Homomorphisms

The definition of isomorphism has two conditions. In this section we will consider the second one. We will study maps that are required only to preserve structure, maps that are not also required to be correspondences.

Experience shows that these maps are tremendously useful. For one thing we shall see in the second subsection below that while isomorphisms describe how spaces are the same, we can think of these maps as describing how spaces are alike.

Definition

Definition 1.1 A function between vector spaces $h : V \to W$ that preserves addition

if ${\vec{v}}_{1}, {\vec{v}}_{2} \in V$ then $h ({\vec{v}}_{1} + {\vec{v}}_{2}) = h ({\vec{v}}_{1}) + h ({\vec{v}}_{2})$

and scalar multiplication

if $\vec{v} \in V$ and $r \in ℝ$ then $h (r \cdot \vec{v}) = r \cdot h (\vec{v})$

is a homomorphism or linear map.

Example 1.2 The projection map $π : ℝ^{3} \to ℝ^{2}$

$(\begin{matrix} x \\ y \\ z \end{matrix}) \overset{π}{⟼} (\begin{matrix} x \\ y \end{matrix})$

is a homomorphism. It preserves addition

$π ((\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix})) = π ((\begin{matrix} x_{1} + x_{2} \\ y_{1} + y_{2} \\ z_{1} + z_{2} \end{matrix})) = (\begin{matrix} x_{1} + x_{2} \\ y_{1} + y_{2} \end{matrix}) = π ((\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix})) + π ((\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}))$

and scalar multiplication.

$π (r \cdot (\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix})) = π ((\begin{matrix} r x_{1} \\ r y_{1} \\ r z_{1} \end{matrix})) = (\begin{matrix} r x_{1} \\ r y_{1} \end{matrix}) = r \cdot π ((\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}))$

This is not an isomorphism since it is not one-to-one. For instance, both $\vec{0}$ and ${\vec{e}}_{3}$ in $ℝ^{3}$ map to the zero vector in $ℝ^{2}$ .

Example 1.3 The domain and codomain can be other than spaces of column vectors. Both of these are homomorphisms; the verifications are straightforward.

$f_{1} : 𝒫_{2} \to 𝒫_{3}$ given by

$a_{0} + a_{1} x + a_{2} x^{2} \mapsto a_{0} x + (a_{1} / 2) x^{2} + (a_{2} / 3) x^{3}$
$f_{2} : M_{2 \times 2} \to ℝ$ given by

$(\begin{matrix} a & b \\ c & d \end{matrix}) \mapsto a + d$

Example 1.4 Between any two spaces there is a zero homomorphism, mapping every vector in the domain to the zero vector in the codomain.

We shall use the two terms ‘homomorphism’ and ‘linear map’ interchangably.

Example 1.5 These two suggest why we say ‘linear map’.

The map $g : ℝ^{3} \to ℝ$ given by

$(\begin{matrix} x \\ y \\ z \end{matrix}) \overset{g}{⟼} 3 x + 2 y - 4.5 z$

is linear, that is, is a homomorphism. The check is easy. In contrast, the map $\hat{g} : ℝ^{3} \to ℝ$ given by

$(\begin{matrix} x \\ y \\ z \end{matrix}) \overset{\hat{g}}{⟼} 3 x + 2 y - 4.5 z + 1$

is not linear. To show this we need only produce a single linear combination that the map does not preserve. Here is one.

$\hat{g} ((\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})) = 4 \hat{g} ((\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})) + \hat{g} ((\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})) = 5$
The first of these two maps $t_{1}, t_{2} : ℝ^{3} \to ℝ^{2}$ is linear while the second is not.

$(\begin{matrix} x \\ y \\ z \end{matrix}) \overset{t_{1}}{⟼} (\begin{matrix} 5 x - 2 y \\ x + y \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \overset{t_{2}}{⟼} (\begin{matrix} 5 x - 2 y \\ x y \end{matrix})$

Finding a linear combination that the second map does not preserve is easy.

So one way to think of ‘homomorphism’ is that we are generalizing ‘isomorphism’ (by dropping the condition that the map is a correspondence), motivated by the observation that many of the properties of isomorphisms have only to do with the map’s structure-preservation property. The next two results are examples of this motivation. In the prior section we saw a proof for each that only uses preservation of addition and preservation of scalar multiplication, and therefore applies to homomorphisms.

Lemma 1.6 A linear map sends the zero vector to the zero vector.

Lemma 1.7 The following are equivalent for any map $f : V \to W$ between vector spaces.

$f$ is a homomorphism
$f (c_{1} \cdot {\vec{v}}_{1} + c_{2} \cdot {\vec{v}}_{2}) = c_{1} \cdot f ({\vec{v}}_{1}) + c_{2} \cdot f ({\vec{v}}_{2})$ for any $c_{1}, c_{2} \in ℝ$ and ${\vec{v}}_{1}, {\vec{v}}_{2} \in V$
$f (c_{1} \cdot {\vec{v}}_{1} + \dots + c_{n} \cdot {\vec{v}}_{n}) = c_{1} \cdot f ({\vec{v}}_{1}) + \dots + c_{n} \cdot f ({\vec{v}}_{n})$ for any $c_{1}, \dots, c_{n} \in ℝ$ and ${\vec{v}}_{1}, \dots, {\vec{v}}_{n} \in V$

Example 1.8 The function $f : ℝ^{2} \to ℝ^{4}$ given by

$(\begin{matrix} x \\ y \end{matrix}) \overset{f}{⟼} (\begin{matrix} x / 2 \\ 0 \\ x + y \\ 3 y \end{matrix})$

is linear since it satisfies item (2).

$(\begin{matrix} r_{1} (x_{1} / 2) + r_{2} (x_{2} / 2) \\ 0 \\ r_{1} (x_{1} + y_{1}) + r_{2} (x_{2} + y_{2}) \\ r_{1} (3 y_{1}) + r_{2} (3 y_{2}) \end{matrix}) = r_{1} (\begin{matrix} x_{1} / 2 \\ 0 \\ x_{1} + y_{1} \\ 3 y_{1} \end{matrix}) + r_{2} (\begin{matrix} x_{2} / 2 \\ 0 \\ x_{2} + y_{2} \\ 3 y_{2} \end{matrix})$

However, some things that hold for isomorphisms fail to hold for homomorphisms. One example is in the proof of Lemma I.2.4, which shows that an isomorphism between spaces gives a correspondence between their bases. Homomorphisms do not give any such correspondence; Example 1.2 shows this and another example is the zero map between two nontrivial spaces. Instead, for homomorphisms we have a weaker but still very useful result.

Theorem 1.9 A homomorphism is determined by its action on a basis: if $V$ is a vector space with basis $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ , if $W$ is a vector space, and if ${\vec{w}}_{1}, \dots, {\vec{w}}_{n} \in W$ (these codomain elements need not be distinct) then there exists a homomorphism from $V$ to $W$ sending each ${\vec{β}}_{i}$ to ${\vec{w}}_{i}$ , and that homomorphism is unique.

Proof For any input $\vec{v} \in V$ let its expression with respect to the basis be $\vec{v} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ . Define the associated output by using the same coordinates $h (\vec{v}) = c_{1} {\vec{w}}_{1} + \dots + c_{n} {\vec{w}}_{n}$ . This is well defined because, with respect to the basis, the representation of each domain vector $\vec{v}$ is unique.

This map is a homomorphism because it preserves linear combinations: where $\vec{v_{1}} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ and $\vec{v_{2}} = d_{1} {\vec{β}}_{1} + \dots + d_{n} {\vec{β}}_{n}$ , here is the calculation.

$\begin{aligned} h (r_{1} {\vec{v}}_{1} + r_{2} {\vec{v}}_{2}) & = h ((r_{1} c_{1} + r_{2} d_{1}) {\vec{β}}_{1} + \dots + (r_{1} c_{n} + r_{2} d_{n}) {\vec{β}}_{n}) \\ = (r_{1} c_{1} + r_{2} d_{1}) {\vec{w}}_{1} + \dots + (r_{1} c_{n} + r_{2} d_{n}) {\vec{w}}_{n} \\ = r_{1} h ({\vec{v}}_{1}) + r_{2} h ({\vec{v}}_{2}) \end{aligned}$

This map is unique because if $\hat{h} : V \to W$ is another homomorphism satisfying that $\hat{h} ({\vec{β}}_{i}) = {\vec{w}}_{i}$ for each $i$ then $h$ and $\hat{h}$ have the same effect on all of the vectors in the domain.

$\begin{matrix} \hat{h} (\vec{v}) = \hat{h} (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}) = c_{1} \hat{h} ({\vec{β}}_{1}) + \dots + c_{n} \hat{h} ({\vec{β}}_{n}) \\ = c_{1} {\vec{w}}_{1} + \dots + c_{n} {\vec{w}}_{n} = h (\vec{v}) \end{matrix}$

They have the same action so they are the same function.

QED

Definition 1.10 Let $V$ and $W$ be vector spaces and let $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ be a basis for $V$ . A function defined on that basis $f : B \to W$ is extended linearly to a function $\hat{f} : V \to W$ if for all $\vec{v} \in V$ such that $\vec{v} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ , the action of the map is $\hat{f} (\vec{v}) = c_{1} \cdot f ({\vec{β}}_{1}) + \dots + c_{n} \cdot f ({\vec{β}}_{n})$ .

Example 1.11 If we specify a map $h : ℝ^{2} \to ℝ^{2}$ that acts on the standard basis $ℰ_{2}$ in this way

$h ((\begin{matrix} 1 \\ 0 \end{matrix})) = (\begin{matrix} - 1 \\ 1 \end{matrix}) h ((\begin{matrix} 0 \\ 1 \end{matrix})) = (\begin{matrix} - 4 \\ 4 \end{matrix})$

then we have also specified the action of $h$ on any other member of the domain. For instance, the value of $h$ on this argument

$h ((\begin{matrix} 3 \\ - 2 \end{matrix})) = h (3 \cdot (\begin{matrix} 1 \\ 0 \end{matrix}) - 2 \cdot (\begin{matrix} 0 \\ 1 \end{matrix})) = 3 \cdot h ((\begin{matrix} 1 \\ 0 \end{matrix})) - 2 \cdot h ((\begin{matrix} 0 \\ 1 \end{matrix})) = (\begin{matrix} 5 \\ - 5 \end{matrix})$

is a direct consequence of the value of $h$ on the basis vectors.

Later in this chapter we shall develop a convenient scheme for computations like this one, using matrices.

Definition 1.12 A linear map from a space into itself $t : V \to V$ is a linear transformation.

Remark 1.13 In this book we use ‘linear transformation’ only in the case where the codomain equals the domain. Be aware that some sources instead use it as a synonym for ‘linear map’. Still another synonym is ‘linear operator’.

Example 1.14 The map on $ℝ^{2}$ that projects all vectors down to the $x$ -axis is a linear transformation.

$(\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} x \\ 0 \end{matrix})$

Example 1.15 The derivative map $d / d x : 𝒫_{n} \to 𝒫_{n}$

$a_{0} + a_{1} x + \dots + a_{n} x^{n} \overset{d / d x}{⟼} a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + \dots + n a_{n} x^{n - 1}$

is a linear transformation as this result from calculus shows: $d (c_{1} f + c_{2} g) / d x = c_{1} (d f / d x) + c_{2} (d g / d x)$ .

Example 1.16 The matrix transpose operation

$(\begin{matrix} a & b \\ c & d \end{matrix}) \mapsto (\begin{matrix} a & c \\ b & d \end{matrix})$

is a linear transformation of $ℳ_{2 \times 2}$ . (Transpose is one-to-one and onto and so is in fact an automorphism.)

We finish this subsection about maps by recalling that we can linearly combine maps. For instance, for these maps from $ℝ^{2}$ to itself

$(\begin{matrix} x \\ y \end{matrix}) \overset{f}{⟼} (\begin{matrix} 2 x \\ 3 x - 2 y \end{matrix}) and (\begin{matrix} x \\ y \end{matrix}) \overset{g}{⟼} (\begin{matrix} 0 \\ 5 x \end{matrix})$

the linear combination $5 f - 2 g$ is also a transformation of $ℝ^{2}$ .

$(\begin{matrix} x \\ y \end{matrix}) \overset{5 f - 2 g}{⟼} (\begin{matrix} 10 x \\ 5 x - 10 y \end{matrix})$

Lemma 1.17 For vector spaces $V$ and $W$ , the set of linear functions from $V$ to $W$ is itself a vector space, a subspace of the space of all functions from $V$ to $W$ .

We denote the space of linear maps from $V$ to $W$ by $ℒ (V, W)$ .

Proof This set is non-empty because it contains the zero homomorphism. So to show that it is a subspace we need only check that it is closed under the operations. Let $f, g : V \to W$ be linear. Then the operation of function addition is preserved

$\begin{aligned} (f + g) (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2}) & = f (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2}) + g (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2}) \\ = c_{1} f ({\vec{v}}_{1}) + c_{2} f ({\vec{v}}_{2}) + c_{1} g ({\vec{v}}_{1}) + c_{2} g ({\vec{v}}_{2}) \\ = c_{1} (f + g) ({\vec{v}}_{1}) + c_{2} (f + g) ({\vec{v}}_{2}) \end{aligned}$

as is the operation of scalar multiplication of a function.

$\begin{aligned} (r \cdot f) (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2}) & = r (c_{1} f ({\vec{v}}_{1}) + c_{2} f ({\vec{v}}_{2})) \\ = c_{1} (r \cdot f) ({\vec{v}}_{1}) + c_{2} (r \cdot f) ({\vec{v}}_{2}) \end{aligned}$

Hence $ℒ (V, W)$ is a subspace.

QED

We started this section by defining ‘homomorphism’ as a generalization of ‘isomorphism’, by isolating the structure preservation property. Some of the points about isomorphisms carried over unchanged, while we adapted others.

Note, however, that the idea of ‘homomorphism’ is in no way somehow secondary to that of ‘isomorphism’. In the rest of this chapter we shall work mostly with homomorphisms. This is partly because any statement made about homomorphisms is automatically true about isomorphisms but more because, while the isomorphism concept is more natural, our experience will show that the homomorphism concept is more fruitful and more central to progress.

Exercises

Exercise 1.18 Worked answer

Recommended. Decide if each $h : ℝ^{3} \to ℝ^{2}$ is linear.
1. $h ((\begin{matrix} x \\ y \\ z \end{matrix})) = (\begin{matrix} x \\ x + y + z \end{matrix})$
2. $h ((\begin{matrix} x \\ y \\ z \end{matrix})) = (\begin{matrix} 0 \\ 0 \end{matrix})$
3. $h ((\begin{matrix} x \\ y \\ z \end{matrix})) = (\begin{matrix} 1 \\ 1 \end{matrix})$
4. $h ((\begin{matrix} x \\ y \\ z \end{matrix})) = (\begin{matrix} 2 x + y \\ 3 y - 4 z \end{matrix})$
Back to Exercise 1.18

Answer.
1. Yes. The verification is straightforward.
  
  $\begin{aligned} h (c_{1} \cdot (\begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array}) + c_{2} \cdot (\begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array})) & = h ((\begin{array}{c} c_{1} x_{1} + c_{2} x_{2} \\ c_{1} y_{1} + c_{2} y_{2} \\ c_{1} z_{1} + c_{2} z_{2} \end{array})) \\ = (\begin{array}{c} c_{1} x_{1} + c_{2} x_{2} \\ c_{1} x_{1} + c_{2} x_{2} + c_{1} y_{1} + c_{2} y_{2} + c_{1} z_{1} + c_{2} z_{2} \end{array}) \\ = c_{1} \cdot (\begin{array}{c} x_{1} \\ x_{1} + y_{1} + z_{1} \end{array}) + c_{2} \cdot (\begin{array}{c} x_{2} \\ c_{2} + y_{2} + z_{2} \end{array}) \\ = c_{1} \cdot h ((\begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array})) + c_{2} \cdot h ((\begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array})) \end{aligned}$
2. Yes. The verification is easy.
  
  $\begin{aligned} h (c_{1} \cdot (\begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array}) + c_{2} \cdot (\begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array})) & = h ((\begin{array}{c} c_{1} x_{1} + c_{2} x_{2} \\ c_{1} y_{1} + c_{2} y_{2} \\ c_{1} z_{1} + c_{2} z_{2} \end{array})) \\ = (\begin{array}{c} 0 \\ 0 \end{array}) \\ = c_{1} \cdot h ((\begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array})) + c_{2} \cdot h ((\begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array})) \end{aligned}$
3. No. An example of an addition that is not respected is this.
  
  $h ((\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})) = (\begin{matrix} 1 \\ 1 \end{matrix}) \neq h ((\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})) + h ((\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}))$
4. Yes. The verification is straightforward.
  
  $\begin{aligned} h (c_{1} \cdot (\begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array}) + c_{2} \cdot (\begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array})) & = h ((\begin{array}{c} c_{1} x_{1} + c_{2} x_{2} \\ c_{1} y_{1} + c_{2} y_{2} \\ c_{1} z_{1} + c_{2} z_{2} \end{array})) \\ = (\begin{array}{c} 2 (c_{1} x_{1} + c_{2} x_{2}) + (c_{1} y_{1} + c_{2} y_{2}) \\ 3 (c_{1} y_{1} + c_{2} y_{2}) - 4 (c_{1} z_{1} + c_{2} z_{2}) \end{array}) \\ = c_{1} \cdot (\begin{array}{c} 2 x_{1} + y_{1} \\ 3 y_{1} - 4 z_{1} \end{array}) + c_{2} \cdot (\begin{array}{c} 2 x_{2} + y_{2} \\ 3 y_{2} - 4 z_{2} \end{array}) \\ = c_{1} \cdot h ((\begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array})) + c_{2} \cdot h ((\begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array})) \end{aligned}$
Exercise 1.19 Worked answer

Recommended. Decide if each map $h : ℳ_{2 \times 2} \to ℝ$ is linear.
1. $h ((\begin{matrix} a & b \\ c & d \end{matrix})) = a + d$
2. $h ((\begin{matrix} a & b \\ c & d \end{matrix})) = a d - b c$
3. $h ((\begin{matrix} a & b \\ c & d \end{matrix})) = 2 a + 3 b + c - d$
4. $h ((\begin{matrix} a & b \\ c & d \end{matrix})) = a^{2} + b^{2}$
Back to Exercise 1.19

Answer. For each, we must either check that the map preserves linear combinations or give an example of a linear combination that is not.
1. Yes. The check that it preserves combinations is routine.
  
  $\begin{aligned} h (r_{1} \cdot (\begin{array}{c} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array}) + r_{2} \cdot (\begin{array}{c} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array})) & = h ((\begin{array}{c} r_{1} a_{1} + r_{2} a_{2} & r_{1} b_{1} + r_{2} b_{2} \\ r_{1} c_{1} + r_{2} c_{2} & r_{1} d_{1} + r_{2} d_{2} \end{array})) \\ = (r_{1} a_{1} + r_{2} a_{2}) + (r_{1} d_{1} + r_{2} d_{2}) \\ = r_{1} (a_{1} + d_{1}) + r_{2} (a_{2} + d_{2}) \\ = r_{1} \cdot h ((\begin{array}{c} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array})) + r_{2} \cdot h ((\begin{array}{c} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array})) \end{aligned}$
2. No. For instance, not preserved is multiplication by the scalar $2$ .
  
  $h (2 \cdot (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})) = h ((\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix})) = 4 while 2 \cdot h ((\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})) = 2 \cdot 1 = 2$
3. Yes. This is the check that it preserves combinations of two members of the domain.
  
  $\begin{matrix} h (r_{1} \cdot (\begin{matrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix}) + r_{2} \cdot (\begin{matrix} a_{2} & b_{2} \\ c_{2} & d_{2} \end{matrix})) \\ \begin{aligned} = h ((\begin{array}{c} r_{1} a_{1} + r_{2} a_{2} & r_{1} b_{1} + r_{2} b_{2} \\ r_{1} c_{1} + r_{2} c_{2} & r_{1} d_{1} + r_{2} d_{2} \end{array})) \\ = 2 (r_{1} a_{1} + r_{2} a_{2}) + 3 (r_{1} b_{1} + r_{2} b_{2}) + (r_{1} c_{1} + r_{2} c_{2}) - (r_{1} d_{1} + r_{2} d_{2}) \\ = r_{1} (2 a_{1} + 3 b_{1} + c_{1} - d_{1}) + r_{2} (2 a_{2} + 3 b_{2} + c_{2} - d_{2}) \\ = r_{1} \cdot h ((\begin{array}{c} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array}) + r_{2} \cdot h ((\begin{array}{c} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array})) \end{aligned} \end{matrix}$
4. No. An example of a combination that is not preserved is that doing it one way gives this
  
  $h ((\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) + (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})) = h ((\begin{matrix} 2 & 0 \\ 0 & 0 \end{matrix})) = 4$
  
  while the other way gives this.
  
  $h ((\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})) + h ((\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})) = 1 + 1 = 2$
Exercise 1.20 Worked answer

Recommended. Show that these are homomorphisms. Are they inverse to each other?
1. $d / d x : 𝒫_{3} \to 𝒫_{2}$ given by $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}$ maps to $a_{1} + 2 a_{2} x + 3 a_{3} x^{2}$
2. $\int : 𝒫_{2} \to 𝒫_{3}$ given by $b_{0} + b_{1} x + b_{2} x^{2}$ maps to $b_{0} x + (b_{1} / 2) x^{2} + (b_{2} / 3) x^{3}$
Back to Exercise 1.20

Answer. The check that each is a homomorphisms is routine. Here is the check for the differentiation map.

$\begin{matrix} \frac{d}{d x} (r \cdot (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}) + s \cdot (b_{0} + b_{1} x + b_{2} x^{2} + b_{3} x^{3})) \\ \begin{aligned} = \frac{d}{d x} ((r a_{0} + s b_{0}) + (r a_{1} + s b_{1}) x + (r a_{2} + s b_{2}) x^{2} + (r a_{3} + s b_{3}) x^{3}) \\ = (r a_{1} + s b_{1}) + 2 (r a_{2} + s b_{2}) x + 3 (r a_{3} + s b_{3}) x^{2} \\ = r \cdot (a_{1} + 2 a_{2} x + 3 a_{3} x^{2}) + s \cdot (b_{1} + 2 b_{2} x + 3 b_{3} x^{2}) \\ = r \cdot \frac{d}{d x} (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}) + s \cdot \frac{d}{d x} (b_{0} + b_{1} x + b_{2} x^{2} + b_{3} x^{3}) \end{aligned} \end{matrix}$

(An alternate proof is to simply note that this is a property of differentiation that is familiar from calculus.)

These two maps are not inverses as this composition does not act as the identity map on this element of the domain.

$1 \in 𝒫_{3} \overset{d / d x}{⟼} 0 \in 𝒫_{2} \overset{\int}{⟼} 0 \in 𝒫_{3}$
Exercise 1.21 Worked answer

Is (perpendicular) projection from $ℝ^{3}$ to the $x z$ -plane a homomorphism? Projection to the $y z$ -plane? To the $x$ -axis? The $y$ -axis? The $z$ -axis? Projection to the origin?

Back to Exercise 1.21

Answer. Each of these projections is a homomorphism. Projection to the $x z$ -plane and to the $y z$ -plane are these maps.

$(\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} x \\ 0 \\ z \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} 0 \\ y \\ z \end{matrix})$

Projection to the $x$ -axis, to the $y$ -axis, and to the $z$ -axis are these maps.

$(\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} x \\ 0 \\ 0 \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} 0 \\ y \\ 0 \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} 0 \\ 0 \\ z \end{matrix})$

And projection to the origin is this map.

$(\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$

Verification that each is a homomorphism is straightforward. (The last one, of course, is the zero transformation on $ℝ^{3}$ .)
Exercise 1.22 Worked answer

Verify that each map is a homomorphism.
1. $h : 𝒫_{2} \to ℝ^{2}$ given by
  
  $a x^{2} + b x + c \mapsto (\begin{matrix} a + b \\ a + c \end{matrix})$
2. $f : ℝ^{2} \to ℝ^{3}$ given by
  
  $(\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} 0 \\ x - y \\ 3 y \end{matrix})$
Back to Exercise 1.22

Answer.
1. This verifies that the map preserves linear combinations. By Lemma 1.7 that suffices to show that it is a homomorphism.
  
  $\begin{matrix} h (d_{1} (a_{1} x^{2} + b_{1} x + c_{1}) + d_{2} (a_{2} x^{2} + b_{2} x + c_{2})) \\ \begin{aligned} = h ((d_{1} a_{1} + d_{2} a_{2}) x^{2} + (d_{1} b_{1} + d_{2} b_{2}) x + (d_{1} c_{1} + d_{2} c_{2})) \\ = (\begin{array}{c} (d_{1} a_{1} + d_{2} a_{2}) + (d_{1} b_{1} + d_{2} b_{2}) \\ (d_{1} a_{1} + d_{2} a_{2}) + (d_{1} c_{1} + d_{2} c_{2}) \end{array}) \\ = (\begin{array}{c} d_{1} a_{1} + d_{1} b_{1} \\ d_{1} a_{1} + d_{1} c_{1} \end{array}) + (\begin{array}{c} d_{2} a_{2} + d_{2} b_{2} \\ d_{2} a_{2} + d_{2} c_{2} \end{array}) \\ = d_{1} (\begin{array}{c} a_{1} + b_{1} \\ a_{1} + c_{1} \end{array}) + d_{2} (\begin{array}{c} a_{2} + b_{2} \\ a_{2} + c_{2} \end{array}) \\ = d_{1} \cdot h (a_{1} x^{2} + b_{1} x + c_{1}) + d_{2} \cdot h (a_{2} x^{2} + b_{2} x + c_{2}) \end{aligned} \end{matrix}$
2. It preserves linear combinations.
  
  $\begin{aligned} f (a_{1} (\begin{array}{c} x_{1} \\ y_{1} \end{array}) + a_{2} (\begin{array}{c} x_{2} \\ y_{2} \end{array})) & = f ((\begin{array}{c} a_{1} x_{1} + a_{2} x_{2} \\ a_{1} y_{1} + a_{2} y_{2} \end{array})) \\ = (\begin{array}{c} 0 \\ (a_{1} x_{1} + a_{2} x_{2}) - (a_{1} y_{1} + a_{2} y_{2}) \\ 3 (a_{1} y_{1} + a_{2} y_{2}) \end{array}) \\ = a_{1} (\begin{array}{c} 0 \\ x_{1} - y_{1} \\ 3 y_{1} \end{array}) + a_{2} (\begin{array}{c} 0 \\ x_{2} - y_{2} \\ 3 y_{2} \end{array}) \\ = a_{1} f ((\begin{array}{c} x_{1} \\ y_{1} \end{array})) + a_{2} f ((\begin{array}{c} x_{2} \\ y_{2} \end{array})) \end{aligned}$
Exercise 1.23 Worked answer

Show that, while the maps from Example 1.3 preserve linear operations, they are not isomorphisms.

Back to Exercise 1.23

Answer. The first is not onto; for instance, there is no polynomial that is sent the constant polynomial $p (x) = 1$ . The second is not one-to-one; both of these members of the domain

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

map to the same member of the codomain, $1 \in ℝ$ .
Exercise 1.24 Worked answer

Is an identity map a linear transformation?

Back to Exercise 1.24

Answer. Yes; in any space $id (c \cdot \vec{v} + d \cdot \vec{w}) = c \cdot \vec{v} + d \cdot \vec{w} = c \cdot id (\vec{v}) + d \cdot id (\vec{w})$ .
Exercise 1.25 Worked answer

Recommended. Stating that a function is ‘linear’ is different than stating that its graph is a line.
1. The function $f_{1} : ℝ \to ℝ$ given by $f_{1} (x) = 2 x - 1$ has a graph that is a line. Show that it is not a linear function.
2. The function $f_{2} : ℝ^{2} \to ℝ$ given by
  
  $(\begin{matrix} x \\ y \end{matrix}) \mapsto x + 2 y$
  
  does not have a graph that is a line. Show that it is a linear function.
Back to Exercise 1.25

Answer.
1. This map does not preserve structure since $f (1 + 1) = 3$ , while $f (1) + f (1) = 2$ .
2. The check is routine.
  
  $\begin{aligned} f (r_{1} \cdot (\begin{array}{c} x_{1} \\ y_{1} \end{array}) + r_{2} \cdot (\begin{array}{c} x_{2} \\ y_{2} \end{array})) & = f ((\begin{array}{c} r_{1} x_{1} + r_{2} x_{2} \\ r_{1} y_{1} + r_{2} y_{2} \end{array})) \\ = (r_{1} x_{1} + r_{2} x_{2}) + 2 (r_{1} y_{1} + r_{2} y_{2}) \\ = r_{1} \cdot (x_{1} + 2 y_{1}) + r_{2} \cdot (x_{2} + 2 y_{2}) \\ = r_{1} \cdot f ((\begin{array}{c} x_{1} \\ y_{1} \end{array})) + r_{2} \cdot f ((\begin{array}{c} x_{2} \\ y_{2} \end{array})) \end{aligned}$
Exercise 1.26 Worked answer

Recommended. Part of the definition of a linear function is that it respects addition. Does a linear function respect subtraction?

Back to Exercise 1.26

Answer. Yes. Where $h : V \to W$ is linear, $h (\vec{u} - \vec{v}) = h (\vec{u} + (- 1) \cdot \vec{v}) = h (\vec{u}) + (- 1) \cdot h (\vec{v}) = h (\vec{u}) - h (\vec{v})$ .
Exercise 1.27 Worked answer

Assume that $h$ is a linear transformation of $V$ and that $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ is a basis of $V$ . Prove each statement.
1. If $h ({\vec{β}}_{i}) = \vec{0}$ for each basis vector then $h$ is the zero map.
2. If $h ({\vec{β}}_{i}) = {\vec{β}}_{i}$ for each basis vector then $h$ is the identity map.
3. If there is a scalar $r$ such that $h ({\vec{β}}_{i}) = r \cdot {\vec{β}}_{i}$ for each basis vector then $h (\vec{v}) = r \cdot \vec{v}$ for all vectors in $V$ .
Back to Exercise 1.27

Answer.
1. Let $\vec{v} \in V$ be represented with respect to the basis as $\vec{v} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ . Then $h (\vec{v}) = h (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}) = c_{1} h ({\vec{β}}_{1}) + \dots + c_{n} h ({\vec{β}}_{n}) = c_{1} \cdot \vec{0} + \dots + c_{n} \cdot \vec{0} = \vec{0}$ .
2. This argument is similar to the prior one. Let $\vec{v} \in V$ be represented with respect to the basis as $\vec{v} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ . Then $h (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}) = c_{1} h ({\vec{β}}_{1}) + \dots + c_{n} h ({\vec{β}}_{n}) = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n} = \vec{v}$ .
3. As above, only $c_{1} h ({\vec{β}}_{1}) + \dots + c_{n} h ({\vec{β}}_{n}) = c_{1} r {\vec{β}}_{1} + \dots + c_{n} r {\vec{β}}_{n} = r (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}) = r \vec{v}$ .
Exercise 1.28 Worked answer

Consider the vector space $ℝ^{+}$ where vector addition and scalar multiplication are not the ones inherited from $ℝ$ but rather are these: $a + b$ is the product of $a$ and $b$ , and $r \cdot a$ is the $r$ -th power of $a$ . (This was shown to be a vector space in an earlier exercise.) Verify that the natural logarithm map $\ln : ℝ^{+} \to ℝ$ is a homomorphism between these two spaces. Is it an isomorphism?

Back to Exercise 1.28

Answer. That it is a homomorphism follows from the familiar rules that the logarithm of a product is the sum of the logarithms $\ln (a b) = \ln (a) + \ln (b)$ and that the logarithm of a power is the multiple of the logarithm $\ln (a^{r}) = r \ln (a)$ . This map is an isomorphism because it has an inverse, namely, the exponential map, so it is a correspondence, and therefore it is an isomorphism.
Exercise 1.29 Worked answer

Consider this transformation of the plane $ℝ^{2}$ .

$(\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} x / 2 \\ y / 3 \end{matrix})$

Find the image under this map of this ellipse.

${(\begin{matrix} x \\ y \end{matrix}) ∣ (x^{2} / 4) + (y^{2} / 9) = 1}$

Back to Exercise 1.29

Answer. Where $\hat{x} = x / 2$ and $\hat{y} = y / 3$ , the image set is

${(\begin{matrix} \hat{x} \\ \hat{y} \end{matrix}) ∣ \frac{(2 \hat{x})^{2}}{4} + \frac{(3 \hat{y})^{2}}{9} = 1} = {(\begin{matrix} \hat{x} \\ \hat{y} \end{matrix}) ∣ {\hat{x}}^{2} + {\hat{y}}^{2} = 1}$

the unit circle in the $\hat{x} \hat{y}$ -plane.
Exercise 1.30 Worked answer

Recommended. Imagine a rope wound around the earth’s equator so that it fits snugly (suppose that the earth is a sphere). How much extra rope must we add so that around the entire world the rope will now be six feet off the ground?

Back to Exercise 1.30

Answer. The circumference function $r \mapsto 2 π r$ is linear. Thus we have $2 π \cdot (r_{earth} + 6) - 2 π \cdot (r_{earth}) = 12 π$ . Observe that it takes the same amount of extra rope to raise the circle from tightly wound around a basketball to six feet above that basketball as it does to raise it from tightly wound around the earth to six feet above the earth.
Exercise 1.31 Worked answer

Recommended. Verify that this map $h : ℝ^{3} \to ℝ$

$(\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} x \\ y \\ z \end{matrix}) \cdot (\begin{matrix} 3 \\ - 1 \\ - 1 \end{matrix}) = 3 x - y - z$

is linear. Generalize.

Back to Exercise 1.31

Answer. Verifying that it is linear is routine.

$\begin{aligned} h (c_{1} \cdot (\begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array}) + c_{2} \cdot (\begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array})) & = h ((\begin{array}{c} c_{1} x_{1} + c_{2} x_{2} \\ c_{1} y_{1} + c_{2} y_{2} \\ c_{1} z_{1} + c_{2} z_{2} \end{array})) \\ = 3 (c_{1} x_{1} + c_{2} x_{2}) - (c_{1} y_{1} + c_{2} y_{2}) - (c_{1} z_{1} + c_{2} z_{2}) \\ = c_{1} \cdot (3 x_{1} - y_{1} - z_{1}) + c_{2} \cdot (3 x_{2} - y_{2} - z_{2}) \\ = c_{1} \cdot h ((\begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array})) + c_{2} \cdot h ((\begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array})) \end{aligned}$

The natural guess at a generalization is that for any fixed $\vec{k} \in ℝ^{3}$ the map $\vec{v} \mapsto \vec{v} \cdot \vec{k}$ is linear. This statement is true. It follows from properties of the dot product we have seen earlier: $(\vec{v} + \vec{u}) \cdot \vec{k} = \vec{v} \cdot \vec{k} + \vec{u} \cdot \vec{k}$ and $(r \vec{v}) \cdot \vec{k} = r (\vec{v} \cdot \vec{k})$ . (The natural guess at a generalization of this generalization, that the map from $ℝ^{n}$ to $ℝ$ whose action consists of taking the dot product of its argument with a fixed vector $\vec{k} \in ℝ^{n}$ is linear, is also true.)
Exercise 1.32 Worked answer

Show that every homomorphism from $ℝ^{1}$ to $ℝ^{1}$ acts via multiplication by a scalar. Conclude that every nontrivial linear transformation of $ℝ^{1}$ is an isomorphism. Is that true for transformations of $ℝ^{2}$ ? $ℝ^{n}$ ?

Back to Exercise 1.32

Answer. Let $h : ℝ^{1} \to ℝ^{1}$ be linear. A linear map is determined by its action on a basis, so fix the basis $⟨ 1 ⟩$ for $ℝ^{1}$ . For any $r \in ℝ^{1}$ we have that $h (r) = h (r \cdot 1) = r \cdot h (1)$ and so $h$ acts on any argument $r$ by multiplying it by the constant $h (1)$ . If $h (1)$ is not zero then the map is a correspondence—its inverse is division by $h (1)$ —so any nontrivial transformation of $ℝ^{1}$ is an isomorphism.

This projection map is an example that shows that not every transformation of $ℝ^{n}$ acts via multiplication by a constant when $n > 1$ , including when $n = 2$ .

$(\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{matrix}) \mapsto (\begin{matrix} x_{1} \\ 0 \\ ⋮ \\ 0 \end{matrix})$
Exercise 1.33 Worked answer

Show that for any scalars $a_{1, 1}, \dots, a_{m, n}$ this map $h : ℝ^{n} \to ℝ^{m}$ is a homomorphism.

$(\begin{matrix} x_{1} \\ ⋮ \\ x_{n} \end{matrix}) \mapsto (\begin{matrix} a_{1, 1} x_{1} + \dots + a_{1, n} x_{n} \\ ⋮ \\ a_{m, 1} x_{1} + \dots + a_{m, n} x_{n} \end{matrix})$

Back to Exercise 1.33

Answer. Where $c$ and $d$ are scalars, we have this.

$\begin{matrix} h (c \cdot (\begin{matrix} x_{1} \\ ⋮ \\ x_{n} \end{matrix}) + d \cdot (\begin{matrix} y_{1} \\ ⋮ \\ y_{n} \end{matrix})) \\ \begin{aligned} = h ((\begin{array}{c} c x_{1} + d y_{1} \\ ⋮ \\ c x_{n} + d y_{n} \end{array})) \\ = (\begin{array}{c} a_{1, 1} (c x_{1} + d y_{1}) + \dots + a_{1, n} (c x_{n} + d y_{n}) \\ ⋮ \\ a_{m, 1} (c x_{1} + d y_{1}) + \dots + a_{m, n} (c x_{n} + d y_{n}) \end{array}) \\ = c \cdot (\begin{array}{c} a_{1, 1} x_{1} + \dots + a_{1, n} x_{n} \\ ⋮ \\ a_{m, 1} x_{1} + \dots + a_{m, n} x_{n} \end{array}) + d \cdot (\begin{array}{c} a_{1, 1} y_{1} + \dots + a_{1, n} y_{n} \\ ⋮ \\ a_{m, 1} y_{1} + \dots + a_{m, n} y_{n} \end{array}) \\ = c \cdot h ((\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array})) + d \cdot h ((\begin{array}{c} y_{1} \\ ⋮ \\ y_{n} \end{array})) \end{aligned} \end{matrix}$
Exercise 1.34 Worked answer

Consider the space of polynomials $𝒫_{n}$ .
1. Show that for each $i$ , the $i$ -th derivative operator $d^{i} / d x^{i}$ is a linear transformation of that space.
2. Conclude that for any scalars $c_{k}, \dots, c_{0}$ this map is a linear transformation of that space.
  
  $f \mapsto c_{k} \frac{d^{k}}{d x^{k}} f + c_{k - 1} \frac{d^{k - 1}}{d x^{k - 1}} f + \dots + c_{1} \frac{d}{d x} f + c_{0} f$
Back to Exercise 1.34

Answer.
1. Each power $i$ of the derivative operator is linear because of these rules familiar from calculus.
  
  $\frac{d^{i}}{d x^{i}} (f (x) + g (x)) = \frac{d^{i}}{d x^{i}} f (x) + \frac{d^{i}}{d x^{i}} g (x) \frac{d^{i}}{d x^{i}} r \cdot f (x) = r \cdot \frac{d^{i}}{d x^{i}} f (x)$
2. Any linear combination of linear maps is also a linear map. Thus, the given map is a linear transformation of $𝒫_{n}$ .
Exercise 1.35 Worked answer

Lemma 1.17 shows that a sum of linear functions is linear and that a scalar multiple of a linear function is linear. Show also that a composition of linear functions is linear.

Back to Exercise 1.35

Answer. (This argument has already appeared, as part of the proof that isomorphism is an equivalence.) Let $f : U \to V$ and $g : V \to W$ be linear. The composition preserves linear combinations

$\begin{matrix} g \circ f (c_{1} {\vec{u}}_{1} + c_{2} {\vec{u}}_{2}) = g (f (c_{1} {\vec{u}}_{1} + c_{2} {\vec{u}}_{2})) = g (c_{1} f ({\vec{u}}_{1}) + c_{2} f ({\vec{u}}_{2})) \\ = c_{1} \cdot g (f ({\vec{u}}_{1})) + c_{2} \cdot g (f ({\vec{u}}_{2})) = c_{1} \cdot g \circ f ({\vec{u}}_{1}) + c_{2} \cdot g \circ f ({\vec{u}}_{2}) \end{matrix}$

where ${\vec{u}}_{1}, {\vec{u}}_{2} \in U$ and scalars $c_{1}, c_{2}$
Exercise 1.36 Worked answer

Where $f : V \to W$ is linear, suppose that $f ({\vec{v}}_{1}) = {\vec{w}}_{1}$ , …, $f ({\vec{v}}_{n}) = {\vec{w}}_{n}$ for some vectors ${\vec{w}}_{1}$ , …, ${\vec{w}}_{n}$ from $W$ .
1. If the set of $\vec{w}$ ’s is independent, must the set of $\vec{v}$ ’s also be independent?
2. If the set of $\vec{v}$ ’s is independent, must the set of $\vec{w}$ ’s also be independent?
3. If the set of $\vec{w}$ ’s spans $W$ , must the set of $\vec{v}$ ’s span $V$ ?
4. If the set of $\vec{v}$ ’s spans $V$ , must the set of $\vec{w}$ ’s span $W$ ?
Back to Exercise 1.36

Answer.
1. Yes. The set of $\vec{w}$ ’s cannot be linearly independent if the set of $\vec{v}$ ’s is linearly dependent because any nontrivial relationship in the domain ${\vec{0}}_{V} = c_{1} {\vec{v}}_{1} + \dots + c_{n} {\vec{v}}_{n}$ would give a nontrivial relationship in the range $f ({\vec{0}}_{V}) = {\vec{0}}_{W} = f (c_{1} {\vec{v}}_{1} + \dots + c_{n} {\vec{v}}_{n}) = c_{1} f ({\vec{v}}_{1}) + \dots + c_{n} f ({\vec{v}}_{n}) = c_{1} \vec{w} + \dots + c_{n} {\vec{w}}_{n}$ .
2. Not necessarily. For instance, the transformation of $ℝ^{2}$ given by
  
  $(\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} x + y \\ x + y \end{matrix})$
  
  sends this linearly independent set in the domain to a linearly dependent image.
  
  ${{\vec{v}}_{1}, {\vec{v}}_{2}} = {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \end{matrix})} \mapsto {(\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 2 \\ 2 \end{matrix})} = {{\vec{w}}_{1}, {\vec{w}}_{2}}$
3. Not necessarily. An example is the projection map $π : ℝ^{3} \to ℝ^{2}$
  
  $(\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} x \\ y \end{matrix})$
  
  and this set that does not span the domain but maps to a set that does span the codomain.
  
  ${(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})} \overset{π}{⟼} {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})}$
4. Not necessarily. For instance, the injection map $ι : ℝ^{2} \to ℝ^{3}$ sends the standard basis $ℰ_{2}$ for the domain to a set that does not span the codomain. (Remark. However, the set of $\vec{w}$ ’s does span the range. A proof is easy.)
Exercise 1.37 Worked answer

Generalize Example 1.16 by proving that for every appropriate domain and codomain the matrix transpose map is linear. What are the appropriate domains and codomains?

Back to Exercise 1.37

Answer. Recall that the entry in row $i$ and column $j$ of the transpose of $M$ is the entry $m_{j, i}$ from row $j$ and column $i$ of $M$ . Now, the check is routine. Start with the transpose of the combination.

${[r \cdot (\begin{matrix} ⋮ \\ \dots & a_{i, j} & \dots \\ ⋮ \end{matrix}) + s \cdot (\begin{matrix} ⋮ \\ \dots & b_{i, j} & \dots \\ ⋮ \end{matrix})]}^{𝖳}$

Combine and take the transpose.

$= {(\begin{matrix} ⋮ \\ \dots & r a_{i, j} + s b_{i, j} & \dots \\ ⋮ \end{matrix})}^{𝖳} = (\begin{matrix} ⋮ \\ \dots & r a_{j, i} + s b_{j, i} & \dots \\ ⋮ \end{matrix})$

Then bring out the scalars, and un-transpose.

$\begin{matrix} = r \cdot (\begin{matrix} ⋮ \\ \dots & a_{j, i} & \dots \\ ⋮ \end{matrix}) + s \cdot (\begin{matrix} ⋮ \\ \dots & b_{j, i} & \dots \\ ⋮ \end{matrix}) \\ = r \cdot {(\begin{matrix} ⋮ \\ \dots & a_{j, i} & \dots \\ ⋮ \end{matrix})}^{𝖳} + s \cdot {(\begin{matrix} ⋮ \\ \dots & b_{j, i} & \dots \\ ⋮ \end{matrix})}^{𝖳} \end{matrix}$

The domain is $ℳ_{m \times n}$ while the codomain is $ℳ_{n \times m}$ .
Exercise 1.38 Worked answer
1. Where $\vec{u}, \vec{v} \in ℝ^{n}$ , by definition the line segment connecting them is the set $ℓ = {t \cdot \vec{u} + (1 - t) \cdot \vec{v} ∣ t \in [0. .1]}$ . Show that the image, under a homomorphism $h$ , of the segment between $\vec{u}$ and $\vec{v}$ is the segment between $h (\vec{u})$ and $h (\vec{v})$ .
2. A subset of $ℝ^{n}$ is convex if, for any two points in that set, the line segment joining them lies entirely in that set. (The inside of a sphere is convex while the skin of a sphere is not.) Prove that linear maps from $ℝ^{n}$ to $ℝ^{m}$ preserve the property of set convexity.
Back to Exercise 1.38

Answer.
1. For any homomorphism $h : ℝ^{n} \to ℝ^{m}$ we have
  
  $h (ℓ) = {h (t \cdot \vec{u} + (1 - t) \cdot \vec{v}) ∣ t \in [0. .1]} = {t \cdot h (\vec{u}) + (1 - t) \cdot h (\vec{v}) ∣ t \in [0. .1]}$
  
  which is the line segment from $h (\vec{u})$ to $h (\vec{v})$ .
2. We must show that if a subset of the domain is convex then its image, as a subset of the range, is also convex. Suppose that $C \subseteq ℝ^{n}$ is convex and consider its image $h (C)$ . To show $h (C)$ is convex we must show that for any two of its members, ${\vec{d}}_{1}$ and ${\vec{d}}_{2}$ , the line segment connecting them
  
  $ℓ = {t \cdot {\vec{d}}_{1} + (1 - t) \cdot {\vec{d}}_{2} ∣ t \in [0. .1]}$
  
  is a subset of $h (C)$ .
  
  Fix any member $\hat{t} \cdot {\vec{d}}_{1} + (1 - \hat{t}) \cdot {\vec{d}}_{2}$ of that line segment. Because the endpoints of $ℓ$ are in the image of $C$ , there are members of $C$ that map to them, say $h ({\vec{c}}_{1}) = {\vec{d}}_{1}$ and $h ({\vec{c}}_{2}) = {\vec{d}}_{2}$ . Now, where $\hat{t}$ is the scalar that we fixed in the first sentence of this paragraph, observe that $h (\hat{t} \cdot {\vec{c}}_{1} + (1 - \hat{t}) \cdot {\vec{c}}_{2}) = \hat{t} \cdot h ({\vec{c}}_{1}) + (1 - \hat{t}) \cdot h ({\vec{c}}_{2}) = \hat{t} \cdot {\vec{d}}_{1} + (1 - \hat{t}) \cdot {\vec{d}}_{2}$ Thus, any member of $ℓ$ is a member of $h (C)$ , and so $h (C)$ is convex.
Exercise 1.39 Worked answer

Recommended. Let $h : ℝ^{n} \to ℝ^{m}$ be a homomorphism.
1. Show that the image under $h$ of a line in $ℝ^{n}$ is a (possibly degenerate) line in $ℝ^{m}$ .
2. What happens to a $k$ -dimensional linear surface?
Back to Exercise 1.39

Answer.
1. For ${\vec{v}}_{0}, {\vec{v}}_{1} \in ℝ^{n}$ , the line through ${\vec{v}}_{0}$ with direction ${\vec{v}}_{1}$ is the set ${{\vec{v}}_{0} + t \cdot {\vec{v}}_{1} ∣ t \in ℝ}$ . The image under $h$ of that line ${h ({\vec{v}}_{0} + t \cdot {\vec{v}}_{1}) ∣ t \in ℝ} = {h ({\vec{v}}_{0}) + t \cdot h ({\vec{v}}_{1}) ∣ t \in ℝ}$ is the line through $h ({\vec{v}}_{0})$ with direction $h ({\vec{v}}_{1})$ . If $h ({\vec{v}}_{1})$ is the zero vector then this line is degenerate.
2. A $k$ -dimensional linear surface in $ℝ^{n}$ maps to a $k$ -dimensional linear surface in $ℝ^{m}$ (possibly it is degenerate). The proof is just like that the one for the line.
Exercise 1.40 Worked answer

Prove that the restriction of a homomorphism to a subspace of its domain is another homomorphism.

Back to Exercise 1.40

Answer. Suppose that $h : V \to W$ is a homomorphism and suppose that $S$ is a subspace of $V$ . Consider the map $\hat{h} : S \to W$ defined by $\hat{h} (\vec{s}) = h (\vec{s})$ . (The only difference between $\hat{h}$ and $h$ is the difference in domain.) Then this new map is linear: $\hat{h} (c_{1} \cdot {\vec{s}}_{1} + c_{2} \cdot {\vec{s}}_{2}) = h (c_{1} {\vec{s}}_{1} + c_{2} {\vec{s}}_{2}) = c_{1} h ({\vec{s}}_{1}) + c_{2} h ({\vec{s}}_{2}) = c_{1} \cdot \hat{h} ({\vec{s}}_{1}) + c_{2} \cdot \hat{h} ({\vec{s}}_{2})$ .
Exercise 1.41 Worked answer

Assume that $h : V \to W$ is linear.
1. Show that the range space of this map ${h (\vec{v}) ∣ \vec{v} \in V}$ is a subspace of the codomain $W$ .
2. Show that the null space of this map ${\vec{v} \in V ∣ h (\vec{v}) = {\vec{0}}_{W}}$ is a subspace of the domain $V$ .
3. Show that if $U$ is a subspace of the domain $V$ then its image ${h (\vec{u}) ∣ \vec{u} \in U}$ is a subspace of the codomain $W$ . This generalizes the first item.
4. Generalize the second item.
Back to Exercise 1.41

Answer. This will appear as a lemma in the next subsection.
1. The range is nonempty because $V$ is nonempty. To finish we need to show that it is closed under combinations. A combination of range vectors has the form, where ${\vec{v}}_{1}, \dots, {\vec{v}}_{n} \in V$ ,
  
  $c_{1} \cdot h ({\vec{v}}_{1}) + \dots + c_{n} \cdot h ({\vec{v}}_{n}) = h (c_{1} {\vec{v}}_{1}) + \dots + h (c_{n} {\vec{v}}_{n}) = h (c_{1} \cdot {\vec{v}}_{1} + \dots + c_{n} \cdot {\vec{v}}_{n}),$
  
  which is itself in the range as $c_{1} \cdot {\vec{v}}_{1} + \dots + c_{n} \cdot {\vec{v}}_{n}$ is a member of domain $V$ . Therefore the range is a subspace.
2. The null space is nonempty since it contains ${\vec{0}}_{V}$ , as ${\vec{0}}_{V}$ maps to ${\vec{0}}_{W}$ . It is closed under linear combinations because, where ${\vec{v}}_{1}, \dots, {\vec{v}}_{n} \in V$ are elements of the inverse image ${\vec{v} \in V ∣ h (\vec{v}) = {\vec{0}}_{W}}$ , for $c_{1}, \dots, c_{n} \in ℝ$
  
  ${\vec{0}}_{W} = c_{1} \cdot h ({\vec{v}}_{1}) + \dots + c_{n} \cdot h ({\vec{v}}_{n}) = h (c_{1} \cdot {\vec{v}}_{1} + \dots + c_{n} \cdot {\vec{v}}_{n})$
  
  and so $c_{1} \cdot {\vec{v}}_{1} + \dots + c_{n} \cdot {\vec{v}}_{n}$ is also in the inverse image of ${\vec{0}}_{W}$ .
3. This image of $U$ nonempty because $U$ is nonempty. For closure under combinations, where ${\vec{u}}_{1}, \dots, {\vec{u}}_{n} \in U$ ,
  
  $c_{1} \cdot h ({\vec{u}}_{1}) + \dots + c_{n} \cdot h ({\vec{u}}_{n}) = h (c_{1} \cdot {\vec{u}}_{1}) + \dots + h (c_{n} \cdot {\vec{u}}_{n}) = h (c_{1} \cdot {\vec{u}}_{1} + \dots + c_{n} \cdot {\vec{u}}_{n})$
  
  which is itself in $h (U)$ as $c_{1} \cdot {\vec{u}}_{1} + \dots + c_{n} \cdot {\vec{u}}_{n}$ is in $U$ . Thus this set is a subspace.
4. The natural generalization is that the inverse image of a subspace of is a subspace.
  
  Suppose that $X$ is a subspace of $W$ . Note that ${\vec{0}}_{W} \in X$ so that the set ${\vec{v} \in V ∣ h (\vec{v}) \in X}$ is not empty. To show that this set is closed under combinations, let ${\vec{v}}_{1}, \dots, {\vec{v}}_{n}$ be elements of $V$ such that $h ({\vec{v}}_{1}) = {\vec{x}}_{1}$ , …, $h ({\vec{v}}_{n}) = {\vec{x}}_{n}$ and note that
  
  $h (c_{1} \cdot {\vec{v}}_{1} + \dots + c_{n} \cdot {\vec{v}}_{n}) = c_{1} \cdot h ({\vec{v}}_{1}) + \dots + c_{n} \cdot h ({\vec{v}}_{n}) = c_{1} \cdot {\vec{x}}_{1} + \dots + c_{n} \cdot {\vec{x}}_{n}$
  
  so a linear combination of elements of $h^{- 1} (X)$ is also in $h^{- 1} (X)$ .
Exercise 1.42 Worked answer

Consider the set of isomorphisms from a vector space to itself. Is this a subspace of the space $ℒ (V, V)$ of homomorphisms from the space to itself?

Back to Exercise 1.42

Answer. No; the set of isomorphisms does not contain the zero map (unless the space is trivial).
Exercise 1.43 Worked answer

Does Theorem 1.9 need that $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ is a basis? That is, can we still get a well-defined and unique homomorphism if we drop either the condition that the set of $\vec{β}$ ’s be linearly independent, or the condition that it span the domain?

Back to Exercise 1.43

Answer. If $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ doesn’t span the space then the map needn’t be unique. For instance, if we try to define a map from $ℝ^{2}$ to itself by specifying only that ${\vec{e}}_{1}$ maps to itself, then there is more than one homomorphism possible; both the identity map and the projection map onto the first component fit this condition.

If we drop the condition that $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ is linearly independent then we risk an inconsistent specification (i.e, there could be no such map). An example is if we consider $⟨ {\vec{e}}_{2}, {\vec{e}}_{1}, 2 {\vec{e}}_{1} ⟩$ , and try to define a map from $ℝ^{2}$ to itself that sends ${\vec{e}}_{2}$ to itself, and sends both ${\vec{e}}_{1}$ and $2 {\vec{e}}_{1}$ to ${\vec{e}}_{1}$ . No homomorphism can satisfy these three conditions.
Exercise 1.44 Worked answer

Let $V$ be a vector space and assume that the maps $f_{1}, f_{2} : V \to ℝ^{1}$ are linear.
1. Define a map $F : V \to ℝ^{2}$ whose component functions are the given linear ones.
  
  $\vec{v} \mapsto (\begin{matrix} f_{1} (\vec{v}) \\ f_{2} (\vec{v}) \end{matrix})$
  
  Show that $F$ is linear.
2. Does the converse hold—is any linear map from $V$ to $ℝ^{2}$ made up of two linear component maps to $ℝ^{1}$ ?
3. Generalize.
Back to Exercise 1.44

Answer.
1. Briefly, the check of linearity is this.
  
  $\begin{matrix} F (r_{1} \cdot {\vec{v}}_{1} + r_{2} \cdot {\vec{v}}_{2}) = (\begin{matrix} f_{1} (r_{1} {\vec{v}}_{1} + r_{2} {\vec{v}}_{2}) \\ f_{2} (r_{1} {\vec{v}}_{1} + r_{2} {\vec{v}}_{2}) \end{matrix}) \\ = r_{1} (\begin{matrix} f_{1} ({\vec{v}}_{1}) \\ f_{2} ({\vec{v}}_{1}) \end{matrix}) + r_{2} (\begin{matrix} f_{1} ({\vec{v}}_{2}) \\ f_{2} ({\vec{v}}_{2}) \end{matrix}) = r_{1} \cdot F ({\vec{v}}_{1}) + r_{2} \cdot F ({\vec{v}}_{2}) \end{matrix}$
2. Yes. Let $π_{1} : ℝ^{2} \to ℝ^{1}$ and $π_{2} : ℝ^{2} \to ℝ^{1}$ be the projections
  
  $(\begin{matrix} x \\ y \end{matrix}) \overset{π_{1}}{⟼} x and (\begin{matrix} x \\ y \end{matrix}) \overset{π_{2}}{⟼} y$
  
  onto the two axes. Now, where $f_{1} (\vec{v}) = π_{1} (F (\vec{v}))$ and $f_{2} (\vec{v}) = π_{2} (F (\vec{v}))$ we have the desired component functions.
  
  $F (\vec{v}) = (\begin{matrix} f_{1} (\vec{v}) \\ f_{2} (\vec{v}) \end{matrix})$
  
  They are linear because they are the composition of linear functions, and the fact that the composition of linear functions is linear was part of the proof that isomorphism is an equivalence relation (alternatively, the check that they are linear is straightforward).
3. In general, a map from a vector space $V$ to an $ℝ^{n}$ is linear if and only if each of the component functions is linear. The verification is as in the prior item.

Range Space and Null Space

Isomorphisms and homomorphisms both preserve structure. The difference is that homomorphisms have fewer restrictions, since they needn’t be onto and needn’t be one-to-one. We will examine what can happen with homomorphisms that cannot happen with isomorphisms.

First consider the fact that homomorphisms need not be onto. Of course, each function is onto some set, namely its range. For example, the injection map $ι : ℝ^{2} \to ℝ^{3}$

$(\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} x \\ y \\ 0 \end{matrix})$

is a homomorphism, and is not onto $ℝ^{3}$ . But it is onto the $x y$ -plane.

Lemma 2.1 Under a homomorphism, the image of any subspace of the domain is a subspace of the codomain. In particular, the image of the entire space, the range of the homomorphism, is a subspace of the codomain.

Proof Let $h : V \to W$ be linear and let $S$ be a subspace of the domain $V$ . The image $h (S)$ is a subset of the codomain $W$ , which is nonempty because $S$ is nonempty. Thus, to show that $h (S)$ is a subspace of $W$ we need only show that it is closed under linear combinations of two vectors. If $h ({\vec{s}}_{1})$ and $h ({\vec{s}}_{2})$ are members of $h (S)$ then $c_{1} \cdot h ({\vec{s}}_{1}) + c_{2} \cdot h ({\vec{s}}_{2}) = h (c_{1} \cdot {\vec{s}}_{1}) + h (c_{2} \cdot {\vec{s}}_{2}) = h (c_{1} \cdot {\vec{s}}_{1} + c_{2} \cdot {\vec{s}}_{2})$ is also a member of $h (S)$ because it is the image of $c_{1} \cdot {\vec{s}}_{1} + c_{2} \cdot {\vec{s}}_{2}$ from $S$ .

QED

Definition 2.2 The range space of a homomorphism $h : V \to W$ is

$ℛ (h) = {h (\vec{v}) ∣ \vec{v} \in V}$

sometimes denoted $h (V)$ . The dimension of the range space is the map’s rank.

We shall soon see the connection between the rank of a map and the rank of a matrix.

Example 2.3 For the derivative map $d / d x : 𝒫_{3} \to 𝒫_{3}$ given by $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \mapsto a_{1} + 2 a_{2} x + 3 a_{3} x^{2}$ the range space $ℛ (d / d x)$ is the set of quadratic polynomials ${r + s x + t x^{2} ∣ r, s, t \in ℝ}$ . Thus, this map’s rank is $3$ .

Example 2.4 With this homomorphism $h : M_{2 \times 2} \to 𝒫_{3}$

$(\begin{matrix} a & b \\ c & d \end{matrix}) \mapsto (a + b + 2 d) + c x^{2} + c x^{3}$

an image vector in the range can have any constant term, must have an $x$ coefficient of zero, and must have the same coefficient of $x^{2}$ as of $x^{3}$ . That is, the range space is $ℛ (h) = {r + s x^{2} + s x^{3} ∣ r, s \in ℝ}$ and so the rank is $2$ .

The prior result shows that, in passing from the definition of isomorphism to the more general definition of homomorphism, omitting the onto requirement doesn’t make an essential difference. Any homomorphism is onto some space, namely its range.

However, omitting the one-to-one condition does make a difference. A homomorphism may have many elements of the domain that map to one element of the codomain. Below is a bean sketch of a many-to-one map between sets.¹ It shows three elements of the codomain that are each the image of many members of the domain. (Rather than picture lots of individual $\mapsto$ arrows, each association of many inputs with one output shows only one such arrow.)

A many-to-one map between two bean-shaped sets: three groups of domain points in the left set map to three individual points in the right set. One mapping arrow represents each inverse-image group.

Recall that for any function $h : V \to W$ , the set of elements of $V$ that map to $\vec{w} \in W$ is the inverse image $h^{- 1} (\vec{w}) = {\vec{v} \in V ∣ h (\vec{v}) = \vec{w}}$ . Above, the left side shows three inverse image sets.

Example 2.5 Consider the projection $π : ℝ^{3} \to ℝ^{2}$

$(\begin{matrix} x \\ y \\ z \end{matrix}) \overset{π}{⟼} (\begin{matrix} x \\ y \end{matrix})$

which is a homomorphism that is many-to-one. An inverse image set is a vertical line of vectors in the domain.

Projection from R3 to R2: a vertical line of domain points maps to one codomain vector w. Several domain vectors end on that line and share the same first two coordinates.

One example is this.

$π^{- 1} ((\begin{matrix} 1 \\ 3 \end{matrix})) = {(\begin{matrix} 1 \\ 3 \\ z \end{matrix}) ∣ z \in ℝ}$

Example 2.6 This homomorphism $h : ℝ^{2} \to ℝ^{1}$

$(\begin{matrix} x \\ y \end{matrix}) \overset{h}{⟼} x + y$

is also many-to-one. For a fixed $w \in ℝ^{1}$ the inverse image $h^{- 1} (w)$

The map from R2 to R1 sends all plane vectors ending on the diagonal line x+y=w to the same point w on the number line.

is the set of plane vectors whose components add to $w$ .

In generalizing from isomorphisms to homomorphisms by dropping the one-to-one condition we lose the property that, intuitively, the domain is “the same” as the range. We lose, that is, that the domain corresponds perfectly to the range. The examples below illustrate that what we retain is that a homomorphism describes how the domain is “analogous to” or “like” the range.

Example 2.7 We think of $ℝ^{3}$ as like $ℝ^{2}$ except that vectors have an extra component. That is, we think of the vector with components $x$ , $y$ , and $z$ as like the vector with components $x$ and $y$ . Defining the projection map $π$ makes precise which members of the domain we are thinking of as related to which members of the codomain.

To understanding how the preservation conditions in the definition of homomorphism show that the domain elements are like the codomain elements, start by picturing $ℝ^{2}$ as the $x y$ -plane inside of $ℝ^{3}$ (the $x y$ plane inside of $ℝ^{3}$ is a set of three-tall vectors with a third component of zero and so does not precisely equal the set of two-tall vectors $ℝ^{2}$ , but this embedding makes the picture much clearer). The preservation of addition property says that vectors in $ℝ^{3}$ act like their shadows in the plane.


$(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix})$ above $(\begin{matrix} x_{1} \\ y_{1} \end{matrix})$	plus	$(\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix})$ above $(\begin{matrix} x_{2} \\ y_{2} \end{matrix})$	equals	$(\begin{matrix} x_{1} + x_{2} \\ y_{1} + y_{2} \\ z_{1} + z_{2} \end{matrix})$ above $(\begin{matrix} x_{1} + x_{2} \\ y_{1} + y_{2} \end{matrix})$

Thinking of $π (\vec{v})$ as the “shadow” of $\vec{v}$ in the plane gives this restatement: the sum of the shadows $π ({\vec{v}}_{1}) + π ({\vec{v}}_{2})$ equals the shadow of the sum $π ({\vec{v}}_{1} + {\vec{v}}_{2})$ . Preservation of scalar multiplication is similar.

Drawing the codomain $ℝ^{2}$ on the right gives a picture that is uglier but is more faithful to the bean sketch above.

Projection from three-dimensional space on the left to two-dimensional space on the right. Three vertical inverse-image lines correspond to the vectors labelled w1, w2 and w1+w2; representative spatial vectors illustrate addition of these classes.

Again, the domain vectors that map to ${\vec{w}}_{1}$ lie in a vertical line; one is drawn, in gray. Call any member of this inverse image $π^{- 1} ({\vec{w}}_{1})$ a “ ${\vec{w}}_{1}$ vector.” Similarly, there is a vertical line of “ ${\vec{w}}_{2}$ vectors” and a vertical line of “ ${\vec{w}}_{1} + {\vec{w}}_{2}$ vectors.” Now, saying that $π$ is a homomorphism is recognizing that if $π ({\vec{v}}_{1}) = {\vec{w}}_{1}$ and $π ({\vec{v}}_{2}) = {\vec{w}}_{2}$ then $π ({\vec{v}}_{1} + {\vec{v}}_{2}) = π ({\vec{v}}_{1}) + π ({\vec{v}}_{2}) = {\vec{w}}_{1} + {\vec{w}}_{2}$ . That is, the classes add: any ${\vec{w}}_{1}$ vector plus any ${\vec{w}}_{2}$ vector equals a ${\vec{w}}_{1} + {\vec{w}}_{2}$ vector. Scalar multiplication is similar.

So although $ℝ^{3}$ and $ℝ^{2}$ are not isomorphic $π$ describes a way in which they are alike: vectors in $ℝ^{3}$ add as do the associated vectors in $ℝ^{2}$ —vectors add as their shadows add.

Example 2.8 A homomorphism can express an analogy between spaces that is more subtle than the prior one. For the map from Example 2.6

$(\begin{matrix} x \\ y \end{matrix}) \overset{h}{⟼} x + y$

fix two numbers in the range $w_{1}, w_{2} \in ℝ$ . A ${\vec{v}}_{1}$ that maps to $w_{1}$ has components that add to $w_{1}$ , so the inverse image $h^{- 1} (w_{1})$ is the set of vectors with endpoint on the diagonal line $x + y = w_{1}$ . Think of these as “ $w_{1}$ vectors.” Similarly we have “ $w_{2}$ vectors” and “ $w_{1} + w_{2}$ vectors.” The addition preservation property says this.


a “ $w_{1}$ vector”	plus	a “ $w_{2}$ vector”	equals	a “ $w_{1} + w_{2}$ vector”

Restated, if we add a $w_{1}$ vector to a $w_{2}$ vector then $h$ maps the result to a $w_{1} + w_{2}$ vector. Briefly, the sum of the images is the image of the sum. Even more briefly, $h ({\vec{v}}_{1}) + h ({\vec{v}}_{2}) = h ({\vec{v}}_{1} + {\vec{v}}_{2})$ .

Example 2.9 The inverse images can be structures other than lines. For the linear map $h : ℝ^{3} \to ℝ^{2}$

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

the inverse image sets are planes $x = 0$ , $x = 1$ , etc., perpendicular to the $x$ -axis.

Three parallel planes perpendicular to the x-axis illustrate inverse images under the map (x,y,z) to (x,x). Each plane fixes x and contains all values of y and z.

We won’t describe how every homomorphism that we will use is an analogy because the formal sense that we make of “alike in that …” is ‘a homomorphism exists such that …’. Nonetheless, the idea that a homomorphism between two spaces expresses how the domain’s vectors fall into classes that act like the range’s vectors is a good way to view homomorphisms.

Another reason that we won’t treat all of the homomorphisms that we see as above is that many vector spaces are hard to draw, e.g., a space of polynomials. But there is nothing wrong with leveraging spaces that we can draw: from the three examples 2.7, 2.8, and 2.9 we draw two insights.

The first insight is that in all three examples the inverse image of the range’s zero vector is a line or plane through the origin. It is therefore a subspace of the domain.

Lemma 2.10 For any homomorphism the inverse image of a subspace of the range is a subspace of the domain. In particular, the inverse image of the trivial subspace of the range is a subspace of the domain.

(The examples above consider inverse images of single vectors but this result is about inverse images of sets $h^{- 1} (S) = {\vec{v} \in V ∣ h (\vec{v}) \in S}$ . We use the same term for both by taking the inverse image of a single element $h^{- 1} (\vec{w})$ to be the inverse image of the one-element set $h^{- 1} ({\vec{w}})$ .)

Proof Let $h : V \to W$ be a homomorphism and let $S$ be a subspace of the range space of $h$ . Consider the inverse image of $S$ . It is nonempty because it contains ${\vec{0}}_{V}$ , since $h ({\vec{0}}_{V}) = {\vec{0}}_{W}$ and ${\vec{0}}_{W}$ is an element of $S$ as $S$ is a subspace. To finish we show that $h^{- 1} (S)$ is closed under linear combinations. Let ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ be two of its elements, so that $h ({\vec{v}}_{1})$ and $h ({\vec{v}}_{2})$ are elements of $S$ . Then $c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2}$ is an element of the inverse image $h^{- 1} (S)$ because $h (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2}) = c_{1} h ({\vec{v}}_{1}) + c_{2} h ({\vec{v}}_{2})$ is a member of $S$ .

QED

Definition 2.11 The null space or kernel of a linear map $h : V \to W$ is the inverse image of ${\vec{0}}_{W}$ .

$𝒩 (h) = h^{- 1} ({\vec{0}}_{W}) = {\vec{v} \in V ∣ h (\vec{v}) = {\vec{0}}_{W}}$

The dimension of the null space is the map’s nullity.

A many-to-one map between two bean-shaped sets, with the domain zero 0V among a group of domain points and the codomain zero 0W at their image. This inverse-image group represents the null space.

Example 2.12 The map from Example 2.3 has this null space $𝒩 (d / d x) = {a_{0} + 0 x + 0 x^{2} + 0 x^{3} ∣ a_{0} \in ℝ}$ so its nullity is $1$ .

Example 2.13 The map from Example 2.4 has this null space, and nullity $2$ .

$𝒩 (h) = {(\begin{matrix} a & b \\ 0 & - (a + b) / 2 \end{matrix}) ∣ a, b \in ℝ}$

Now for the second insight from the above examples. In Example 2.7 each of the vertical lines squashes down to a single point—in passing from the domain to the range, $π$ takes all of these one-dimensional vertical lines and maps them to a point, leaving the range smaller than the domain by one dimension. Similarly, in Example 2.8 the two-dimensional domain compresses to a one-dimensional range by breaking the domain into the diagonal lines and maps each of those to a single member of the range. Finally, in Example 2.9 the domain breaks into planes which get squashed to a point and so the map starts with a three-dimensional domain but ends two smaller, with a one-dimensional range. (The codomain is two-dimensional but the range is one-dimensional and the dimension of the range is what matters.)

Theorem 2.14 A linear map’s rank plus its nullity equals the dimension of its domain.

Proof Let $h : V \to W$ be linear and let $B_{N} = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{k} ⟩$ be a basis for the null space. Expand that to a basis $B_{V} = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{k}, {\vec{β}}_{k + 1}, \dots, {\vec{β}}_{n} ⟩$ for the entire domain, using Corollary Two.III.2.12. We shall show that $B_{R} = ⟨ h ({\vec{β}}_{k + 1}), \dots, h ({\vec{β}}_{n}) ⟩$ is a basis for the range space. Then counting the size of the bases gives the result.

To see that $B_{R}$ is linearly independent, consider ${\vec{0}}_{W} = c_{k + 1} h ({\vec{β}}_{k + 1}) + \dots + c_{n} h ({\vec{β}}_{n})$ . We have ${\vec{0}}_{W} = h (c_{k + 1} {\vec{β}}_{k + 1} + \dots + c_{n} {\vec{β}}_{n})$ and so $c_{k + 1} {\vec{β}}_{k + 1} + \dots + c_{n} {\vec{β}}_{n}$ is in the null space of $h$ . As $B_{N}$ is a basis for the null space there are scalars $c_{1}, \dots, c_{k}$ satisfying this relationship.

$c_{1} {\vec{β}}_{1} + \dots + c_{k} {\vec{β}}_{k} = c_{k + 1} {\vec{β}}_{k + 1} + \dots + c_{n} {\vec{β}}_{n}$

But this is an equation among members of $B_{V}$ , which is a basis for $V$ , so each $c_{i}$ equals $0$ . Therefore $B_{R}$ is linearly independent.

To show that $B_{R}$ spans the range space consider a member of the range space $h (\vec{v})$ . Express $\vec{v}$ as a linear combination $\vec{v} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ of members of $B_{V}$ . This gives $h (\vec{v}) = h (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}) = c_{1} h ({\vec{β}}_{1}) + \dots + c_{k} h ({\vec{β}}_{k}) + c_{k + 1} h ({\vec{β}}_{k + 1}) + \dots + c_{n} h ({\vec{β}}_{n})$ and since ${\vec{β}}_{1}$ , …, ${\vec{β}}_{k}$ are in the null space, we have that $h (\vec{v}) = \vec{0} + \dots + \vec{0} + c_{k + 1} h ({\vec{β}}_{k + 1}) + \dots + c_{n} h ({\vec{β}}_{n})$ . Thus, $h (\vec{v})$ is a linear combination of members of $B_{R}$ , and so $B_{R}$ spans the range space.

QED

Example 2.15 Where $h : ℝ^{3} \to ℝ^{4}$ is

$(\begin{matrix} x \\ y \\ z \end{matrix}) \overset{h}{⟼} (\begin{matrix} x \\ 0 \\ y \\ 0 \end{matrix})$

the range space and null space are

$ℛ (h) = {(\begin{matrix} a \\ 0 \\ b \\ 0 \end{matrix}) ∣ a, b \in ℝ} and 𝒩 (h) = {(\begin{matrix} 0 \\ 0 \\ z \end{matrix}) ∣ z \in ℝ}$

and so the rank of $h$ is $2$ while the nullity is $1$ .

Example 2.16 If $t : ℝ \to ℝ$ is the linear transformation $x \mapsto - 4 x,$ then the range is $ℛ (t) = ℝ$ . The rank is $1$ and the nullity is $0$ .

Corollary 2.17 The rank of a linear map is less than or equal to the dimension of the domain. Equality holds if and only if the nullity of the map is $0$ .

We know that an isomorphism exists between two spaces if and only if the dimension of the range equals the dimension of the domain. We have now seen that for a homomorphism to exist a necessary condition is that the dimension of the range must be less than or equal to the dimension of the domain. For instance, there is no homomorphism from $ℝ^{2}$ onto $ℝ^{3}$ . There are many homomorphisms from $ℝ^{2}$ into $ℝ^{3}$ , but none onto.

The range space of a linear map can be of dimension strictly less than the dimension of the domain and so linearly independent sets in the domain may map to linearly dependent sets in the range. (Example 2.3’s derivative transformation on $𝒫_{3}$ has a domain of dimension $4$ but a range of dimension $3$ and the derivative sends ${1, x, x^{2}, x^{3}}$ to ${0, 1, 2 x, 3 x^{2}}$ ). That is, under a homomorphism independence may be lost. In contrast, dependence stays.

Lemma 2.18 Under a linear map, the image of a linearly dependent set is linearly dependent.

Proof Suppose that $c_{1} {\vec{v}}_{1} + \dots + c_{n} {\vec{v}}_{n} = {\vec{0}}_{V}$ with some $c_{i}$ nonzero. Apply $h$ to both sides: $h (c_{1} {\vec{v}}_{1} + \dots + c_{n} {\vec{v}}_{n}) = c_{1} h ({\vec{v}}_{1}) + \dots + c_{n} h ({\vec{v}}_{n})$ and $h ({\vec{0}}_{V}) = {\vec{0}}_{W}$ . Thus we have $c_{1} h ({\vec{v}}_{1}) + \dots + c_{n} h ({\vec{v}}_{n}) = {\vec{0}}_{W}$ with some $c_{i}$ nonzero.

QED

When is independence not lost? The obvious sufficient condition is when the homomorphism is an isomorphism. This condition is also necessary; see Exercise 2.37. We will finish this subsection comparing homomorphisms with isomorphisms by observing that a one-to-one homomorphism is an isomorphism from its domain onto its range.

Example 2.19 This one-to-one homomorphism $ι : ℝ^{2} \to ℝ^{3}$

$(\begin{matrix} x \\ y \end{matrix}) \overset{ι}{⟼} (\begin{matrix} x \\ y \\ 0 \end{matrix})$

gives a correspondence between $ℝ^{2}$ and the $x y$ -plane subset of $ℝ^{3}$ .

Theorem 2.20 Where $V$ is an $n$ -dimensional vector space, these are equivalent statements about a linear map $h : V \to W$ .

$h$ is one-to-one
$h$ has an inverse from its range to its domain that is a linear map
$𝒩 (h) = {\vec{0}}$ , that is, $nullity (h) = 0$
$rank (h) = n$
if $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ is a basis for $V$ then $⟨ h ({\vec{β}}_{1}), \dots, h ({\vec{β}}_{n}) ⟩$ is a basis for $ℛ (h)$

Proof We will first show that $(1) ⟺ (2)$ . We will then show that $(1) ⟹ (3) ⟹ (4) ⟹ (5) ⟹ (2)$ .

For $(1) ⟹ (2)$ , suppose that the linear map $h$ is one-to-one, and therefore has an inverse $h^{- 1} : ℛ (h) \to V$ . The domain of that inverse is the range of $h$ and thus a linear combination of two members of it has the form $c_{1} h ({\vec{v}}_{1}) + c_{2} h ({\vec{v}}_{2})$ . On that combination, the inverse $h^{- 1}$ gives this.

$\begin{aligned} h^{- 1} (c_{1} h ({\vec{v}}_{1}) + c_{2} h ({\vec{v}}_{2})) & = h^{- 1} (h (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2})) \\ = h^{- 1} \circ h (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2}) \\ = c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2} \\ = c_{1} \cdot h^{- 1} (h ({\vec{v}}_{1})) + c_{2} \cdot h^{- 1} (h ({\vec{v}}_{2})) \end{aligned}$

Thus if a linear map has an inverse then the inverse must be linear. But this also gives the $(2) ⟹ (1)$ implication, because the inverse itself must be one-to-one.

Of the remaining implications, $(1) ⟹ (3)$ holds because any homomorphism maps ${\vec{0}}_{V}$ to ${\vec{0}}_{W}$ , but a one-to-one map sends at most one member of $V$ to ${\vec{0}}_{W}$ .

Next, $(3) ⟹ (4)$ is true since rank plus nullity equals the dimension of the domain.

For $(4) ⟹ (5)$ , to show that $⟨ h ({\vec{β}}_{1}), \dots, h ({\vec{β}}_{n}) ⟩$ is a basis for the range space we need only show that it is a spanning set, because by assumption the range has dimension $n$ . Consider $h (\vec{v}) \in ℛ (h)$ . Expressing $\vec{v}$ as a linear combination of basis elements produces $h (\vec{v}) = h (c_{1} {\vec{β}}_{1} + c_{2} {\vec{β}}_{2} + \dots + c_{n} {\vec{β}}_{n})$ , which gives that $h (\vec{v}) = c_{1} h ({\vec{β}}_{1}) + \dots + c_{n} h ({\vec{β}}_{n})$ , as desired.

Finally, for the $(5) ⟹ (2)$ implication, assume that $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ is a basis for $V$ so that $⟨ h ({\vec{β}}_{1}), \dots, h ({\vec{β}}_{n}) ⟩$ is a basis for $ℛ (h)$ . Then every $\vec{w} \in ℛ (h)$ has the unique representation $\vec{w} = c_{1} h ({\vec{β}}_{1}) + \dots + c_{n} h ({\vec{β}}_{n})$ . Define a map from $ℛ (h)$ to $V$ by

$\vec{w} \mapsto c_{1} {\vec{β}}_{1} + c_{2} {\vec{β}}_{2} + \dots + c_{n} {\vec{β}}_{n}$

(uniqueness of the representation makes this well-defined). Checking that it is linear and that it is the inverse of $h$ are easy.

QED

We have seen that a linear map expresses how the structure of the domain is like that of the range. We can think of such a map as organizing the domain space into inverse images of points in the range. In the special case that the map is one-to-one, each inverse image is a single point and the map is an isomorphism between the domain and the range.

Exercises

Exercise 2.21 Worked answer

Recommended. Let $h : 𝒫_{3} \to 𝒫_{4}$ be given by $p (x) \mapsto x \cdot p (x)$ . Which of these are in the null space? Which are in the range space?
1. $x^{3}$
2. $0$
3. $7$
4. $12 x - 0.5 x^{3}$
5. $1 + 3 x^{2} - x^{3}$
Back to Exercise 2.21

Answer. First, to answer whether a polynomial is in the null space, we have to consider it as a member of the domain $𝒫_{3}$ . To answer whether it is in the range space, we consider it as a member of the codomain $𝒫_{4}$ . That is, for $p (x) = x^{4}$ , the question of whether it is in the range space is sensible but the question of whether it is in the null space is not because it is not even in the domain.
1. The polynomial $x^{3} \in 𝒫_{3}$ is not in the null space because $h (x^{3}) = x^{4}$ is not the zero polynomial in $𝒫_{4}$ . The polynomial $x^{3} \in 𝒫_{4}$ is in the range space because $x^{2} \in 𝒫_{3}$ is mapped by $h$ to $x^{3}$ .
2. The answer to both questions is, “Yes, because $h (0) = 0$ .” The polynomial $0 \in 𝒫_{3}$ is in the null space because it is mapped by $h$ to the zero polynomial in $𝒫_{4}$ . The polynomial $0 \in 𝒫_{4}$ is in the range space because it is the image, under $h$ , of $0 \in 𝒫_{3}$ .
3. The polynomial $7 \in 𝒫_{3}$ is not in the null space because $h (7) = 7 x$ is not the zero polynomial in $𝒫_{4}$ . The polynomial $7 \in 𝒫_{4}$ is not in the range space because there is no member of the domain that when multiplied by $x$ gives the constant polynomial $p (x) = 7$ .
4. The polynomial $12 x - 0.5 x^{3} \in 𝒫_{3}$ is not in the null space because $h (12 x - 0.5 x^{3}) = 12 x^{2} - 0.5 x^{4}$ . The polynomial $12 x - 0.5 x^{3} \in 𝒫_{4}$ is in the range space because it is the image of $12 - 0.5 x^{2}$ .
5. The polynomial $1 + 3 x^{2} - x^{3} \in 𝒫_{3}$ is not in the null space because $h (1 + 3 x^{2} - x^{3}) = x + 3 x^{3} - x^{4}$ . The polynomial $1 + 3 x^{2} - x^{3} \in 𝒫_{4}$ is not in the range space because of the constant term.
Exercise 2.22 Worked answer

Find the range space and the rank of each homomorphism.
1. $h : 𝒫_{3} \to ℝ^{2}$ given by
  
  $a x^{2} + b x + c \mapsto (\begin{matrix} a + b \\ a + c \end{matrix})$
2. $f : ℝ^{2} \to ℝ^{3}$ given by
  
  $(\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} 0 \\ x - y \\ 3 y \end{matrix})$
Back to Exercise 2.22

Answer.
1. The range of $h$ is all of the codomain $ℝ^{2}$ because given
  
  $(\begin{matrix} x \\ y \end{matrix}) \in ℝ^{2}$
  
  it is the image under $h$ of the domain vector $0 x^{2} + b x + c$ . So the rank of $h$ is $2$ .
2. The range is the $y z$ plane. Any
  
  $(\begin{matrix} 0 \\ a \\ b \end{matrix})$
  
  is the image under $f$ of this domain vector.
  
  $(\begin{matrix} a + b / 3 \\ b / 3 \end{matrix})$
  
  So the rank of the map is $2$ .
Exercise 2.23 Worked answer

Recommended. Find the range space and rank of each map.
1. $h : ℝ^{2} \to 𝒫_{3}$ given by
  
  $(\begin{matrix} a \\ b \end{matrix}) \mapsto a + a x + a x^{2}$
2. $h : ℳ_{2 \times 2} \to ℝ$ given by
  
  $(\begin{matrix} a & b \\ c & d \end{matrix}) \mapsto a + d$
3. $h : ℳ_{2 \times 2} \to 𝒫_{2}$ given by
  
  $(\begin{matrix} a & b \\ c & d \end{matrix}) \mapsto a + b + c + d x^{2}$
4. the zero map $Z : ℝ^{3} \to ℝ^{4}$
Back to Exercise 2.23

Answer.
1. The range space is
  
  $ℛ (h) = {a + a x + a x^{2} \in 𝒫_{3} ∣ a, b \in ℝ} = {a \cdot (1 + x + x^{2}) ∣ a \in ℝ}$
  
  and so the rank is one.
2. The range space
  
  $ℛ (h) = {a + d ∣ a, b, c, d \in ℝ}$
  
  is all of $ℝ$ (we can get any real number by taking $d$ to be $0$ and taking $a$ to be the desired number). Thus, the rank is one.
3. The range space is $ℛ (h) = {r + s x^{2} ∣ r, s \in ℝ}$ . The rank is two.
4. The range space is the trivial subspace of $ℝ^{4}$ so the rank is zero.
Exercise 2.24 Worked answer

Recommended. For each linear map in the prior exercise, find the null space and nullity.
Back to Exercise 2.24

Answer.
1. The null space is
  
  $\begin{aligned} 𝒩 (h) & = {(\begin{array}{c} a \\ b \end{array}) \in ℝ^{2} ∣ a + a x + a x^{2} + 0 x^{3} = 0 + 0 x + 0 x^{2} + 0 x^{3}} \\ = {(\begin{array}{c} 0 \\ b \end{array}) ∣ b \in ℝ} \end{aligned}$
  
  and so the nullity is one.
2. The null space is this.
  
  $𝒩 (h) = {(\begin{matrix} a & b \\ c & d \end{matrix}) ∣ a + d = 0} = {(\begin{matrix} - d & b \\ c & d \end{matrix}) ∣ b, c, d \in ℝ}$
  
  Thus the nullity is three.
3. The null space
  
  $𝒩 (h) = {(\begin{matrix} a & b \\ c & d \end{matrix}) ∣ a + b + c = 0, d = 0} = {(\begin{matrix} - b - c & b \\ c & 0 \end{matrix}) ∣ b, c \in ℝ}$
  
  is a dimension $2$ space, so the nullity is two.
4. Every vector in the domain is mapped to the zero vector so the nullspace is $𝒩 (h) = ℝ^{3}$ .
Exercise 2.25 Worked answer

Recommended. Find the nullity of each map below.
1. $h : ℝ^{5} \to ℝ^{8}$ of rank five
2. $h : 𝒫_{3} \to 𝒫_{3}$ of rank one
3. $h : ℝ^{6} \to ℝ^{3}$ , an onto map
4. $h : ℳ_{3 \times 3} \to ℳ_{3 \times 3}$ , onto
Back to Exercise 2.25

Answer. For each, use the result that the rank plus the nullity equals the dimension of the domain.
1. $0$
2. $3$
3. $3$
4. $0$
Exercise 2.26 Worked answer

Recommended. What is the null space of the differentiation transformation $d / d x : 𝒫_{n} \to 𝒫_{n}$ ? What is the null space of the second derivative, as a transformation of $𝒫_{n}$ ? The $k$ -th derivative?

Back to Exercise 2.26

Answer. Because

$\frac{d}{d x} (a_{0} + a_{1} x + \dots + a_{n} x^{n}) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + \dots + n a_{n} x^{n - 1}$

we have this.

$\begin{aligned} 𝒩 (\frac{d}{d x}) & = {a_{0} + \dots + a_{n} x^{n} ∣ a_{1} + 2 a_{2} x + \dots + n a_{n} x^{n - 1} = 0 + 0 x + \dots + 0 x^{n - 1}} \\ = {a_{0} + \dots + a_{n} x^{n} ∣ a_{1} = 0, and a_{2} = 0, \dots, a_{n} = 0} \\ = {a_{0} + 0 x + 0 x^{2} + \dots + 0 x^{n} ∣ a_{0} \in ℝ} \end{aligned}$

In the same way,

$𝒩 (\frac{d^{k}}{d x^{k}}) = {a_{0} + a_{1} x + \dots + a_{k - 1} x^{k - 1} ∣ a_{0}, \dots, a_{k - 1} \in ℝ}$

for $k \leq n$ .
Exercise 2.27 Worked answer

For the map $h : ℝ^{3} \to ℝ^{2}$ given by

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

find the range space, rank, null space, and nullity.

Back to Exercise 2.27

Answer. To see that the range space is all of $ℝ^{2}$ note that for any $u, v \in ℝ$ this system has a solution.

$\begin{array}{rcrcrcr} x & + & y & = & u \\ x & + & z & = & v \end{array} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & + & y & = & u \\ - y & + & z & = & - u + v \end{array}$

(In fact, because there is a free variable, $z$ , it has infinitely many solutions.) Thus the rank, the dimension of the range space, is $2$ .

Since the rank plus the nullity equals the dimension of the domain, we know that the nullity is $1$ . Finding the null space verifies that:

$\begin{array}{rcrcrcr} x & + & y & = & 0 \\ x & + & z & = & 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & + & y & = & 0 \\ - y & + & z & = & 0 \end{array}$

the nullspace is this set

${(\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix}) \cdot z ∣ z \in ℝ}$

which is one-dimensional.
Exercise 2.28 Worked answer

Example 2.7 restates the first condition in the definition of homomorphism as ‘the shadow of a sum is the sum of the shadows’. Restate the second condition in the same style.

Back to Exercise 2.28

Answer. The shadow of a scalar multiple is the scalar multiple of the shadow.
Exercise 2.29 Worked answer

For the homomorphism $h : 𝒫_{3} \to 𝒫_{3}$ given by $h (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}) = a_{0} + (a_{0} + a_{1}) x + (a_{2} + a_{3}) x^{3}$ find these.
1. $𝒩 (h)$
2. $h^{- 1} (2 - x^{3})$
3. $h^{- 1} (1 + x^{2})$
Back to Exercise 2.29

Answer.
1. Setting $a_{0} + (a_{0} + a_{1}) x + (a_{2} + a_{3}) x^{3} = 0 + 0 x + 0 x^{2} + 0 x^{3}$ gives $a_{0} = 0$ and $a_{0} + a_{1} = 0$ and $a_{2} + a_{3} = 0$ , so the null space is ${- a_{3} x^{2} + a_{3} x^{3} ∣ a_{3} \in ℝ}$ .
2. Setting $a_{0} + (a_{0} + a_{1}) x + (a_{2} + a_{3}) x^{3} = 2 + 0 x + 0 x^{2} - x^{3}$ gives that $a_{0} = 2$ , and $a_{1} = - 2$ , and $a_{2} + a_{3} = - 1$ . Taking $a_{3}$ as a parameter, and renaming it $a_{3} = a$ gives this set description ${2 - 2 x + (- 1 - a) x^{2} + a x^{3} ∣ a \in ℝ} = {(2 - 2 x - x^{2}) + a \cdot (- x^{2} + x^{3}) ∣ a \in ℝ}$ .
3. This set is empty because the range of $h$ includes only those polynomials with a $0 x^{2}$ term.
Exercise 2.30 Worked answer

Recommended. For the map $f : ℝ^{2} \to ℝ$ given by

$f ((\begin{matrix} x \\ y \end{matrix})) = 2 x + y$

sketch these inverse image sets: $f^{- 1} (- 3)$ , $f^{- 1} (0)$ , and $f^{- 1} (1)$ .

Back to Exercise 2.30

Answer. All inverse images are lines with slope $- 2$ .
Exercise 2.31 Worked answer

Recommended. Each of these transformations of $𝒫_{3}$ is one-to-one. For each, find the inverse.
1. $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \mapsto a_{0} + a_{1} x + 2 a_{2} x^{2} + 3 a_{3} x^{3}$
2. $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \mapsto a_{0} + a_{2} x + a_{1} x^{2} + a_{3} x^{3}$
3. $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \mapsto a_{1} + a_{2} x + a_{3} x^{2} + a_{0} x^{3}$
4. $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \mapsto a_{0} + (a_{0} + a_{1}) x + (a_{0} + a_{1} + a_{2}) x^{2} + (a_{0} + a_{1} + a_{2} + a_{3}) x^{3}$
Back to Exercise 2.31

Answer. These are the inverses.
1. $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \mapsto a_{0} + a_{1} x + (a_{2} / 2) x^{2} + (a_{3} / 3) x^{3}$
2. $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \mapsto a_{0} + a_{2} x + a_{1} x^{2} + a_{3} x^{3}$
3. $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \mapsto a_{3} + a_{0} x + a_{1} x^{2} + a_{2} x^{3}$
4. $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \mapsto a_{0} + (a_{1} - a_{0}) x + (a_{2} - a_{1}) x^{2} + (a_{3} - a_{2}) x^{3}$
For instance, for the second one, the map given in the question sends $0 + 1 x + 2 x^{2} + 3 x^{3} \mapsto 0 + 2 x + 1 x^{2} + 3 x^{3}$ and then the inverse above sends $0 + 2 x + 1 x^{2} + 3 x^{3} \mapsto 0 + 1 x + 2 x^{2} + 3 x^{3}$ . So this map is actually self-inverse.
Exercise 2.32 Worked answer

Describe the null space and range space of a transformation given by $\vec{v} \mapsto 2 \vec{v}$ .

Back to Exercise 2.32

Answer. For any vector space $V$ , the null space

${\vec{v} \in V ∣ 2 \vec{v} = \vec{0}}$

is trivial, while the range space

${\vec{w} \in V ∣ \vec{w} = 2 \vec{v} for some \vec{v} \in V}$

is all of $V$ , because every vector $\vec{w}$ is twice some other vector, specifically, it is twice $(1 / 2) \vec{w}$ . (Thus, this transformation is actually an automorphism.)
Exercise 2.33 Worked answer

List all pairs $(rank (h), nullity (h))$ that are possible for linear maps from $ℝ^{5}$ to $ℝ^{3}$ .

Back to Exercise 2.33

Answer. Because the rank plus the nullity equals the dimension of the domain (here, five), and the rank is at most three, the possible pairs are: $(3, 2)$ , $(2, 3)$ , $(1, 4)$ , and $(0, 5)$ . Coming up with linear maps that show that each pair is indeed possible is easy.
Exercise 2.34 Worked answer

Does the differentiation map $d / d x : 𝒫_{n} \to 𝒫_{n}$ have an inverse?

Back to Exercise 2.34

Answer. No (unless $𝒫_{n}$ is trivial), because the two polynomials $f_{0} (x) = 0$ and $f_{1} (x) = 1$ have the same derivative; a map must be one-to-one to have an inverse.
Exercise 2.35 Worked answer

Recommended. Find the nullity of this map $h : 𝒫_{n} \to ℝ$ .

$a_{0} + a_{1} x + \dots + a_{n} x^{n} \mapsto \int_{x = 0}^{x = 1} a_{0} + a_{1} x + \dots + a_{n} x^{n} d x$

Back to Exercise 2.35

Answer. The null space is this.

$\begin{matrix} {a_{0} + a_{1} x + \dots + a_{n} x^{n} ∣ a_{0} (1) + \frac{a_{1}}{2} (1^{2}) + \dots + \frac{a_{n}}{n + 1} (1^{n + 1}) = 0} \\ = {a_{0} + a_{1} x + \dots + a_{n} x^{n} ∣ a_{0} + (a_{1} / 2) + \dots + (a_{n} / n + 1) = 0} \end{matrix}$

Thus the nullity is $n$ .
Exercise 2.36 Worked answer
1. Prove that a homomorphism is onto if and only if its rank equals the dimension of its codomain.
2. Conclude that a homomorphism between vector spaces with the same dimension is one-to-one if and only if it is onto.
Back to Exercise 2.36

Answer.
1. One direction is obvious: if the homomorphism is onto then its range is the codomain and so its rank equals the dimension of its codomain. For the other direction assume that the map’s rank equals the dimension of the codomain. Then the map’s range is a subspace of the codomain, and has dimension equal to the dimension of the codomain. Therefore, the map’s range must equal the codomain, and the map is onto. (The ‘therefore’ is because there is a linearly independent subset of the range that is of size equal to the dimension of the codomain, but any such linearly independent subset of the codomain must be a basis for the codomain, and so the range equals the codomain.)
2. By Theorem 2.20, a homomorphism is one-to-one if and only if its nullity is zero. Because rank plus nullity equals the dimension of the domain, it follows that a homomorphism is one-to-one if and only if its rank equals the dimension of its domain. But this domain and codomain have the same dimension, so the map is one-to-one if and only if it is onto.
Exercise 2.37 Worked answer

Show that a linear map is one-to-one if and only if it preserves linear independence.

Back to Exercise 2.37

Answer. We are proving that $h : V \to W$ is one-to-one if and only if for every linearly independent subset $S$ of $V$ the subset $h (S) = {h (\vec{s}) ∣ \vec{s} \in S}$ of $W$ is linearly independent.

One half is easy—by Theorem 2.20, if $h$ is not one-to-one then its null space is nontrivial, that is, it contains more than just the zero vector. So where $\vec{v} \neq {\vec{0}}_{V}$ is in that null space, the singleton set ${\vec{v}}$ is independent while its image ${h (\vec{v})} = {{\vec{0}}_{W}}$ is not.

For the other half, assume that $h$ is one-to-one and so by Theorem 2.20 has a trivial null space. Then for any ${\vec{v}}_{1}, \dots, {\vec{v}}_{n} \in V$ , the relation

${\vec{0}}_{W} = c_{1} \cdot h ({\vec{v}}_{1}) + \dots + c_{n} \cdot h ({\vec{v}}_{n}) = h (c_{1} \cdot {\vec{v}}_{1} + \dots + c_{n} \cdot {\vec{v}}_{n})$

implies the relation $c_{1} \cdot {\vec{v}}_{1} + \dots + c_{n} \cdot {\vec{v}}_{n} = {\vec{0}}_{V}$ . Hence, if a subset of $V$ is independent then so is its image in $W$ .

Remark. The statement is that a linear map is one-to-one if and only if it preserves independence for all sets (that is, if a set is independent then its image is also independent). A map that is not one-to-one may well preserve some independent sets. One example is this map from $ℝ^{3}$ to $ℝ^{2}$ .

$(\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} x + y + z \\ 0 \end{matrix})$

Linear independence is preserved for this set

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

and (in a somewhat more tricky example) also for this set

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

(recall that in a set, repeated elements do not appear twice). However, there are sets whose independence is not preserved under this map

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

and so not all sets have independence preserved.
Exercise 2.38 Worked answer

Corollary 2.17 says that for there to be an onto homomorphism from a vector space $V$ to a vector space $W$ , it is necessary that the dimension of $W$ be less than or equal to the dimension of $V$ . Prove that this condition is also sufficient; use Theorem 1.9 to show that if the dimension of $W$ is less than or equal to the dimension of $V$ , then there is a homomorphism from $V$ to $W$ that is onto.

Back to Exercise 2.38

Answer. (We use the notation from Theorem 1.9.) Fix a basis $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ for $V$ and a basis $⟨ {\vec{w}}_{1}, \dots, {\vec{w}}_{k} ⟩$ for $W$ . If the dimension $k$ of $W$ is less than or equal to the dimension $n$ of $V$ then the theorem gives a linear map from $V$ to $W$ determined in this way.

${\vec{β}}_{1} \mapsto {\vec{w}}_{1}, \dots, {\vec{β}}_{k} \mapsto {\vec{w}}_{k} and {\vec{β}}_{k + 1} \mapsto {\vec{w}}_{k}, \dots, {\vec{β}}_{n} \mapsto {\vec{w}}_{k}$

We need only to verify that this map is onto.

We can write any member of $W$ as a linear combination of basis elements $c_{1} \cdot {\vec{w}}_{1} + \dots + c_{k} \cdot {\vec{w}}_{k}$ . This vector is the image, under the map described above, of $c_{1} \cdot {\vec{β}}_{1} + \dots + c_{k} \cdot {\vec{β}}_{k} + 0 \cdot {\vec{β}}_{k + 1} \dots + 0 \cdot {\vec{β}}_{n}$ . Thus the map is onto.
Exercise 2.39 Worked answer

Recommended. Recall that the null space is a subset of the domain and the range space is a subset of the codomain. Are they necessarily distinct? Is there a homomorphism that has a nontrivial intersection of its null space and its range space?

Back to Exercise 2.39

Answer. Yes. For the transformation of $ℝ^{2}$ given by

$(\begin{matrix} x \\ y \end{matrix}) \overset{h}{⟼} (\begin{matrix} 0 \\ x \end{matrix})$

we have this.

$𝒩 (h) = {(\begin{matrix} 0 \\ y \end{matrix}) ∣ y \in ℝ} = ℛ (h)$

Remark. We will see more of this in the fifth chapter.
Exercise 2.40 Worked answer

Prove that the image of a span equals the span of the images. That is, where $h : V \to W$ is linear, prove that if $S$ is a subset of $V$ then $h ([S])$ equals $[h (S)]$ . This generalizes Lemma 2.1 since it shows that if $U$ is any subspace of $V$ then its image ${h (\vec{u}) ∣ \vec{u} \in U}$ is a subspace of $W$ , because the span of the set $U$ is $U$ .

Back to Exercise 2.40

Answer. This is a simple calculation.

$\begin{aligned} h ([S]) & = {h (c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n}) ∣ c_{1}, \dots, c_{n} \in ℝ and {\vec{s}}_{1}, \dots, {\vec{s}}_{n} \in S} \\ = {c_{1} h ({\vec{s}}_{1}) + \dots + c_{n} h ({\vec{s}}_{n}) ∣ c_{1}, \dots, c_{n} \in ℝ and {\vec{s}}_{1}, \dots, {\vec{s}}_{n} \in S} \\ = [h (S)] \end{aligned}$
Exercise 2.41 Worked answer
1. Prove that for any linear map $h : V \to W$ and any $\vec{w} \in W$ , the set $h^{- 1} (\vec{w})$ has the form
  
  $h^{- 1} (\vec{w}) = {\vec{v} + \vec{n} ∣ \vec{v}, \vec{n} \in V and \vec{n} \in 𝒩 (h) and h (\vec{v}) = \vec{w}}$
  
  (if $h$ is not onto and $\vec{w}$ is not in the range of $h$ then this set is empty since its third condition cannot be satisfied). Such a set is a coset of $𝒩 (h)$ and we denote it as $\vec{v} + 𝒩 (h)$ .
2. Consider the map $t : ℝ^{2} \to ℝ^{2}$ given by
  
  $(\begin{matrix} x \\ y \end{matrix}) \overset{t}{⟼} (\begin{matrix} a x + b y \\ c x + d y \end{matrix})$
  
  for some scalars $a$ , $b$ , $c$ , and $d$ . Prove that $t$ is linear.
3. Conclude from the prior two items that for any linear system of the form
  
  $\begin{array}{rcrcr} a x & + & b y & = & e \\ c x & + & d y & = & f \end{array}$
  
  we can write the solution set (the vectors are members of $ℝ^{2}$ )
  
  ${\vec{p} + \vec{h} ∣ \vec{h} satisfies the associated homogeneous system}$
  
  where $\vec{p}$ is a particular solution of that linear system (if there is no particular solution then the above set is empty).
4. Show that this map $h : ℝ^{n} \to ℝ^{m}$ is linear
  
  $(\begin{matrix} x_{1} \\ ⋮ \\ x_{n} \end{matrix}) \mapsto (\begin{matrix} a_{1, 1} x_{1} + \dots + a_{1, n} x_{n} \\ ⋮ \\ a_{m, 1} x_{1} + \dots + a_{m, n} x_{n} \end{matrix})$
  
  for any scalars $a_{1, 1}$ , …, $a_{m, n}$ . Extend the conclusion made in the prior item.
5. Show that the $k$ -th derivative map is a linear transformation of $𝒫_{n}$ for each $k$ . Prove that this map is a linear transformation of the space
  
  $f \mapsto \frac{d^{k}}{d x^{k}} f + c_{k - 1} \frac{d^{k - 1}}{d x^{k - 1}} f + \dots + c_{1} \frac{d}{d x} f + c_{0} f$
  
  for any scalars $c_{k}$ , …, $c_{0}$ . Draw a conclusion as above.
Back to Exercise 2.41

Answer.
1. We will show the sets are equal $h^{- 1} (\vec{w}) = {\vec{v} + \vec{n} ∣ \vec{n} \in 𝒩 (h)}$ by mutual inclusion. For the ${\vec{v} + \vec{n} ∣ \vec{n} \in 𝒩 (h)} \subseteq h^{- 1} (\vec{w})$ direction, just note that $h (\vec{v} + \vec{n}) = h (\vec{v}) + h (\vec{n})$ equals $\vec{w}$ , and so any member of the first set is a member of the second. For the $h^{- 1} (\vec{w}) \subseteq {\vec{v} + \vec{n} ∣ \vec{n} \in 𝒩 (h)}$ direction, consider $\vec{u} \in h^{- 1} (\vec{w})$ . Because $h$ is linear, $h (\vec{u}) = h (\vec{v})$ implies that $h (\vec{u} - \vec{v}) = \vec{0}$ . We can write $\vec{u} - \vec{v}$ as $\vec{n}$ , and then we have that $\vec{u} \in {\vec{v} + \vec{n} ∣ \vec{n} \in 𝒩 (h)}$ , as desired, because $\vec{u} = \vec{v} + (\vec{u} - \vec{v})$ .
2. This check is routine.
3. This is immediate.
4. For the linearity check, briefly, where $c, d$ are scalars and $\vec{x}, \vec{y} \in ℝ^{n}$ have components $x_{1}, \dots, x_{n}$ and $y_{1}, \dots, y_{n}$ , we have this.
  
  $\begin{aligned} h (c \cdot \vec{x} + d \cdot \vec{y}) & = (\begin{array}{c} a_{1, 1} (c x_{1} + d y_{1}) + \dots + a_{1, n} (c x_{n} + d y_{n}) \\ ⋮ \\ a_{m, 1} (c x_{1} + d y_{1}) + \dots + a_{m, n} (c x_{n} + d y_{n}) \end{array}) \\ = (\begin{array}{c} a_{1, 1} c x_{1} + \dots + a_{1, n} c x_{n} \\ ⋮ \\ a_{m, 1} c x_{1} + \dots + a_{m, n} c x_{n} \end{array}) + (\begin{array}{c} a_{1, 1} d y_{1} + \dots + a_{1, n} d y_{n} \\ ⋮ \\ a_{m, 1} d y_{1} + \dots + a_{m, n} d y_{n} \end{array}) \\ = c \cdot h (\vec{x}) + d \cdot h (\vec{y}) \end{aligned}$
  
  The appropriate conclusion is that $General = Particular + Homogeneous$ .
5. Each power of the derivative is linear because of the rules
  
  $\frac{d^{k}}{d x^{k}} (f (x) + g (x)) = \frac{d^{k}}{d x^{k}} f (x) + \frac{d^{k}}{d x^{k}} g (x) and \frac{d^{k}}{d x^{k}} r f (x) = r \frac{d^{k}}{d x^{k}} f (x)$
  
  from calculus. Thus the given map is a linear transformation of the space because any linear combination of linear maps is also a linear map by Lemma 1.17. The appropriate conclusion is $General = Particular + Homogeneous$ , where the associated homogeneous differential equation has a constant of $0$ .
Exercise 2.42 Worked answer

Prove that for any transformation $t : V \to V$ that is rank one, the map given by composing the operator with itself $t \circ t : V \to V$ satisfies $t \circ t = r \cdot t$ for some real number $r$ .

Back to Exercise 2.42

Answer. Because the rank of $t$ is one, the range space of $t$ is a one-dimensional set. Taking $⟨ h (\vec{v}) ⟩$ as a basis (for some appropriate $\vec{v}$ ), we have that for every $\vec{w} \in V$ , the image $h (\vec{w}) \in V$ is a multiple of this basis vector—associated with each $\vec{w}$ there is a scalar $c_{\vec{w}}$ such that $t (\vec{w}) = c_{\vec{w}} t (\vec{v})$ . Apply $t$ to both sides of that equation and take $r$ to be $c_{t (\vec{v})}$

$t \circ t (\vec{w}) = t (c_{\vec{w}} \cdot t (\vec{v})) = c_{\vec{w}} \cdot t \circ t (\vec{v}) = c_{\vec{w}} \cdot c_{t (\vec{v})} \cdot t (\vec{v}) = c_{\vec{w}} \cdot r \cdot t (\vec{v}) = r \cdot c_{\vec{w}} \cdot t (\vec{v}) = r \cdot t (\vec{w})$

to get the desired conclusion.
Exercise 2.43 Worked answer

Let $h : V \to ℝ$ be a homomorphism, but not the zero homomorphism. Prove that if $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ is a basis for the null space and if $\vec{v} \in V$ is not in the null space then $⟨ \vec{v}, {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ is a basis for the entire domain $V$ .

Back to Exercise 2.43

Answer. By assumption, $h$ is not the zero map and so a vector $\vec{v} \in V$ exists that is not in the null space. Note that $⟨ h (\vec{v}) ⟩$ is a basis for $ℝ$ , because it is a size-one linearly independent subset of $ℝ$ . Consequently $h$ is onto, as for any $r \in ℝ$ we have $r = c \cdot h (\vec{v})$ for some scalar $c$ , and so $r = h (c \vec{v})$ .

Thus the rank of $h$ is one. Because the nullity is $n$ , the dimension of the domain of $h$ , the vector space $V$ , is $n + 1$ . We can finish by showing ${\vec{v}, {\vec{β}}_{1}, \dots, {\vec{β}}_{n}}$ is linearly independent, as it is a size $n + 1$ subset of a dimension $n + 1$ space. Because ${{\vec{β}}_{1}, \dots, {\vec{β}}_{n}}$ is linearly independent we need only show that $\vec{v}$ is not a linear combination of the other vectors. But $c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n} = \vec{v}$ would give $- \vec{v} + c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n} = \vec{0}$ and applying $h$ to both sides would give a contradiction.
Exercise 2.44 Worked answer

Show that for any space $V$ of dimension $n$ , the dual space

$ℒ (V, ℝ) = {h : V \to ℝ ∣ h is linear}$

is isomorphic to $ℝ^{n}$ . It is often denoted $V^{*}$ . Conclude that $V^{*} ≅ V$ .

Back to Exercise 2.44

Answer. Fix a basis $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ for $V$ . We shall prove that this map

$h \overset{Φ}{⟼} (\begin{matrix} h ({\vec{β}}_{1}) \\ ⋮ \\ h ({\vec{β}}_{n}) \end{matrix})$

is an isomorphism from $V^{*}$ to $ℝ^{n}$ .

To see that $Φ$ is one-to-one, assume that $h_{1}$ and $h_{2}$ are members of $V^{*}$ such that $Φ (h_{1}) = Φ (h_{2})$ . Then

$(\begin{matrix} h_{1} ({\vec{β}}_{1}) \\ ⋮ \\ h_{1} ({\vec{β}}_{n}) \end{matrix}) = (\begin{matrix} h_{2} ({\vec{β}}_{1}) \\ ⋮ \\ h_{2} ({\vec{β}}_{n}) \end{matrix})$

and consequently, $h_{1} ({\vec{β}}_{1}) = h_{2} ({\vec{β}}_{1})$ , etc. But a homomorphism is determined by its action on a basis, so $h_{1} = h_{2}$ , and therefore $Φ$ is one-to-one.

To see that $Φ$ is onto, consider

$(\begin{matrix} x_{1} \\ ⋮ \\ x_{n} \end{matrix})$

for $x_{1}, \dots, x_{n} \in ℝ$ . This function $h$ from $V$ to $ℝ$

$c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n} \overset{h}{⟼} c_{1} x_{1} + \dots + c_{n} x_{n}$

is linear and $Φ$ maps it to the given vector in $ℝ^{n}$ , so $Φ$ is onto.

The map $Φ$ also preserves structure: where

$\begin{aligned} c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n} & \overset{h_{1}}{⟼} c_{1} h_{1} ({\vec{β}}_{1}) + \dots + c_{n} h_{1} ({\vec{β}}_{n}) \\ c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n} & \overset{h_{2}}{⟼} c_{1} h_{2} ({\vec{β}}_{1}) + \dots + c_{n} h_{2} ({\vec{β}}_{n}) \end{aligned}$

we have

$\begin{matrix} (r_{1} h_{1} + r_{2} h_{2}) (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}) \\ \begin{aligned} = c_{1} (r_{1} h_{1} ({\vec{β}}_{1}) + r_{2} h_{2} ({\vec{β}}_{1})) + \dots + c_{n} (r_{1} h_{1} ({\vec{β}}_{n}) + r_{2} h_{2} ({\vec{β}}_{n})) \\ = r_{1} (c_{1} h_{1} ({\vec{β}}_{1}) + \dots + c_{n} h_{1} ({\vec{β}}_{n})) + r_{2} (c_{1} h_{2} ({\vec{β}}_{1}) + \dots + c_{n} h_{2} ({\vec{β}}_{n})) \end{aligned} \end{matrix}$

so $Φ (r_{1} h_{1} + r_{2} h_{2}) = r_{1} Φ (h_{1}) + r_{2} Φ (h_{2})$ .
Exercise 2.45 Worked answer

Show that any linear map is the sum of maps of rank one.

Back to Exercise 2.45

Answer. Let $h : V \to W$ be linear and fix a basis $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ for $V$ . Consider these $n$ maps from $V$ to $W$

$h_{1} (\vec{v}) = c_{1} \cdot h ({\vec{β}}_{1}), h_{2} (\vec{v}) = c_{2} \cdot h ({\vec{β}}_{2}), \dots, h_{n} (\vec{v}) = c_{n} \cdot h ({\vec{β}}_{n})$

for any $\vec{v} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ . Clearly $h$ is the sum of the $h_{i}$ ’s. We need only check that each $h_{i}$ is linear: where $\vec{u} = d_{1} {\vec{β}}_{1} + \dots + d_{n} {\vec{β}}_{n}$ we have $h_{i} (r \vec{v} + s \vec{u}) = r c_{i} + s d_{i} = r h_{i} (\vec{v}) + s h_{i} (\vec{u})$ .
Exercise 2.46 Worked answer

Is ‘is homomorphic to’ an equivalence relation? (Hint: the difficulty is to decide on an appropriate meaning for the quoted phrase.)

Back to Exercise 2.46

Answer. Either yes (trivially) or no (nearly trivially).

If we take $V$ ‘is homomorphic to’ $W$ to mean there is a homomorphism from $V$ into (but not necessarily onto) $W$ , then every space is homomorphic to every other space as a zero map always exists.

If we take $V$ ‘is homomorphic to’ $W$ to mean there is an onto homomorphism from $V$ to $W$ then the relation is not an equivalence. For instance, there is an onto homomorphism from $ℝ^{3}$ to $ℝ^{2}$ (projection is one) but no homomorphism from $ℝ^{2}$ onto $ℝ^{3}$ by Corollary 2.17, so the relation is not reflexive.²
Exercise 2.47 Worked answer

Show that the range spaces and null spaces of powers of linear maps $t : V \to V$ form descending

$V \supseteq ℛ (t) \supseteq ℛ (t^{2}) \supseteq \dots$

and ascending

${\vec{0}} \subseteq 𝒩 (t) \subseteq 𝒩 (t^{2}) \subseteq \dots$

chains. Also show that if $k$ is such that $ℛ (t^{k}) = ℛ (t^{k + 1})$ then all following range spaces are equal: $ℛ (t^{k}) = ℛ (t^{k + 1}) = ℛ (t^{k + 2}) \dots$ . Similarly, if $𝒩 (t^{k}) = 𝒩 (t^{k + 1})$ then $𝒩 (t^{k}) = 𝒩 (t^{k + 1}) = 𝒩 (t^{k + 2}) = \dots$ .

Back to Exercise 2.47

Answer. That they form the chains is obvious. For the rest, we show here that $ℛ (t^{j + 1}) = ℛ (t^{j})$ implies that $ℛ (t^{j + 2}) = ℛ (t^{j + 1})$ . Induction then applies.

Assume that $ℛ (t^{j + 1}) = ℛ (t^{j})$ . Then $t : ℛ (t^{j + 1}) \to ℛ (t^{j + 2})$ is the same map, with the same domain, as $t : ℛ (t^{j}) \to ℛ (t^{j + 1})$ . Thus it has the same range: $ℛ (t^{j + 2}) = ℛ (t^{j + 1})$ .

References cited in this section

More information on many-to-one maps is in the appendix.↩︎
More information on equivalence relations is in the appendix.↩︎