v_{1} + \dots + h_{m, n} v_{n} \end{matrix})$
  
  Fix $\vec{w} \in W$ and consider the linear system defined by the above equation.
  
  $\begin{array}{rcrcrcr} h_{1, 1} v_{1} & + & \dots & + & h_{1, n} v_{n} & = & w_{1} \\ h_{2, 1} v_{1} & + & \dots & + & h_{2, n} v_{n} & = & w_{2} \\ ⋮ \\ h_{n, 1} v_{1} & + & \dots & + & h_{n, n} v_{n} & = & w_{n} \end{array}$
  
  (Again, here the $w_{i}$ are fixed and the $v_{j}$ are unknowns.) Now, $H$ is nonsingular if and only if for all $w_{1}$ , …, $w_{n}$ this system has a solution and the solution is unique. By the first two parts of this exercise this is true if and only if the map $h$ is onto and one-to-one. This in turn is true if and only if $h$ is an isomorphism.
Exercise 2.26 Worked answer

Recommended. Because the rank of a matrix equals the rank of any map it represents, if one matrix represents two different maps $H = {Rep}_{B, D} (h) = {Rep}_{\hat{B}, \hat{D}} (\hat{h})$ (where $h, \hat{h} : V \to W$ ) then the dimension of the range space of $h$ equals the dimension of the range space of $\hat{h}$ . Must these equal-dimensional range spaces actually be the same?

Back to Exercise 2.26

Answer. No, the range spaces may differ. Example 2.3 shows this.
Exercise 2.27 Worked answer

Let $V$ be an $n$ -dimensional space with bases $B$ and $D$ . Consider a map that sends, for $\vec{v} \in V$ , the column vector representing $\vec{v}$ with respect to $B$ to the column vector representing $\vec{v}$ with respect to $D$ . Show that map is a linear transformation of $ℝ^{n}$ .

Back to Exercise 2.27

Answer. Recall that the representation map

$V \overset{{Rep}_{B}}{⟼} ℝ^{n}$

is an isomorphism. Thus, its inverse map ${Rep}_{B}^{- 1} : ℝ^{n} \to V$ is also an isomorphism. The desired transformation of $ℝ^{n}$ is then this composition.

$ℝ^{n} \overset{{Rep}_{B}^{- 1}}{⟼} V \overset{{Rep}_{D}}{⟼} ℝ^{n}$

Because a composition of isomorphisms is also an isomorphism, this map ${Rep}_{D} \circ {Rep}_{B}^{- 1}$ is an isomorphism.
Exercise 2.28 Worked answer

Example 2.3 shows that changing the pair of bases can change the map that a matrix represents, even though the domain and codomain remain the same. Could the map ever not change? Is there a matrix $H$ , vector spaces $V$ and $W$ , and associated pairs of bases $B_{1}, D_{1}$ and $B_{2}, D_{2}$ (with $B_{1} \neq B_{2}$ or $D_{1} \neq D_{2}$ or both) such that the map represented by $H$ with respect to $B_{1}, D_{1}$ equals the map represented by $H$ with respect to $B_{2}, D_{2}$ ?

Back to Exercise 2.28

Answer. Yes. Consider

$H = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$

representing a map from $ℝ^{2}$ to $ℝ^{2}$ . With respect to the standard bases $B_{1} = ℰ_{2}, D_{1} = ℰ_{2}$ this matrix represents the identity map. With respect to

$B_{2} = D_{2} = ⟨ (\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ - 1 \end{matrix}) ⟩$

this matrix again represents the identity. In fact, as long as the starting and ending bases are equal—as long as $B_{i} = D_{i}$ —then the map represented by $H$ is the identity.
Exercise 2.29 Worked answer

Recommended. A square matrix is a diagonal matrix if it is all zeroes except possibly for the entries on its upper-left to lower-right diagonal—its $1, 1$ entry, its $2, 2$ entry, etc. Show that a linear map is an isomorphism if there are bases such that, with respect to those bases, the map is represented by a diagonal matrix with no zeroes on the diagonal.

Back to Exercise 2.29

Answer. This is immediate from Lemma 2.9.
Exercise 2.30 Worked answer

Describe geometrically the action on $ℝ^{2}$ of the map represented with respect to the standard bases $ℰ_{2}, ℰ_{2}$ by this matrix.

$(\begin{matrix} 3 & 0 \\ 0 & 2 \end{matrix})$

Do the same for these.

$(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix})$

Back to Exercise 2.30

Answer. The first map

$(\begin{matrix} x \\ y \end{matrix}) = {(\begin{matrix} x \\ y \end{matrix})}_{ℰ_{2}} \mapsto {(\begin{matrix} 3 x \\ 2 y \end{matrix})}_{ℰ_{2}} = (\begin{matrix} 3 x \\ 2 y \end{matrix})$

stretches vectors by a factor of three in the $x$ direction and by a factor of two in the $y$ direction. The second map

$(\begin{matrix} x \\ y \end{matrix}) = {(\begin{matrix} x \\ y \end{matrix})}_{ℰ_{2}} \mapsto {(\begin{matrix} x \\ 0 \end{matrix})}_{ℰ_{2}} = (\begin{matrix} x \\ 0 \end{matrix})$

projects vectors onto the $x$ axis. The third

$(\begin{matrix} x \\ y \end{matrix}) = {(\begin{matrix} x \\ y \end{matrix})}_{ℰ_{2}} \mapsto {(\begin{matrix} y \\ x \end{matrix})}_{ℰ_{2}} = (\begin{matrix} y \\ x \end{matrix})$

interchanges first and second components (that is, it is a reflection about the line $y = x$ ). The last

$(\begin{matrix} x \\ y \end{matrix}) = {(\begin{matrix} x \\ y \end{matrix})}_{ℰ_{2}} \mapsto {(\begin{matrix} x + 3 y \\ y \end{matrix})}_{ℰ_{2}} = (\begin{matrix} x + 3 y \\ y \end{matrix})$

stretches vectors parallel to the $y$ axis, by an amount equal to three times their distance from that axis (this is a skew.)
Exercise 2.31 Worked answer

The fact that for any linear map the rank plus the nullity equals the dimension of the domain shows that a necessary condition for the existence of a homomorphism between two spaces, onto the second space, is that there be no gain in dimension. That is, where $h : V \to W$ is onto, the dimension of $W$ must be less than or equal to the dimension of $V$ .
1. Show that this (strong) converse holds: no gain in dimension implies that there is a homomorphism and, further, any matrix with the correct size and correct rank represents such a map.
2. Are there bases for $ℝ^{3}$ such that this matrix
  
  $H = (\begin{matrix} 1 & 0 & 0 \\ 2 & 0 & 0 \\ 0 & 1 & 0 \end{matrix})$
  
  represents a map from $ℝ^{3}$ to $ℝ^{3}$ whose range is the $x y$ plane subspace of $ℝ^{3}$ ?
Back to Exercise 2.31

Answer.
1. This is immediate from Theorem 2.4.
2. Yes. This is immediate from the prior item.
  
  To give a specific example, we can start with $ℰ_{3}$ as the basis for the domain, and then we require a basis $D$ for the codomain $ℝ^{3}$ . The matrix $H$ gives the action of the map as this
  
  $\begin{matrix} (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})}_{ℰ_{3}} \mapsto {(\begin{matrix} 1 \\ 2 \\ 0 \end{matrix})}_{D} (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) = {(\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})}_{ℰ_{3}} \mapsto {(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}_{D} \\ (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) = {(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}_{ℰ_{3}} \mapsto {(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})}_{D} \end{matrix}$
  
  and there is no harm in finding a basis $D$ so that
  
  ${Rep}_{D} ((\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})) = {(\begin{matrix} 1 \\ 2 \\ 0 \end{matrix})}_{D} and {Rep}_{D} ((\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})) = {(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}_{D}$
  
  that is, so that the map represented by $H$ with respect to $ℰ_{3}, D$ is projection down onto the $x y$ plane. The second condition gives that the third member of $D$ is ${\vec{e}}_{2}$ . The first condition gives that the first member of $D$ plus twice the second equals ${\vec{e}}_{1}$ , and so this basis will do.
  
  $D = ⟨ (\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} - 1 / 2 \\ 0 \\ - 1 / 2 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) ⟩$
Exercise 2.32 Worked answer

Let $V$ be an $n$ -dimensional space and suppose that $\vec{x} \in ℝ^{n}$ . Fix a basis $B$ for $V$ and consider the map $h_{\vec{x}} : V \to ℝ$ given $\vec{v} \mapsto \vec{x} \cdot {Rep}_{B} (\vec{v})$ by the dot product.
1. Show that this map is linear.
2. Show that for any linear map $g : V \to ℝ$ there is an $\vec{x} \in ℝ^{n}$ such that $g = h_{\vec{x}}$ .
3. In the prior item we fixed the basis and varied the $\vec{x}$ to get all possible linear maps. Can we get all possible linear maps by fixing an $\vec{x}$ and varying the basis?
Back to Exercise 2.32

Answer.
1. Recall that the representation map ${Rep}_{B} : V \to ℝ^{n}$ is linear (it is actually an isomorphism, but we do not need that it is one-to-one or onto here). Considering the column vector $x$ to be a $n \times 1$ matrix gives that the map from $ℝ^{n}$ to $ℝ$ that takes a column vector to its dot product with $\vec{x}$ is linear (this is a matrix-vector product and so Theorem 2.2 applies). Thus the map under consideration $h_{\vec{x}}$ is linear because it is the composition of two linear maps.
  
  $\vec{v} \mapsto {Rep}_{B} (\vec{v}) \mapsto \vec{x} \cdot {Rep}_{B} (\vec{v})$
2. Any linear map $g : V \to ℝ$ is represented by some matrix
  
  $(\begin{matrix} g_{1} & g_{2} & \dots & g_{n} \end{matrix})$
  
  (the matrix has $n$ columns because $V$ is $n$ -dimensional and it has only one row because $ℝ$ is one-dimensional). Then taking $\vec{x}$ to be the column vector that is the transpose of this matrix
  
  $\vec{x} = (\begin{matrix} g_{1} \\ ⋮ \\ g_{n} \end{matrix})$
  
  has the desired action.
  
  $\vec{v} = (\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix}) \mapsto (\begin{matrix} g_{1} \\ ⋮ \\ g_{n} \end{matrix}) \cdot (\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix}) = g_{1} v_{1} + \dots + g_{n} v_{n}$
3. No. If $\vec{x}$ has any nonzero entries then $h_{\vec{x}}$ cannot be the zero map (and if $\vec{x}$ is the zero vector then $h_{\vec{x}}$ can only be the zero map).
Exercise 2.33 Worked answer

Let $V, W, X$ be vector spaces with bases $B, C, D$ .
1. Suppose that $h : V \to W$ is represented with respect to $B, C$ by the matrix $H$ . Give the matrix representing the scalar multiple $r h$ (where $r \in ℝ$ ) with respect to $B, C$ by expressing it in terms of $H$ .
2. Suppose that $h, g : V \to W$ are represented with respect to $B, C$ by $H$ and $G$ . Give the matrix representing $h + g$ with respect to $B, C$ by expressing it in terms of $H$ and $G$ .
3. Suppose that $h : V \to W$ is represented with respect to $B, C$ by $H$ and $g : W \to X$ is represented with respect to $C, D$ by $G$ . Give the matrix representing $g \circ h$ with respect to $B, D$ by expressing it in terms of $H$ and $G$ .
Back to Exercise 2.33

Answer. See the following section.

Computing Linear Maps

Representing Linear Maps with Matrices

Exercises

Any Matrix Represents a Linear Map

Exercises

References cited in this section