4.6 Products and quotients

A vector space can be built out of others in two basic ways. A product places several spaces side by side: a vector of $𝑉 \times 𝑊$ is a pair, one vector from each. A quotient goes the other way. It takes a subspace $𝑈$ of $𝑉$ and collapses it to a single point, so that two vectors of $𝑉$ become equal when they differ by an element of $𝑈$ . Recall from Corollary 2.6.31 that a linear map $𝑇$ slices its domain into parallel copies of $ker 𝑇$ and sends each slice to one vector of $im 𝑇$ . The quotient $𝑉 / ker 𝑇$ turns the set of slices into a vector space on which $𝑇$ becomes injective, and this is the first isomorphism theorem (Theorem 4.6.34). It gives a third proof of rank–nullity. Products, for their part, give the cleanest way to say when a sum of subspaces is direct.

Neither construction appears in the 21-242 lectures (they are in LADR, Section 3E). Quotients come back in Section 7.5, where they give a second route to triangularization, and in Section 13.1.

Throughout, $𝑉$ and $𝑊$ are vector spaces over $𝐹$ .

4.6.1Product spaces

Definition 4.6.1 (Product space). Let $𝑉_{1}, \dots, 𝑉_{𝑚}$ be vector spaces over $𝐹$ . Their product is the set of $𝑚$ -tuples

𝑉_{1} \times \dots \times 𝑉_{𝑚} = {(𝐯_{1}, \dots, 𝐯_{𝑚}) | 𝐯_{𝑗} \in 𝑉_{𝑗} for 𝑗 = 1, \dots, 𝑚},

with the componentwise operations

\begin{aligned} (𝐯_{1}, \dots, 𝐯_{𝑚}) + (𝐰_{1}, \dots, 𝐰_{𝑚}) & = (𝐯_{1} + 𝐰_{1}, \dots, 𝐯_{𝑚} + 𝐰_{𝑚}), \\ 𝑎 (𝐯_{1}, \dots, 𝐯_{𝑚}) & = (𝑎 𝐯_{1}, \dots, 𝑎 𝐯_{𝑚}) . \end{aligned}

For a single space $𝑉$ we write $𝑉^{𝑚} = 𝑉 \times \dots \times 𝑉$ ( $𝑚$ factors).

Proposition 4.6.2 (Products are vector spaces).

𝑉_{1} \times \dots \times 𝑉_{𝑚}

is a vector space over

𝐹

. Its zero vector is

(𝟎, \dots, 𝟎)

, and

- (𝐯_{1}, \dots, 𝐯_{𝑚}) =

(- 𝐯_{1}, \dots, - 𝐯_{𝑚})

Proof.routine

Uses. Definition 1.2.1 (Vector space), Definition 4.6.1 (Product space)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Two tuples are equal exactly when they are equal in every component.

Hint 2. Write out the

𝑗

th component of both sides of each axiom. It is the same axiom in

𝑉_{𝑗}

Sketch. Every axiom is an identity between tuples, and its

𝑗

th component is that axiom in

𝑉_{𝑗}

. The zero and the negatives are found the same way.

Sums and scalar multiples of tuples are tuples of the right kind, since each

𝑉_{𝑗}

is closed under its operations. Each vector space axiom is an identity between elements of the product, and two tuples are equal when they agree in every component. In the

𝑗

th component, each axiom becomes the same axiom in

𝑉_{𝑗}

. For example, for

𝐱 = (𝐯_{1}, \dots, 𝐯_{𝑚})

and

𝐲 = (𝐰_{1}, \dots, 𝐰_{𝑚})

, the

𝑗

th components of

𝑎 (𝐱 + 𝐲)

and

𝑎 𝐱 + 𝑎 𝐲

are

𝑎 (𝐯_{𝑗} + 𝐰_{𝑗})

and

𝑎 𝐯_{𝑗} + 𝑎 𝐰_{𝑗}

, which are equal by distributivity in

𝑉_{𝑗}

. The other identities (commutativity, associativity, the second distributive law,

(𝑎 𝑏) 𝐱 = 𝑎 (𝑏 𝐱)

, and

1 𝐱 = 𝐱

) are checked identically. Finally,

(𝐯_{1}, \dots, 𝐯_{𝑚}) +

(𝟎, \dots, 𝟎) =

(𝐯_{1}, \dots, 𝐯_{𝑚})

and

(𝐯_{1}, \dots, 𝐯_{𝑚}) +

(- 𝐯_{1}, \dots, - 𝐯_{𝑚}) =

(𝟎, \dots, 𝟎)

, componentwise. □

Example 4.6.3 (Products).

(a) $𝐹^{𝑚}$ is $𝐹 \times \dots \times 𝐹$ ( $𝑚$ factors); Definition 1.2.4 is the case $𝑉_{1} = \dots = 𝑉_{𝑚} = 𝐹$ , with the tuples written as columns.
(b)An element of $𝐹^{2} \times 𝐹^{3}$ is a pair $((𝑎, 𝑏), (𝑐, 𝑑, 𝑒))$ , which is not literally an element of $𝐹^{5}$ . But removing the inner parentheses, $((𝑎, 𝑏), (𝑐, 𝑑, 𝑒)) \mapsto$ $(𝑎, 𝑏, 𝑐, 𝑑, 𝑒)$ , is a bijective linear map, so $𝐹^{2} \times 𝐹^{3} ≅ 𝐹^{5}$ (Definition 4.5.1). In the same way $𝐹^{𝑛} \times 𝐹^{𝑝} ≅ 𝐹^{𝑛 + 𝑝}$ , and we identify the two whenever convenient.
(c)In $𝑉 \times 𝑊$ , the subsets $𝑉 \times {𝟎}$ and ${𝟎} \times 𝑊$ are subspaces, and $𝑉 \times$ $𝑊 =$ $(𝑉 \times {𝟎}) \oplus$ $({𝟎} \times 𝑊)$ : every $(𝐯, 𝐰)$ is $(𝐯, 𝟎) + (𝟎, 𝐰)$ , and this is the only way to write it as such a sum, since the components are forced.

Non-example 4.6.4 (Scaling only one component). Let

𝑊 \neq {𝟎}

. On the set

𝑉 \times 𝑊

keep componentwise addition, but scale only the first component:

𝑎 ⊙ (𝐯, 𝐰) =

(𝑎 𝐯, 𝐰)

. This is not a vector space. In any vector space

0 𝐱 = 𝟎

(Proposition 1.2.32), but here

0 ⊙ (𝐯, 𝐰) = (𝟎, 𝐰)

, which is not the zero vector

(𝟎, 𝟎)

when

𝐰 \neq 𝟎

. (The axiom that fails is distributivity over scalar addition:

(1 + 1) ⊙ (𝐯, 𝐰)

has second component

𝐰

, while

1 ⊙ (𝐯, 𝐰) +

1 ⊙ (𝐯, 𝐰)

has second component

𝐰 + 𝐰

For each $𝑗$ there are two linear maps connecting $𝑉_{𝑗}$ with the product. The inclusion $𝜄_{𝑗} : 𝑉_{𝑗} \to$ $𝑉_{1} \times$ $\dots \times$ $𝑉_{𝑚}$ puts a vector in the $𝑗$ th slot and $𝟎$ in the others, and the projection ${Pr}_{𝑗} : 𝑉_{1} \times$ $\dots \times$ $𝑉_{𝑚} \to$ $𝑉_{𝑗}$ reads off the $𝑗$ th component. Both are linear because the operations are componentwise, ${Pr}_{𝑗} 𝜄_{𝑗} = {id}_{𝑉_{𝑗}}$ , and every vector of the product splits as

(𝐯_{1}, \dots, 𝐯_{𝑚}) = 𝜄_{1} (𝐯_{1}) + \dots + 𝜄_{𝑚} (𝐯_{𝑚}) .

(4.6.1)

Theorem 4.6.5 (Dimension of a product). Suppose $𝑉_{1}, \dots, 𝑉_{𝑚}$ are finite-dimensional. Then $𝑉_{1} \times \dots \times 𝑉_{𝑚}$ is finite-dimensional, and

dim (𝑉_{1} \times \dots \times 𝑉_{𝑚}) = dim 𝑉_{1} + \dots + dim 𝑉_{𝑚} .

More precisely, if ${ℬ︀}_{𝑗}$ is a basis of $𝑉_{𝑗}$ for each $𝑗$ , then the list obtained by concatenating $𝜄_{1} ({ℬ︀}_{1}),$ $\dots,$ $𝜄_{𝑚} ({ℬ︀}_{𝑚})$ is a basis of the product.

The first proof checks the basis directly. The second never checks a basis of the product: it peels off one factor with a surjection and counts the kernel by rank–nullity, which is how quotients will be measured below.

Proof (by building a basis).routine

Uses. Definition 1.6.1 (Basis), Proposition 1.6.2 (Bases are independent spanning lists), (4.6.1), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. Take a basis

𝐛_{𝑗, 1},

\dots,

𝐛_{𝑗, 𝑛_{𝑗}}

of each

𝑉_{𝑗}

. Which vectors of the product would you guess form a basis?

Hint 2. For spanning, use (4.6.1) and expand each

𝐯_{𝑗}

in its basis. For independence, look at the

𝑗

th component of a linear relation.

Sketch. By (4.6.1) and the linearity of

𝜄_{𝑗}

, each tuple is a combination of the

𝜄_{𝑗} (𝐛_{𝑗, 𝑖})

. The

𝑗

th component of a relation

\sum_{𝑗, 𝑖} 𝑐_{𝑗, 𝑖} 𝜄_{𝑗} (𝐛_{𝑗, 𝑖}) =

𝟎

\sum_{𝑖} 𝑐_{𝑗, 𝑖} 𝐛_{𝑗, 𝑖} =

𝟎

, so all

𝑐_{𝑗, 𝑖} = 0

Let ${ℬ︀}_{𝑗} =$ $𝐛_{𝑗, 1},$ $\dots,$ $𝐛_{𝑗, 𝑛_{𝑗}}$ be a basis of $𝑉_{𝑗}$ , so $𝑛_{𝑗} = dim 𝑉_{𝑗}$ , and let $ℒ︀$ be the concatenation of the lists $𝜄_{𝑗} ({ℬ︀}_{𝑗})$ . It has $𝑛_{1} + \dots + 𝑛_{𝑚}$ entries, so it suffices to show that $ℒ︀$ is a basis.

Spanning. Let $(𝐯_{1}, \dots, 𝐯_{𝑚})$ be in the product, and write $𝐯_{𝑗} =$ $\sum_{𝑖} 𝑐_{𝑗, 𝑖} 𝐛_{𝑗, 𝑖}$ . By (4.6.1) and the linearity of each $𝜄_{𝑗}$ ,

(𝐯_{1}, \dots, 𝐯_{𝑚}) = \sum_{𝑗 = 1}^{𝑚} 𝜄_{𝑗} (𝐯_{𝑗}) = \sum_{𝑗 = 1}^{𝑚} \sum_{𝑖 = 1}^{𝑛_{𝑗}} 𝑐_{𝑗, 𝑖} 𝜄_{𝑗} (𝐛_{𝑗, 𝑖}) .

Independence. Suppose $\sum_{𝑗, 𝑖} 𝑐_{𝑗, 𝑖} 𝜄_{𝑗} (𝐛_{𝑗, 𝑖}) =$ $𝟎$ . The $𝑗$ th component of $𝜄_{𝑗} (𝐛_{𝑗, 𝑖})$ is $𝐛_{𝑗, 𝑖}$ , and the $𝑗$ th component of $𝜄_{𝑘} (𝐛_{𝑘, 𝑖})$ is $𝟎$ for $𝑘 \neq 𝑗$ . So the $𝑗$ th component of the relation reads $\sum_{𝑖} 𝑐_{𝑗, 𝑖} 𝐛_{𝑗, 𝑖} =$ $𝟎$ , and since ${ℬ︀}_{𝑗}$ is independent, $𝑐_{𝑗, 1} =$ $\dots =$ $𝑐_{𝑗, 𝑛_{𝑗}} =$ $0$ . This holds for every $𝑗$ . □

Proof (by induction on $𝑚$ , with rank–nullity).standard

Uses. Theorem 4.2.11 (Rank–nullity), Proposition 4.2.6 (Injective maps preserve dimension), (4.6.1), Definition 1.6.23 (Finite-dimensional)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. For

𝑚 \geq 2

, the projection

{Pr}_{𝑚}

onto the last factor is surjective. What is its kernel?

Hint 2. The kernel of

{Pr}_{𝑚}

consists of the tuples

(𝐯_{1}, \dots, 𝐯_{𝑚 - 1}, 𝟎)

, a copy of the product of the first

𝑚 - 1

factors. Apply rank–nullity, after checking that the product is finite-dimensional at all.

Sketch. The vectors

𝜄_{𝑗} (𝐛)

𝐛

in a basis of

𝑉_{𝑗}

, span the product by (4.6.1), so it is finite-dimensional.

{Pr}_{𝑚}

is surjective, and

(𝐯_{1}, \dots, 𝐯_{𝑚 - 1}) \mapsto

(𝐯_{1}, \dots, 𝐯_{𝑚 - 1}, 𝟎)

is injective with image

ker {Pr}_{𝑚}

. Rank–nullity gives

dim =

dim (𝑉_{1} \times \dots \times 𝑉_{𝑚 - 1}) +

dim 𝑉_{𝑚}

, and induction finishes.

We prove the dimension formula by induction on $𝑚$ ; the case $𝑚 = 1$ is trivial. Let $𝑚 \geq 2$ and $𝑃 = 𝑉_{1} \times \dots \times 𝑉_{𝑚}$ . By (4.6.1), the vectors $𝜄_{𝑗} (𝐛)$ , with $𝐛$ running through a basis of $𝑉_{𝑗}$ and $𝑗 = 1, \dots, 𝑚$ , span $𝑃$ (expand each $𝐯_{𝑗}$ in its basis), so $𝑃$ is finite-dimensional. The projection ${Pr}_{𝑚} : 𝑃 \to 𝑉_{𝑚}$ is linear and surjective, since ${Pr}_{𝑚} 𝜄_{𝑚} (𝐯) = 𝐯$ . Its kernel is the set of tuples whose last component is $𝟎$ , which is the image of

𝐽 : 𝑉_{1} \times \dots \times 𝑉_{𝑚 - 1} \to 𝑃, (𝐯_{1}, \dots, 𝐯_{𝑚 - 1}) \mapsto (𝐯_{1}, \dots, 𝐯_{𝑚 - 1}, 𝟎) .

$𝐽$ is linear and injective, so by Proposition 4.2.6(a) and the induction hypothesis, $dim ker {Pr}_{𝑚} =$ $dim (𝑉_{1} \times \dots \times 𝑉_{𝑚 - 1}) =$ $dim 𝑉_{1} +$ $\dots +$ $dim 𝑉_{𝑚 - 1}$ . By rank–nullity (Theorem 4.2.11),

dim 𝑃 = dim ker {Pr}_{𝑚} + dim im {Pr}_{𝑚} = (dim 𝑉_{1} + \dots + dim 𝑉_{𝑚 - 1}) + dim 𝑉_{𝑚},

as desired. (This proof gives the dimension only; the explicit basis is the first proof’s.) □

Exercise 4.6.6.routine

(a)Show that if $𝑉^{'}$ is a subspace of $𝑉$ and $𝑊^{'}$ is a subspace of $𝑊$ , then $𝑉^{'} \times 𝑊^{'}$ is a subspace of $𝑉 \times 𝑊$ .
(b)Prove or disprove: every subspace of $𝑉 \times 𝑊$ is of the form $𝑉^{'} \times 𝑊^{'}$ as in (a).

Solution.

Uses. Theorem 1.4.3 (Subspace criterion), Definition 4.6.1 (Product space)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. For (a), check the subspace criterion one component at a time.

Hint 2. For (b), try the smallest interesting case,

𝑉 = 𝑊 = 𝐹

. Is there a line in

𝐹 \times 𝐹 = 𝐹^{2}

that is not a product?

Sketch. (a) Componentwise. (b) False: the diagonal

{(𝑡, 𝑡) | 𝑡 \in 𝐹}

is a subspace of

𝐹 \times 𝐹

. A product

𝑉^{'} \times 𝑊^{'}

containing

(1, 1)

has

1 \in 𝑉^{'}

and

0 \in 𝑊^{'}

, so it contains

(1, 0)

, which is not on the diagonal.

(a) $(𝟎, 𝟎) \in 𝑉^{'} \times 𝑊^{'}$ . If $(𝐯, 𝐰),$ $(𝐯^{'}, 𝐰^{'}) \in$ $𝑉^{'} \times$ $𝑊^{'}$ and $𝑎 \in 𝐹$ , then $𝑎 (𝐯, 𝐰) +$ $(𝐯^{'}, 𝐰^{'}) =$ $(𝑎 𝐯 + 𝐯^{'}, 𝑎 𝐰 + 𝐰^{'})$ , whose components lie in $𝑉^{'}$ and $𝑊^{'}$ because these are subspaces. By Theorem 1.4.3, $𝑉^{'} \times 𝑊^{'}$ is a subspace.

(b) The claim is false. Take $𝑉 = 𝑊 = 𝐹$ and the diagonal $𝐷 =$ ${(𝑡, 𝑡) | 𝑡 \in 𝐹}$ , which is a subspace of $𝐹 \times 𝐹$ (it is $span ((1, 1))$ ). Suppose $𝐷 = 𝑉^{'} \times 𝑊^{'}$ . Since $(1, 1) \in 𝐷$ , we get $1 \in 𝑉^{'}$ and $0 \in 𝑊^{'}$ (as $𝑊^{'}$ is a subspace), so $(1, 0) \in 𝑉^{'} \times 𝑊^{'} = 𝐷$ . But $(1, 0)$ is not of the form $(𝑡, 𝑡)$ .

Maps into and out of a product

A map into a product is the same as a list of maps into the factors, and a map out of a product is the same as a list of maps out of the factors.

Proposition 4.6.7 (Maps into and out of a product).

(a)Let $𝑇_{𝑗} \in L (𝑈, 𝑊_{𝑗})$ for $𝑗 = 1, \dots, 𝑚$ . Then
$(𝑇_{1}, \dots, 𝑇_{𝑚}) : 𝑈 \to 𝑊_{1} \times \dots \times 𝑊_{𝑚}, 𝐮 \mapsto (𝑇_{1} (𝐮), \dots, 𝑇_{𝑚} (𝐮)),$
is linear, and every $𝑇 \in$ $L (𝑈, 𝑊_{1} \times \dots \times 𝑊_{𝑚})$ equals $(𝑇_{1}, \dots, 𝑇_{𝑚})$ for exactly one list of linear maps, namely $𝑇_{𝑗} = {Pr}_{𝑗} 𝑇$ .
(b)Let $𝑆_{𝑗} \in L (𝑉_{𝑗}, 𝑊)$ for $𝑗 = 1, \dots, 𝑚$ . Then
$𝑉_{1} \times \dots \times 𝑉_{𝑚} \to 𝑊, (𝐯_{1}, \dots, 𝐯_{𝑚}) \mapsto 𝑆_{1} (𝐯_{1}) + \dots + 𝑆_{𝑚} (𝐯_{𝑚}),$
is linear, and every $𝑆 \in$ $L (𝑉_{1} \times \dots \times 𝑉_{𝑚}, 𝑊)$ is of this form for exactly one list of linear maps, namely $𝑆_{𝑗} = 𝑆 𝜄_{𝑗}$ .

Proof.routine

Uses. Theorem 2.1.3 (Checking linearity), Proposition 2.1.17 (Compositions of linear maps are linear), (4.6.1)

Techniques. Assemble linear maps from linear pieces (Technique 2.1.26), Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Linearity in both parts is a componentwise computation.

Hint 2. For the converse in (a), compare components of

𝑇 (𝐮)

. In (b), apply

𝑆

to (4.6.1).

Sketch. (a) The

𝑗

th component of

(𝑇_{1}, \dots, 𝑇_{𝑚}) (𝑎 𝐮 + 𝐮^{'})

𝑎 𝑇_{𝑗} (𝐮) +

𝑇_{𝑗} (𝐮^{'})

. A map

𝑇

into the product is determined by the components

{Pr}_{𝑗} 𝑇 (𝐮)

of its values. (b) Linearity is a sum of linear computations, and applying

𝑆

to (4.6.1) gives

𝑆 (𝐯_{1}, \dots, 𝐯_{𝑚}) =

\sum_{𝑗} 𝑆 𝜄_{𝑗} (𝐯_{𝑗})

. Uniqueness: evaluate at

𝜄_{𝑗} (𝐯)

(a) For $𝐮, 𝐮^{'} \in 𝑈$ and $𝑎 \in 𝐹$ , the $𝑗$ th component of $(𝑇_{1}, \dots, 𝑇_{𝑚}) (𝑎 𝐮 + 𝐮^{'})$ is $𝑇_{𝑗} (𝑎 𝐮 + 𝐮^{'}) =$ $𝑎 𝑇_{𝑗} (𝐮) +$ $𝑇_{𝑗} (𝐮^{'})$ , which is the $𝑗$ th component of $𝑎 (𝑇_{1}, \dots, 𝑇_{𝑚}) (𝐮) +$ $(𝑇_{1}, \dots, 𝑇_{𝑚}) (𝐮^{'})$ . So the map is linear (Theorem 2.1.3). Now let $𝑇 \in$ $L (𝑈, 𝑊_{1} \times \dots \times 𝑊_{𝑚})$ . Each ${Pr}_{𝑗} 𝑇$ is linear (Proposition 2.1.17), and the $𝑗$ th component of $𝑇 (𝐮)$ is ${Pr}_{𝑗} 𝑇 (𝐮)$ , so $𝑇 =$ $({Pr}_{1} 𝑇, \dots, {Pr}_{𝑚} 𝑇)$ . Conversely, if $𝑇 = (𝑇_{1}, \dots, 𝑇_{𝑚})$ , then $𝑇_{𝑗} (𝐮)$ is the $𝑗$ th component of $𝑇 (𝐮)$ , that is, $𝑇_{𝑗} = {Pr}_{𝑗} 𝑇$ . So the list is unique.

(b) Write $𝑆^{'}$ for the map defined by the $𝑆_{𝑗}$ . For tuples $𝐱 = {(𝐯_{𝑗})}_{𝑗}$ , $𝐲 = {(𝐰_{𝑗})}_{𝑗}$ and $𝑎 \in 𝐹$ ,

𝑆^{'} (𝑎 𝐱 + 𝐲) = \sum_{𝑗 = 1}^{𝑚} 𝑆_{𝑗} (𝑎 𝐯_{𝑗} + 𝐰_{𝑗}) = 𝑎 \sum_{𝑗 = 1}^{𝑚} 𝑆_{𝑗} (𝐯_{𝑗}) + \sum_{𝑗 = 1}^{𝑚} 𝑆_{𝑗} (𝐰_{𝑗}) = 𝑎 𝑆^{'} (𝐱) + 𝑆^{'} (𝐲) .

Also $𝑆^{'} 𝜄_{𝑗} (𝐯) =$ $𝑆_{𝑗} (𝐯)$ , since the other components of $𝜄_{𝑗} (𝐯)$ are $𝟎$ ; so $𝑆^{'} 𝜄_{𝑗} = 𝑆_{𝑗}$ , which shows that the list is determined by the map. Finally, let $𝑆 \in$ $L (𝑉_{1} \times \dots \times 𝑉_{𝑚}, 𝑊)$ . Applying $𝑆$ to (4.6.1),

𝑆 (𝐯_{1}, \dots, 𝐯_{𝑚}) = \sum_{𝑗 = 1}^{𝑚} 𝑆 (𝜄_{𝑗} (𝐯_{𝑗})) = \sum_{𝑗 = 1}^{𝑚} (𝑆 𝜄_{𝑗}) (𝐯_{𝑗}),

so $𝑆$ is the map defined by the list $𝑆 𝜄_{1}, \dots, 𝑆 𝜄_{𝑚}$ of linear maps. □

The assignments $𝑇 \mapsto$ $({Pr}_{1} 𝑇, \dots, {Pr}_{𝑚} 𝑇)$ and $𝑆 \mapsto$ $(𝑆 𝜄_{1}, \dots, 𝑆 𝜄_{𝑚})$ are linear, so the proposition says that

L (𝑈, 𝑊_{1} \times \dots \times 𝑊_{𝑚}) ≅ L (𝑈, 𝑊_{1}) \times \dots \times L (𝑈, 𝑊_{𝑚}), L (𝑉_{1} \times \dots \times 𝑉_{𝑚}, 𝑊) ≅ L (𝑉_{1}, 𝑊) \times \dots \times L (𝑉_{𝑚}, 𝑊) .

Combining the two for $𝐹^{𝑛} \times 𝐹^{𝑝} = 𝐹^{𝑛 + 𝑝}$ and $𝐹^{𝑚} \times 𝐹^{𝑞} = 𝐹^{𝑚 + 𝑞}$ : a linear map $𝐹^{𝑛 + 𝑝} \to 𝐹^{𝑚 + 𝑞}$ is the same as four maps ${Pr}_{𝑖} 𝑇 𝜄_{𝑗}$ , and its matrix is the block matrix whose blocks are their matrices (Definition 2.4.33).

4.6.2Products and direct sums

A sum of subspaces is the image of a single linear map out of a product, and directness becomes injectivity of that map.

Definition 4.6.8 (Addition map). Let $𝑈_{1}, \dots, 𝑈_{𝑚}$ be subspaces of $𝑉$ . The addition map is

Γ : 𝑈_{1} \times \dots \times 𝑈_{𝑚} \to 𝑉, Γ (𝐮_{1}, \dots, 𝐮_{𝑚}) = 𝐮_{1} + \dots + 𝐮_{𝑚} .

By Proposition 4.6.7(b), applied to the inclusions $𝑈_{𝑗} \to 𝑉$ , the map $Γ$ is linear, and by Definition 4.4.1 its image is $𝑈_{1} + \dots + 𝑈_{𝑚}$ . More is true: for $𝐯 \in 𝑉$ , the elements of the fiber $Γ^{- 1} (𝐯)$ are exactly the decompositions $𝐯 = 𝐮_{1} + \dots + 𝐮_{𝑚}$ of $𝐯$ along $𝑈_{1}, \dots, 𝑈_{𝑚}$ , each recorded as the tuple $(𝐮_{1}, \dots, 𝐮_{𝑚})$ of its components.

Theorem 4.6.9 (Direct sums and the addition map). Let

𝑈_{1}, \dots, 𝑈_{𝑚}

be subspaces of

𝑉

. The sum

𝑈_{1} + \dots + 𝑈_{𝑚}

is direct if and only if

Γ

is injective. In that case

Γ

is an isomorphism from

𝑈_{1} \times \dots \times 𝑈_{𝑚}

onto

𝑈_{1} \oplus \dots \oplus 𝑈_{𝑚}

. In particular, if

𝑉 = 𝑈_{1} \oplus \dots \oplus 𝑈_{𝑚}

, then

𝑉 ≅ 𝑈_{1} \times \dots \times 𝑈_{𝑚}

Proof.routine

Uses. Definition 4.4.4 (Direct sum), Definition 2.6.1 (Injective), Definition 4.6.8 (Addition map), Proposition 2.7.7 (Bijective linear maps are invertible)

Hint 1. What are the preimages of

𝐯

under

Γ

Hint 2. The preimages of

𝐯

are its decompositions. Vectors outside

im Γ

have no preimages at all.

Sketch.

Γ^{- 1} (𝐯)

is the set of decompositions of

𝐯

. The sum is direct when every

𝐯 \in im Γ

has exactly one, that is, when every

𝐯 \in 𝑉

has at most one preimage. That is injectivity. An injective linear map is an isomorphism onto its image.

By the remark before the theorem, the decompositions of

𝐯

along

𝑈_{1}, \dots, 𝑈_{𝑚}

correspond to the elements of

Γ^{- 1} (𝐯)

. By Definition 4.4.4, the sum is direct exactly when every

𝐯 \in 𝑈_{1} + \dots + 𝑈_{𝑚} = im Γ

has exactly one preimage. Vectors outside

im Γ

have none, and vectors in

im Γ

have at least one. So the sum is direct if and only if every

𝐯 \in 𝑉

has at most one preimage under

Γ

, which is injectivity (Definition 2.6.1). In that case

Γ

, regarded as a map onto

𝑈_{1} \oplus \dots \oplus 𝑈_{𝑚}

, is linear and bijective, hence an isomorphism (Proposition 2.7.7). □

By Theorem 2.6.11, injectivity of $Γ$ means $ker Γ = {𝟎}$ : the only decomposition of $𝟎$ is the trivial one. That is criterion (II) of Theorem 4.4.7, now read as a statement about a kernel. The point of the reformulation is that kernels can be counted.

Corollary 4.6.10 (Dimension of a sum). Let $𝑈_{1}, \dots, 𝑈_{𝑚}$ be finite-dimensional subspaces of $𝑉$ . Then

dim (𝑈_{1} + \dots + 𝑈_{𝑚}) \leq dim 𝑈_{1} + \dots + dim 𝑈_{𝑚},

with equality if and only if the sum is direct.

Proof.routine

Uses. Theorem 4.6.5 (Dimension of a product), Theorem 4.2.11 (Rank–nullity), Theorem 4.6.9 (Direct sums and the addition map), Theorem 2.6.11 (Injectivity is decided at $𝟎$ )

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. Apply rank–nullity to

Γ

Hint 2. The domain has dimension

\sum_{𝑗} dim 𝑈_{𝑗}

by Theorem 4.6.5, and the image is the sum. What does

dim ker Γ = 0

mean?

Sketch. Rank–nullity for

Γ

gives

dim (𝑈_{1} + \dots + 𝑈_{𝑚}) =

\sum_{𝑗} dim 𝑈_{𝑗} -

dim ker Γ

. Equality holds iff

ker Γ = {𝟎}

iff

Γ

is injective iff the sum is direct.

The domain of $Γ$ is finite-dimensional of dimension $\sum_{𝑗} dim 𝑈_{𝑗}$ by Theorem 4.6.5, and $im Γ = 𝑈_{1} + \dots + 𝑈_{𝑚}$ . By rank–nullity (Theorem 4.2.11),

dim (𝑈_{1} + \dots + 𝑈_{𝑚}) = dim 𝑈_{1} + \dots + dim 𝑈_{𝑚} - dim ker Γ \leq dim 𝑈_{1} + \dots + dim 𝑈_{𝑚} .

Equality holds if and only if $dim ker Γ = 0$ , that is, $ker Γ = {𝟎}$ . By Theorem 2.6.11 and Theorem 4.6.9, this happens exactly when the sum is direct. □

So $dim ker Γ$ measures how far a sum is from being direct. For two subspaces the kernel can be named, and the count becomes the dimension formula of Theorem 4.4.19.

Proposition 4.6.11 (The kernel of the addition map). Let $𝑈$ and $𝑊$ be subspaces of $𝑉$ , and let $Γ : 𝑈 \times 𝑊 \to 𝑉$ , $(𝐮, 𝐰) \mapsto 𝐮 + 𝐰$ .

(a) $ker Γ =$ ${(𝐮, - 𝐮) | 𝐮 \in 𝑈 \cap 𝑊}$ , and $𝐮 \mapsto (𝐮, - 𝐮)$ is an isomorphism from $𝑈 \cap 𝑊$ onto $ker Γ$ .
(b)If $𝑈$ and $𝑊$ are finite-dimensional, then $dim (𝑈 + 𝑊) =$ $dim 𝑈 +$ $dim 𝑊 -$ $dim (𝑈 \cap 𝑊)$ .

Proof.routine

Uses. Proposition 4.6.7 (Maps into and out of a product), Proposition 4.2.6 (Injective maps preserve dimension), Proposition 4.1.21 (Dimensions of subspaces), Theorem 4.6.5 (Dimension of a product), Theorem 4.2.11 (Rank–nullity)

Techniques. Double containment (Technique 1.3.14), Dimension counting (Technique 4.2.24)

Hint 1. If

𝐮 + 𝐰 = 𝟎

, then

𝐰 = - 𝐮

. Which subspaces does

𝐮

lie in?

Hint 2. For (b), apply rank–nullity to

Γ

, using (a) for the kernel and Theorem 4.6.5 for the domain.

Sketch. (a)

𝐮 + 𝐰 = 𝟎

forces

𝐮 = - 𝐰 \in 𝑈 \cap 𝑊

and

(𝐮, 𝐰) = (𝐮, - 𝐮)

; conversely such pairs are killed. The map

𝐮 \mapsto (𝐮, - 𝐮)

is linear and injective with image

ker Γ

. (b)

dim 𝑈 +

dim 𝑊 =

dim ker Γ +

dim (𝑈 + 𝑊)

and

dim ker Γ =

dim (𝑈 \cap 𝑊)

(a) If $(𝐮, 𝐰) \in ker Γ$ , then $𝐰 = - 𝐮$ . So $𝐮 = - 𝐰 \in 𝑊$ , hence $𝐮 \in 𝑈 \cap 𝑊$ and $(𝐮, 𝐰) = (𝐮, - 𝐮)$ . Conversely, if $𝐮 \in 𝑈 \cap 𝑊$ , then $(𝐮, - 𝐮) \in 𝑈 \times 𝑊$ and $Γ (𝐮, - 𝐮) = 𝟎$ . This proves the description of $ker Γ$ . The map $𝜑 : 𝑈 \cap 𝑊 \to 𝑈 \times 𝑊$ , $𝜑 (𝐮) = (𝐮, - 𝐮)$ , is linear by Proposition 4.6.7(a), it is injective because the first component of $𝜑 (𝐮)$ is $𝐮$ , and its image is $ker Γ$ . So it is an isomorphism onto $ker Γ$ .

(b) $𝑈 \cap 𝑊$ is a subspace of the finite-dimensional space $𝑈$ , hence finite-dimensional (Proposition 4.1.21), and by (a) and Proposition 4.2.6(a), $dim ker Γ =$ $dim (𝑈 \cap 𝑊)$ . The domain $𝑈 \times 𝑊$ has dimension $dim 𝑈 + dim 𝑊$ (Theorem 4.6.5) and $im Γ = 𝑈 + 𝑊$ . Rank–nullity (Theorem 4.2.11) gives

dim 𝑈 + dim 𝑊 = dim (𝑈 \cap 𝑊) + dim (𝑈 + 𝑊),

as desired. □

This proof builds no basis of $𝑈 + 𝑊$ . It counts instead: the pairs in $𝑈 \times 𝑊$ that add up to $𝟎$ form a copy of $𝑈 \cap 𝑊$ , and that copy is exactly what the naive count $dim 𝑈 + dim 𝑊$ counts twice.

4.6.3Cosets

Intuition (Collapsing a subspace to a point). Take $𝑉 = ℝ^{2}$ and let $𝑈$ be the line through $𝟎$ spanned by $(2, 1)$ . The translates $𝐯 + 𝑈$ are the lines parallel to $𝑈$ . They stack up across the plane like the rules on a sheet of paper, and every point of the plane lies on exactly one of them.

Now forget everything about a vector except which of these lines it lies on. Two vectors become indistinguishable exactly when they differ by an element of $𝑈$ , so the whole line $𝑈$ collapses to one point, and so does every line parallel to it. One point per line remains. The lines are labelled by a single number, the value $𝑐 = 𝑥 - 2 𝑦$ that is constant along each of them, so the collapsed plane looks like a copy of $ℝ$ (Figure 4.6.1).

The quotient $𝑉 / 𝑈$ is this collapsed space. Its elements are the lines themselves, and it becomes a vector space once we say how to add and scale lines.

Figure 4.6.1. Left: the line

𝑈 = span ((2, 1))

(red) and some of its translates, the lines

𝑥 - 2 𝑦 = 𝑐

. The vectors

𝐯 = (2, 0)

and

𝐯^{'} = (0, - 1)

lie on the same translate (blue), because

𝐯 - 𝐯^{'} = (2, 1) \in 𝑈 .

Right: in

𝑉 / 𝑈

each line becomes a single point, drawn at position

𝑐

; the line

𝑈

becomes the origin.

Definition 4.6.12 (Coset). Let $𝑈$ be a subspace of $𝑉$ and $𝐯 \in 𝑉$ . The translate

𝐯 + 𝑈 = {𝐯 + 𝐮 | 𝐮 \in 𝑈}

(Notation 3.1.17) is called a coset of $𝑈$ , and $𝐯$ is a representative of it.

A coset is a subset of $𝑉$ , and it has many representatives: by Lemma 4.6.14 below, every element of $𝐯 + 𝑈$ is one. Cosets reappear in Section 4.7 as affine subspaces, where their geometry is studied in its own right.

Example 4.6.13 (Cosets).

(a)In $ℝ^{2}$ let $𝑈 = span ((2, 1))$ . The coset $(𝑣_{1}, 𝑣_{2}) + 𝑈$ is the line $𝑥 - 2 𝑦 = 𝑣_{1} - 2 𝑣_{2}$ . Indeed, its points $(𝑣_{1} + 2 𝑡, 𝑣_{2} + 𝑡)$ satisfy $𝑥 - 2 𝑦 = 𝑣_{1} - 2 𝑣_{2}$ ; conversely, if $𝑥 - 2 𝑦 = 𝑣_{1} - 2 𝑣_{2}$ , then $(𝑥, 𝑦) =$ $(𝑣_{1}, 𝑣_{2}) +$ $(𝑦 - 𝑣_{2}) (2, 1)$ .
(b)In $𝔽_{2}^{3}$ let $𝑈 =$ $span ((1, 1, 0)) =$ ${(0, 0, 0), (1, 1, 0)}$ . The cosets of $𝑈$ are
$𝑈, (1, 0, 0) + 𝑈 = {(1, 0, 0), (0, 1, 0)}, (0, 0, 1) + 𝑈 = {(0, 0, 1), (1, 1, 1)}, (1, 0, 1) + 𝑈 = {(1, 0, 1), (0, 1, 1)} .$
They have two elements each and split the eight vectors of $𝔽_{2}^{3}$ into four pairs.
(c)For $𝑇 \in L (𝑉, 𝑊)$ , the nonempty fibers $𝑇^{- 1} (𝐰)$ are exactly the cosets of $ker 𝑇$ (Corollary 2.6.31). In particular the solution set of a consistent system $𝐴 𝐱 = 𝐲$ is a coset of $ker 𝑇_{𝐴}$ (Proposition 3.1.19).
(d)In $𝐹 [𝑥]$ let $𝑈 = {𝑝 | 𝑝 (0) = 0}$ , the polynomials with constant term $0$ . Then $𝑝 +$ $𝑈 =$ ${𝑞 \in 𝐹 [𝑥] | 𝑞 (0) = 𝑝 (0)}$ : if $𝑞 = 𝑝 + 𝑢$ with $𝑢 \in 𝑈$ , then $𝑞 (0) =$ $𝑝 (0) +$ $𝑢 (0) =$ $𝑝 (0)$ (Proposition 1.2.18), and if $𝑞 (0) = 𝑝 (0)$ , then $𝑞 - 𝑝 \in 𝑈$ . So there is one coset for each scalar.

Lemma 4.6.14 (Equal cosets). Let $𝑈$ be a subspace of $𝑉$ and let $𝐯, 𝐰 \in 𝑉$ . The following are equivalent.

(I) $𝐯 + 𝑈 = 𝐰 + 𝑈$ .
(II) $𝐯 - 𝐰 \in 𝑈$ .
(III) $𝐯 \in 𝐰 + 𝑈$ .
(IV) $(𝐯 + 𝑈) \cap (𝐰 + 𝑈) \neq \emptyset$ .

Proof.routine

Uses. Definition 4.6.12 (Coset), Definition 1.4.1 (Subspace)

Techniques. Proving TFAE (Technique 2.7.1), Double containment (Technique 1.3.14)

Hint 1. Prove the cycle (I)

\Rightarrow

(III)

\Rightarrow

(IV)

\Rightarrow

(II)

\Rightarrow

(I). Only the last step takes work.

Hint 2. For (II)

\Rightarrow

(I), write

𝐯 +

𝐮 =

𝐰 +

((𝐯 - 𝐰) + 𝐮)

to get

𝐯 + 𝑈 \subseteq 𝐰 + 𝑈

. For the other containment, note that

𝐰 - 𝐯 \in 𝑈

as well.

Sketch. (I)

\Rightarrow

(III):

𝐯 = 𝐯 + 𝟎

. (III)

\Rightarrow

(IV):

𝐯

is in both cosets. (IV)

\Rightarrow

(II): a common element

𝐯 + 𝐮 = 𝐰 + 𝐮^{'}

gives

𝐯 - 𝐰 = 𝐮^{'} - 𝐮 \in 𝑈

. (II)

\Rightarrow

(I): double containment, using that

𝑈

is closed under addition and negation.

We prove (I) $\Rightarrow$ (III) $\Rightarrow$ (IV) $\Rightarrow$ (II) $\Rightarrow$ (I).

(I) $\Rightarrow$ (III). $𝐯 =$ $𝐯 +$ $𝟎 \in$ $𝐯 +$ $𝑈 =$ $𝐰 +$ $𝑈$ .

(III) $\Rightarrow$ (IV). $𝐯 \in 𝐯 + 𝑈$ , and $𝐯 \in 𝐰 + 𝑈$ by (III).

(IV) $\Rightarrow$ (II). Let $𝐱$ be in both cosets, say $𝐱 = 𝐯 + 𝐮 = 𝐰 + 𝐮^{'}$ with $𝐮, 𝐮^{'} \in 𝑈$ . Then $𝐯 - 𝐰 = 𝐮^{'} - 𝐮 \in 𝑈$ .

(II) $\Rightarrow$ (I). Let $𝐳 = 𝐯 - 𝐰 \in 𝑈$ . For $𝐮 \in 𝑈$ ,

𝐯 + 𝐮 = 𝐰 + (𝐳 + 𝐮) \in 𝐰 + 𝑈,

since $𝐳 + 𝐮 \in 𝑈$ . So $𝐯 + 𝑈 \subseteq 𝐰 + 𝑈$ . Since $𝐰 - 𝐯 = - 𝐳 \in 𝑈$ , the same argument with $𝐯$ and $𝐰$ exchanged gives $𝐰 + 𝑈 \subseteq 𝐯 + 𝑈$ . □

Corollary 4.6.15 (Cosets partition $𝑉$ ). Let $𝑈$ be a subspace of $𝑉$ .

(a)Every $𝐯 \in 𝑉$ lies in exactly one coset of $𝑈$ , namely $𝐯 + 𝑈$ .
(b)Two cosets of $𝑈$ are either equal or disjoint.
(c) $𝐯 + 𝑈 = 𝑈$ if and only if $𝐯 \in 𝑈$ .

Proof.routine

Uses. Lemma 4.6.14 (Equal cosets)

Hint. Each part is one implication of Lemma 4.6.14. For (c), write

𝑈 = 𝟎 + 𝑈

Sketch. (a)

𝐯 \in 𝐯 + 𝑈

, and

𝐯 \in 𝐰 + 𝑈

forces

𝐰 + 𝑈 = 𝐯 + 𝑈

by (III)

\Rightarrow

(I). (b) is (IV)

\Rightarrow

(I). (c) is (I)

\Leftrightarrow

(II) with

𝐰 = 𝟎

(a) $𝐯 = 𝐯 + 𝟎 \in 𝐯 + 𝑈$ . If also $𝐯 \in 𝐰 + 𝑈$ , then $𝐯 + 𝑈 = 𝐰 + 𝑈$ by (III) $\Rightarrow$ (I) of Lemma 4.6.14.

(b) If two cosets are not disjoint, they are equal by (IV) $\Rightarrow$ (I).

(c) $𝑈 = 𝟎 + 𝑈$ , so by (I) $\Leftrightarrow$ (II), $𝐯 + 𝑈 = 𝑈$ if and only if $𝐯 - 𝟎 = 𝐯 \in 𝑈$ . □

Remark 4.6.16 (Congruence modulo

𝑈

). Write

𝐯 \equiv 𝐰 (mod 𝑈)

when

𝐯 - 𝐰 \in 𝑈

. This is an equivalence relation: it is reflexive because

𝟎 \in 𝑈

, symmetric because

𝑈

is closed under negation, and transitive because

𝑈

is closed under addition. By Lemma 4.6.14 its equivalence classes are exactly the cosets of

𝑈

. (Section A.3 treats equivalence relations in general, but nothing from there is needed here beyond Lemma 4.6.14.) Notice that each of the three properties used a different part of “

𝑈

is a subspace”.

Exercise 4.6.17.routine Let $𝑈 =$ $span ((1, 0, 1), (0, 1, 1)) \subseteq$ $ℝ^{3}$ .

(a)Show that $𝑈 =$ ${(𝑥, 𝑦, 𝑧) | 𝑧 = 𝑥 + 𝑦}$ , and that $𝐯 + 𝑈$ is the plane $𝑧 -$ $𝑥 -$ $𝑦 =$ $𝑣_{3} -$ $𝑣_{1} -$ $𝑣_{2}$ .
(b)Which of the cosets $(1, 2, 3) + 𝑈$ , $(0, 0, 1) + 𝑈$ , $(2, - 1, 2) + 𝑈$ , and $(1, 1, 1) + 𝑈$ are equal?

Solution.

Uses. Lemma 4.6.14 (Equal cosets), Definition 1.3.10 (Span)

Hint 1. A general element of

𝑈

𝑎 (1, 0, 1) +

𝑏 (0, 1, 1)

. Write it out.

Hint 2. By Lemma 4.6.14,

𝐯 + 𝑈 = 𝐰 + 𝑈

exactly when

𝐯 - 𝐰 \in 𝑈

, that is, when

𝐯

and

𝐰

have the same value of

𝑧 - 𝑥 - 𝑦

Sketch. (a)

𝑎 (1, 0, 1) +

𝑏 (0, 1, 1) =

(𝑎, 𝑏, 𝑎 + 𝑏)

. (b) The values of

𝑧 - 𝑥 - 𝑦

are

0, 1, 1, - 1

, so only

(0, 0, 1) +

𝑈 =

(2, - 1, 2) +

𝑈

; also

(1, 2, 3) + 𝑈 = 𝑈

(a) $𝑎 (1, 0, 1) +$ $𝑏 (0, 1, 1) =$ $(𝑎, 𝑏, 𝑎 + 𝑏)$ , and every $(𝑥, 𝑦, 𝑧)$ with $𝑧 = 𝑥 + 𝑦$ has this form with $𝑎 = 𝑥$ , $𝑏 = 𝑦$ . So $𝑈 =$ ${(𝑥, 𝑦, 𝑧) | 𝑧 - 𝑥 - 𝑦 = 0}$ . By Lemma 4.6.14, $(𝑥, 𝑦, 𝑧) \in 𝐯 + 𝑈$ if and only if $(𝑥, 𝑦, 𝑧) - 𝐯 \in 𝑈$ , that is, $(𝑧 - 𝑣_{3}) -$ $(𝑥 - 𝑣_{1}) -$ $(𝑦 - 𝑣_{2}) =$ $0$ , which is $𝑧 -$ $𝑥 -$ $𝑦 =$ $𝑣_{3} -$ $𝑣_{1} -$ $𝑣_{2}$ .

(b) Let $𝑓 (𝑥, 𝑦, 𝑧) =$ $𝑧 -$ $𝑥 -$ $𝑦$ . By (a) and Lemma 4.6.14, two cosets $𝐯 + 𝑈$ and $𝐰 + 𝑈$ are equal if and only if $𝑓 (𝐯) = 𝑓 (𝐰)$ . The values are

𝑓 (1, 2, 3) = 0, 𝑓 (0, 0, 1) = 1, 𝑓 (2, - 1, 2) = 1, 𝑓 (1, 1, 1) = - 1 .

So $(0, 0, 1) +$ $𝑈 =$ $(2, - 1, 2) +$ $𝑈$ , and the other two cosets are different from each other and from it. (The first one is $𝑈$ itself, since $(1, 2, 3) \in 𝑈$ .)

4.6.4The quotient space

We want to add and scale cosets, and the only handle on a coset is a representative. So, as in the picture, add two lines by picking a point on each, adding the points, and taking the line through the sum:

(𝐯 + 𝑈) + (𝐰 + 𝑈) = (𝐯 + 𝐰) + 𝑈, 𝑎 (𝐯 + 𝑈) = 𝑎 𝐯 + 𝑈 .

The right sides are computed from the chosen representatives, and a coset has many. If two choices could produce different cosets, these formulas would not define operations at all. Figure 4.6.2 shows why they do: replacing $𝐯$ by $𝐯^{'} = 𝐯 + 𝐮$ and $𝐰$ by $𝐰^{'} = 𝐰 + 𝐮^{'}$ changes the sum by $𝐮 + 𝐮^{'} \in 𝑈$ , which moves it along its line but not off it.

Figure 4.6.2. Adding cosets of

𝑈 = span ((2, 1))

(red) in

ℝ^{2}

. The green line is

𝑥 - 2 𝑦 = 2

and the ochre line is

𝑥 - 2 𝑦 = 4

. The representatives

𝐯

𝐯^{'}

of the green line and

𝐰

𝐰^{'}

of the ochre line give different sums,

𝐯 + 𝐰 = (- 2, - 4)

and

𝐯^{'} + 𝐰^{'} = (4, - 1)

, but both lie on the blue line

𝑥 - 2 𝑦 = 6

. So the sum of the green and ochre lines is the blue line, whichever points we pick.

Technique 4.6.18 (Check that a definition does not depend on the representative). Objects such as cosets are handled through representatives: a coset is named by any one of its elements, and a function on cosets is defined by a formula in that element, $𝑓 (𝐯 + 𝑈) = 𝑔 (𝐯)$ . Such a formula defines a function only if every name of the same coset gives the same value. So the first thing to prove about it is

𝐯 + 𝑈 = 𝐯^{'} + 𝑈 \Rightarrow 𝑔 (𝐯) = 𝑔 (𝐯^{'}),

which by Lemma 4.6.14 means: if $𝐯 - 𝐯^{'} \in 𝑈$ , then $𝑔 (𝐯) = 𝑔 (𝐯^{'})$ . One says that $𝑓$ is well defined.

Mind the direction. Equal inputs must give equal outputs. The converse, that equal outputs force equal inputs, is injectivity, which is a different property that $𝑓$ may or may not have (Exercise 4.6.43).

Reach for it whenever an object is defined through a choice: of a representative, a basis, a preimage, an extension. The canonical example is Theorem 4.6.22. The same check appeared in Exercise 2.6.43, where a map was defined on preimages, and in Proposition 4.2.15, which shows that the rank of a matrix of $𝑇$ does not depend on the bases. It returns when the determinant of an operator is defined through any one of its matrices (Section 5.4).

Used in 8 places and 2 practice exercises

Theorem 4.6.22, Exercise 4.6.24, Theorem 4.6.30, Theorem 4.6.34, Exercise 4.6.43, Theorem A.3.10, Exercise A.3.19, Exercise A.4.10; 2 practice exercises.

Non-example 4.6.19 (A formula that defines no function). Let

𝑈 = span ((2, 1)) \subseteq ℝ^{2}

. The formula

𝑓 ((𝑣_{1}, 𝑣_{2}) + 𝑈) =

𝑣_{1}

does not define a function on the set of cosets:

(0, 0) + 𝑈

and

(2, 1) + 𝑈

are the same coset, since

(2, 1) - (0, 0) \in 𝑈

, but the formula gives

0

for the first name and

2

for the second. On the other hand,

𝑔 ((𝑣_{1}, 𝑣_{2}) + 𝑈) =

𝑣_{1} -

2 𝑣_{2}

is well defined: if

(𝑣_{1}, 𝑣_{2}) -

(𝑣_{1}^{'}, 𝑣_{2}^{'}) =

𝑡 (2, 1)

, then

(𝑣_{1} - 2 𝑣_{2}) -

(𝑣_{1}^{'} - 2 𝑣_{2}^{'}) =

2 𝑡 -

2 𝑡 =

0

. In the picture,

𝑔

reads off the label

𝑐

of the line

𝑥 - 2 𝑦 = 𝑐

, which is a property of the line, while

𝑓

reads off a property of a chosen point on it.

Definition 4.6.20 (Quotient space). Let $𝑈$ be a subspace of $𝑉$ . The quotient of $𝑉$ by $𝑈$ is the set of cosets

𝑉 / 𝑈 = {𝐯 + 𝑈 | 𝐯 \in 𝑉},

with addition and scalar multiplication

(𝐯 + 𝑈) + (𝐰 + 𝑈) = (𝐯 + 𝐰) + 𝑈, 𝑎 (𝐯 + 𝑈) = 𝑎 𝐯 + 𝑈 .

By Theorem 4.6.22 these operations are well defined and make $𝑉 / 𝑈$ a vector space over $𝐹$ . We read $𝑉 / 𝑈$ as “ $𝑉$ mod $𝑈$ ”.

Notation 4.6.21. When

𝑈

is clear from context, we sometimes write

[𝐯]

for

𝐯 + 𝑈

. Then

[𝐯] = [𝐰]

if and only if

𝐯 - 𝐰 \in 𝑈

, and the operations read

[𝐯] + [𝐰] = [𝐯 + 𝐰]

and

𝑎 [𝐯] = [𝑎 𝐯]

Theorem 4.6.22 (The quotient is a vector space). Let $𝑈$ be a subspace of $𝑉$ .

(a)The operations of Definition 4.6.20 are well defined: if $𝐯 + 𝑈 = 𝐯^{'} + 𝑈$ and $𝐰 + 𝑈 = 𝐰^{'} + 𝑈$ , then
$(𝐯 + 𝐰) + 𝑈 = (𝐯^{'} + 𝐰^{'}) + 𝑈 and 𝑎 𝐯 + 𝑈 = 𝑎 𝐯^{'} + 𝑈 for all 𝑎 \in 𝐹 .$
(b)With these operations, $𝑉 / 𝑈$ is a vector space over $𝐹$ . Its zero vector is $𝟎 + 𝑈 = 𝑈$ , and $- (𝐯 + 𝑈) = (- 𝐯) + 𝑈$ .

Proof.standard

Uses. Lemma 4.6.14 (Equal cosets), Definition 1.4.1 (Subspace), Definition 1.2.1 (Vector space)

Techniques. Check that a definition does not depend on the representative (Technique 4.6.18), Proving from axioms (Technique 1.1.7)

Hint 1. Translate the hypotheses with Lemma 4.6.14:

𝐯 - 𝐯^{'} \in 𝑈

and

𝐰 - 𝐰^{'} \in 𝑈

. What must you show about the outputs?

Hint 2. Show that

(𝐯 + 𝐰) - (𝐯^{'} + 𝐰^{'})

and

𝑎 𝐯 - 𝑎 𝐯^{'}

lie in

𝑈

. Which properties of

𝑈

does this use?

Hint 3. For (b), write every coset with a representative, use the definitions to turn each axiom into a computation in

𝑉

, and use the axiom there.

Sketch. (a)

(𝐯 + 𝐰) -

(𝐯^{'} + 𝐰^{'}) =

(𝐯 - 𝐯^{'}) +

(𝐰 - 𝐰^{'}) \in

𝑈

and

𝑎 𝐯 -

𝑎 𝐯^{'} =

𝑎 (𝐯 - 𝐯^{'}) \in

𝑈

, because

𝑈

is closed under addition and scalar multiplication; conclude with Lemma 4.6.14. (b) Each axiom for cosets is the same axiom for representatives, e.g.

𝑎 ((𝐯 + 𝑈) + (𝐰 + 𝑈)) =

(𝑎 𝐯 + 𝑎 𝐰) +

𝑈 =

𝑎 (𝐯 + 𝑈) +

𝑎 (𝐰 + 𝑈)

(a) By Lemma 4.6.14, the hypotheses say that $𝐯 - 𝐯^{'} \in 𝑈$ and $𝐰 - 𝐰^{'} \in 𝑈$ . Since $𝑈$ is a subspace,

(𝐯 + 𝐰) - (𝐯^{'} + 𝐰^{'}) = (𝐯 - 𝐯^{'}) + (𝐰 - 𝐰^{'}) \in 𝑈 and 𝑎 𝐯 - 𝑎 𝐯^{'} = 𝑎 (𝐯 - 𝐯^{'}) \in 𝑈 .

By Lemma 4.6.14 again, $(𝐯 + 𝐰) +$ $𝑈 =$ $(𝐯^{'} + 𝐰^{'}) +$ $𝑈$ and $𝑎 𝐯 + 𝑈 = 𝑎 𝐯^{'} + 𝑈$ .

(b) By (a), the sum of two cosets and a scalar multiple of a coset are well-defined cosets. Each axiom is an identity between cosets. Writing each coset with a representative and applying the definitions turns both sides into cosets of vectors of $𝑉$ , which agree by the corresponding axiom in $𝑉$ . For instance, for $𝐮, 𝐯, 𝐰 \in 𝑉$ and $𝑎, 𝑏 \in 𝐹$ ,

\begin{aligned} ((𝐮 + 𝑈) + (𝐯 + 𝑈)) + (𝐰 + 𝑈) & = ((𝐮 + 𝐯) + 𝐰) + 𝑈 = (𝐮 + (𝐯 + 𝐰)) + 𝑈 \\ = (𝐮 + 𝑈) + ((𝐯 + 𝑈) + (𝐰 + 𝑈)), \\ 𝑎 ((𝐯 + 𝑈) + (𝐰 + 𝑈)) & = 𝑎 (𝐯 + 𝐰) + 𝑈 = (𝑎 𝐯 + 𝑎 𝐰) + 𝑈 = 𝑎 (𝐯 + 𝑈) + 𝑎 (𝐰 + 𝑈), \\ (𝑎 + 𝑏) (𝐯 + 𝑈) & = (𝑎 + 𝑏) 𝐯 + 𝑈 = (𝑎 𝐯 + 𝑏 𝐯) + 𝑈 = 𝑎 (𝐯 + 𝑈) + 𝑏 (𝐯 + 𝑈), \end{aligned}

and commutativity, $(𝑎 𝑏) (𝐯 + 𝑈) =$ $𝑎 (𝑏 (𝐯 + 𝑈))$ , and $1 (𝐯 + 𝑈) = 𝐯 + 𝑈$ are checked the same way. Finally, $(𝐯 + 𝑈) +$ $(𝟎 + 𝑈) =$ $𝐯 +$ $𝑈$ and $(𝐯 + 𝑈) +$ $((- 𝐯) + 𝑈) =$ $𝟎 +$ $𝑈$ , so $𝟎 + 𝑈 = 𝑈$ is a zero vector and $(- 𝐯) + 𝑈$ is an additive inverse of $𝐯 + 𝑈$ . □

All the content is in (a), and (a) used exactly that $𝑈$ is closed under addition and under scalar multiplication. Part (b) is bookkeeping. Notice also that $𝑈$ , the subspace we collapsed, is now the origin of $𝑉 / 𝑈$ .

Example 4.6.23 (Quotient spaces).

(a)Let $𝑈 = span ((2, 1)) \subseteq ℝ^{2}$ . By Non-example 4.6.19, $𝑔 : ℝ^{2} / 𝑈 \to ℝ$ , $𝑔 ((𝑣_{1}, 𝑣_{2}) + 𝑈) =$ $𝑣_{1} -$ $2 𝑣_{2}$ , is well defined. It is linear, since $𝑔 (𝑎 (𝐯 + 𝑈) + (𝐰 + 𝑈)) =$ $𝑔 ((𝑎 𝐯 + 𝐰) + 𝑈) =$ $𝑎 (𝑣_{1} - 2 𝑣_{2}) +$ $(𝑤_{1} - 2 𝑤_{2})$ . It is injective: if $𝑣_{1} - 2 𝑣_{2} = 0$ , then $𝐯 = 𝑣_{2} (2, 1) \in 𝑈$ , so $𝐯 + 𝑈 = 𝑈$ is the zero coset. It is surjective, since $𝑔 ((𝑐, 0) + 𝑈) = 𝑐$ . So $ℝ^{2} / 𝑈 ≅ ℝ$ , as Intuition (Collapsing a subspace to a point) predicted, and $𝑔$ is the coordinate across the stack of lines.
(b)Let $𝑈 =$ $span ((1, 1, 0)) \subseteq$ $𝔽_{2}^{3}$ as in Example 4.6.13(b). Then $𝔽_{2}^{3} / 𝑈$ has four elements. For instance $((1, 0, 0) + 𝑈) +$ $((0, 0, 1) + 𝑈) =$ $(1, 0, 1) +$ $𝑈$ , and with the other representatives $(0, 1, 0)$ and $(1, 1, 1)$ we get $(1, 0, 1) + 𝑈$ once more. With $(1, 0, 0)$ and $(1, 1, 1)$ we get $(0, 1, 1) + 𝑈$ , which is the same coset, as it must be. Every element is its own negative, and Theorem 4.6.27 below gives $dim (𝔽_{2}^{3} / 𝑈) = 2$ .
(c)The extreme cases: $𝑉 / 𝑉$ has the single element $𝟎 + 𝑉 = 𝑉$ , so it is the zero space. In $𝑉 / {𝟎}$ each coset $𝐯 + {𝟎} = {𝐯}$ has one element, and $𝐯 \mapsto {𝐯}$ is an isomorphism $𝑉 \to 𝑉 / {𝟎}$ .

Exercise 4.6.24.routine Let $𝑈 = span ((1, 1)) \subseteq ℝ^{2}$ . Which of the following formulas define functions on $ℝ^{2} / 𝑈$ ? For those that do, decide whether the function is linear.

(a) $𝑓 (𝐯 + 𝑈) = 𝑣_{1} + 𝑣_{2}$ .
(b) $𝑓 (𝐯 + 𝑈) = 𝑣_{1} - 𝑣_{2}$ .
(c) $𝑓 (𝐯 + 𝑈) =$ ${(𝑣_{1} - 𝑣_{2})}^{2}$ .
(d) $𝑆 (𝐯 + 𝑈) =$ $(𝑣_{2}, 𝑣_{1}) +$ $𝑈$ .
(e) $𝑆 (𝐯 + 𝑈) =$ $(𝑣_{1}, 0) +$ $𝑈$ .

Solution.

Uses. Lemma 4.6.14 (Equal cosets), Definition 4.6.20 (Quotient space)

Techniques. Check that a definition does not depend on the representative (Technique 4.6.18)

Hint 1. By Lemma 4.6.14, two representatives of one coset differ by

𝑡 (1, 1)

. For each formula, compare its values at

𝐯

and at

𝐯 + 𝑡 (1, 1)

Hint 2. To show that a formula is not well defined, one pair of representatives of the same coset suffices, for example

(0, 0)

and

(1, 1)

Sketch. (a) No:

(0, 0)

and

(1, 1)

give

0

and

2

. (b) Yes, linear. (c) Yes (a function of (b)), not linear. (d) Yes, linear; it equals

- id

, since

(𝑣_{2}, 𝑣_{1}) +

(𝑣_{1}, 𝑣_{2}) \in

𝑈

. (e) No:

(0, 0) + 𝑈

and

(1, 1) + 𝑈

go to

𝑈

and

(1, 0) + 𝑈 \neq 𝑈

Two representatives of the same coset differ by $𝑡 (1, 1)$ for some $𝑡 \in ℝ$ (Lemma 4.6.14). Write $𝐯^{'} = 𝐯 + 𝑡 (1, 1)$ .

(a) Not well defined: $(0, 0) + 𝑈 = (1, 1) + 𝑈$ , but the formula gives $0$ and $2$ .

(b) Well defined, since $𝑣_{1}^{'} -$ $𝑣_{2}^{'} =$ $(𝑣_{1} + 𝑡) -$ $(𝑣_{2} + 𝑡) =$ $𝑣_{1} -$ $𝑣_{2}$ . It is linear: $𝑓 (𝑎 (𝐯 + 𝑈) + (𝐰 + 𝑈)) =$ $𝑓 ((𝑎 𝐯 + 𝐰) + 𝑈) =$ $𝑎 (𝑣_{1} - 𝑣_{2}) +$ $(𝑤_{1} - 𝑤_{2})$ .

(c) Well defined, since it is the square of the function in (b). Not linear: $𝑓 ((1, 0) + 𝑈) = 1$ but $𝑓 (2 ((1, 0) + 𝑈)) =$ $𝑓 ((2, 0) + 𝑈) =$ $4 \neq$ $2$ .

(d) Well defined: swapping the coordinates of $𝐯^{'}$ gives $(𝑣_{2}, 𝑣_{1}) + 𝑡 (1, 1)$ , which lies in the same coset as $(𝑣_{2}, 𝑣_{1})$ . It is linear, by the same computation as in (b). In fact $(𝑣_{2}, 𝑣_{1}) +$ $(𝑣_{1}, 𝑣_{2}) =$ $(𝑣_{1} + 𝑣_{2}) (1, 1) \in$ $𝑈$ , so $(𝑣_{2}, 𝑣_{1}) +$ $𝑈 =$ $- (𝐯 + 𝑈)$ , and $𝑆 = - id$ .

(e) Not well defined: $(0, 0) + 𝑈 = (1, 1) + 𝑈$ , but the formula gives $(0, 0) + 𝑈 = 𝑈$ and $(1, 0) + 𝑈$ , which differ because $(1, 0) \notin 𝑈$ .

The quotient map

Definition 4.6.25 (Quotient map). Let

𝑈

be a subspace of

𝑉

. The quotient map is

𝜋 : 𝑉 \to 𝑉 / 𝑈

𝜋 (𝐯) = 𝐯 + 𝑈

Proposition 4.6.26 (The quotient map). The quotient map

𝜋 : 𝑉 \to 𝑉 / 𝑈

is linear and surjective, and

ker 𝜋 = 𝑈

Proof.routine

Uses. Definition 4.6.20 (Quotient space), Corollary 4.6.15 (Cosets partition $𝑉$ ), Theorem 2.1.3 (Checking linearity)

Hint 1. Linearity is the definition of the operations on

𝑉 / 𝑈

, read backwards.

Hint 2. For the kernel, recall that the zero of

𝑉 / 𝑈

𝑈

, and use Corollary 4.6.15(c).

Sketch.

𝜋 (𝑎 𝐯 + 𝐰) =

(𝑎 𝐯 + 𝐰) +

𝑈 =

𝑎 (𝐯 + 𝑈) +

(𝐰 + 𝑈)

. Every coset is

𝜋

of its representatives.

𝜋 (𝐯) = 𝑈

iff

𝐯 \in 𝑈

For $𝐯, 𝐰 \in 𝑉$ and $𝑎 \in 𝐹$ , the definition of the operations gives

𝜋 (𝑎 𝐯 + 𝐰) = (𝑎 𝐯 + 𝐰) + 𝑈 = 𝑎 (𝐯 + 𝑈) + (𝐰 + 𝑈) = 𝑎 𝜋 (𝐯) + 𝜋 (𝐰),

so $𝜋$ is linear (Theorem 2.1.3). Every element of $𝑉 / 𝑈$ has the form $𝐯 + 𝑈 = 𝜋 (𝐯)$ , so $𝜋$ is surjective. The zero of $𝑉 / 𝑈$ is $𝑈$ (Theorem 4.6.22), and $𝜋 (𝐯) = 𝑈$ if and only if $𝐯 \in 𝑈$ (Corollary 4.6.15(c)). So $ker 𝜋 = 𝑈$ . □

In fact the operations on $𝑉 / 𝑈$ were chosen to make $𝜋$ linear. One consequence: every subspace of every vector space is the kernel of some linear map. In finite dimensions Section 4.4 gets this from a complement; the quotient map gives it with no choices and no dimension hypothesis.

4.6.5The dimension of a quotient

In the plane, collapsing a line leaves a line. In general, collapsing $𝑈$ should remove exactly $dim 𝑈$ dimensions. The first proof below finds a basis of $𝑉 / 𝑈$ inside a basis of $𝑉$ , and the second reads the count off rank–nullity for $𝜋$ .

Theorem 4.6.27 (Dimension of a quotient). Let $𝑉$ be finite-dimensional and let $𝑈$ be a subspace of $𝑉$ . Then $𝑉 / 𝑈$ is finite-dimensional, and

dim (𝑉 / 𝑈) = dim 𝑉 - dim 𝑈 .

More precisely, if $𝐮_{1},$ $\dots,$ $𝐮_{𝑘},$ $𝐯_{1},$ $\dots,$ $𝐯_{𝑟}$ is a basis of $𝑉$ such that $𝐮_{1}, \dots, 𝐮_{𝑘}$ is a basis of $𝑈$ , then $𝐯_{1} + 𝑈, \dots, 𝐯_{𝑟} + 𝑈$ is a basis of $𝑉 / 𝑈$ .

Proof (by extending a basis of $𝑈$ ).standard

Uses. Proposition 4.1.21 (Dimensions of subspaces), Theorem 4.1.16 (Extending to a basis), Lemma 4.6.14 (Equal cosets), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Extend a basis of a subspace (Technique 4.1.18)

Hint 1. Bases of the required kind exist by Theorem 4.1.16. Why should the cosets

𝐮_{𝑖} + 𝑈

play no role?

Hint 2. Each

𝐮_{𝑖} + 𝑈

is the zero coset. For spanning, expand an arbitrary

𝐯

in the basis and apply

𝜋

Hint 3. For independence:

\sum_{𝑗} 𝑏_{𝑗} (𝐯_{𝑗} + 𝑈) =

𝑈

means

\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗} \in

𝑈 =

span (𝐮_{1}, \dots, 𝐮_{𝑘})

. Write that as a linear relation among the basis vectors of

𝑉

Sketch. Extend a basis

𝐮_{1}, \dots, 𝐮_{𝑘}

𝑈

to a basis of

𝑉

. Since

𝐯 - \sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗} \in 𝑈

, every coset is

\sum_{𝑗} 𝑏_{𝑗} (𝐯_{𝑗} + 𝑈)

. If

\sum_{𝑗} 𝑏_{𝑗} (𝐯_{𝑗} + 𝑈) =

𝑈

, then

\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗} =

\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖}

, a relation among basis vectors, so all

𝑏_{𝑗} = 0

. Hence

dim (𝑉 / 𝑈) =

𝑟 =

dim 𝑉 -

dim 𝑈

By Proposition 4.1.21(a), $𝑈$ is finite-dimensional, so it has a basis $𝐮_{1}, \dots, 𝐮_{𝑘}$ , and by Theorem 4.1.16 this extends to a basis $𝐮_{1},$ $\dots,$ $𝐮_{𝑘},$ $𝐯_{1},$ $\dots,$ $𝐯_{𝑟}$ of $𝑉$ with $𝑘 + 𝑟 = dim 𝑉$ . So it suffices to prove the second statement, for any such basis. Then $dim (𝑉 / 𝑈) =$ $𝑟 =$ $dim 𝑉 -$ $dim 𝑈$ .

Spanning. Let $𝐯 \in 𝑉$ and write $𝐯 =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} +$ $\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗}$ . Then $𝐯 -$ $\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗} =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} \in$ $𝑈$ , so by Lemma 4.6.14 and the definition of the operations,

𝐯 + 𝑈 = (\sum_{𝑗 = 1}^{𝑟} 𝑏_{𝑗} 𝐯_{𝑗}) + 𝑈 = \sum_{𝑗 = 1}^{𝑟} 𝑏_{𝑗} (𝐯_{𝑗} + 𝑈) .

Independence. Suppose $\sum_{𝑗} 𝑏_{𝑗} (𝐯_{𝑗} + 𝑈) =$ $𝟎 +$ $𝑈$ , that is, $(\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗}) +$ $𝑈 =$ $𝑈$ . By Corollary 4.6.15(c), $\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗} \in 𝑈$ , so $\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗} =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖}$ for some scalars $𝑎_{𝑖}$ . Then

\sum_{𝑖 = 1}^{𝑘} (- 𝑎_{𝑖}) 𝐮_{𝑖} + \sum_{𝑗 = 1}^{𝑟} 𝑏_{𝑗} 𝐯_{𝑗} = 𝟎

is a linear relation among the vectors of a basis of $𝑉$ , so all its coefficients vanish (Theorem 1.5.5). In particular $𝑏_{1} = \dots = 𝑏_{𝑟} = 0$ . □

Proof (by rank–nullity for the quotient map).routine

Uses. Proposition 4.6.26 (The quotient map), Theorem 4.2.11 (Rank–nullity), Proposition 2.1.16 (Linear maps preserve spans and linear relations), Corollary 4.1.11 (Spanning lists are long)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1.

𝜋 : 𝑉 \to 𝑉 / 𝑈

is linear and surjective. What is its kernel?

Hint 2. Rank–nullity applies because

𝑉

is finite-dimensional. For the basis statement, a spanning list of length

dim (𝑉 / 𝑈)

is a basis.

Sketch.

𝜋

is surjective with kernel

𝑈

, so rank–nullity gives

dim 𝑉 =

dim 𝑈 +

dim (𝑉 / 𝑈)

, with

𝑉 / 𝑈 = im 𝜋

finite-dimensional. The images

𝜋 (𝐮_{𝑖}) = 𝟎

and

𝜋 (𝐯_{𝑗})

span

𝑉 / 𝑈

, so the

𝑟 = dim (𝑉 / 𝑈)

cosets

𝐯_{𝑗} + 𝑈

span it, hence form a basis.

By Proposition 4.6.26, $𝜋$ is linear and surjective with $ker 𝜋 = 𝑈$ . Since $𝑉$ is finite-dimensional, rank–nullity (Theorem 4.2.11(b)) says that $im 𝜋 = 𝑉 / 𝑈$ is finite-dimensional and $dim 𝑉 =$ $dim 𝑈 +$ $dim (𝑉 / 𝑈)$ .

For the second statement, the list $𝜋 (𝐮_{1}),$ $\dots,$ $𝜋 (𝐮_{𝑘}),$ $𝜋 (𝐯_{1}),$ $\dots,$ $𝜋 (𝐯_{𝑟})$ spans $im 𝜋 = 𝑉 / 𝑈$ (Proposition 2.1.16(a)). Its first $𝑘$ entries are $𝟎$ , so $𝐯_{1} + 𝑈, \dots, 𝐯_{𝑟} + 𝑈$ spans $𝑉 / 𝑈$ as well. It has length $𝑟 =$ $dim 𝑉 -$ $dim 𝑈 =$ $dim (𝑉 / 𝑈)$ , so it is a basis (Corollary 4.1.11(b)). □

The second proof is shorter, but it leans on rank–nullity. The first proof uses only the extension of bases, and that matters in Section 4.6.7, where the theorem is used to reprove rank–nullity.

The converse direction needs no finiteness assumption on $𝑉$ : a basis of $𝑈$ together with lifts of a basis of $𝑉 / 𝑈$ is a basis of $𝑉$ . This is what makes quotients useful for induction on dimension, where a statement about $𝑉$ is split into statements about $𝑈$ and $𝑉 / 𝑈$ .

Proposition 4.6.28 (Lifting a basis of a quotient). Let

𝑈

be a subspace of

𝑉

, let

𝐮_{1}, \dots, 𝐮_{𝑘}

be a basis of

𝑈

, and let

𝐯_{1}, \dots, 𝐯_{𝑟} \in 𝑉

be such that

𝐯_{1} + 𝑈, \dots, 𝐯_{𝑟} + 𝑈

is a basis of

𝑉 / 𝑈

. Then

𝐮_{1},

\dots,

𝐮_{𝑘},

𝐯_{1},

\dots,

𝐯_{𝑟}

is a basis of

𝑉

. In particular, if

𝑈

and

𝑉 / 𝑈

are finite-dimensional, then so is

𝑉

, and

dim 𝑉 =

dim 𝑈 +

dim (𝑉 / 𝑈)

Proof.standard

Uses. Proposition 4.6.26 (The quotient map), Lemma 4.6.14 (Equal cosets), Corollary 1.6.24 (Bases exist in finite dimensions)

Techniques. Apply an operator to a relation to kill terms (Technique 4.1.25)

Hint 1. For spanning, expand

𝐯 + 𝑈

in the basis of

𝑉 / 𝑈

and pull back with Lemma 4.6.14.

Hint 2. For independence, apply

𝜋

to a linear relation. Which terms does it kill?

Sketch. If

𝐯 +

𝑈 =

\sum_{𝑗} 𝑏_{𝑗} (𝐯_{𝑗} + 𝑈)

, then

𝐯 -

\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗} \in

𝑈 =

span (𝐮_{1}, \dots, 𝐮_{𝑘})

. Applying

𝜋

to a relation

\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} +

\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗} =

𝟎

kills the

𝐮_{𝑖}

and leaves a relation among the

𝐯_{𝑗} + 𝑈

, so all

𝑏_{𝑗} = 0

, and then all

𝑎_{𝑖} = 0

Spanning. Let $𝐯 \in 𝑉$ . Since $𝐯_{1} + 𝑈, \dots, 𝐯_{𝑟} + 𝑈$ spans $𝑉 / 𝑈$ , there are scalars $𝑏_{𝑗}$ with

𝐯 + 𝑈 = \sum_{𝑗 = 1}^{𝑟} 𝑏_{𝑗} (𝐯_{𝑗} + 𝑈) = (\sum_{𝑗 = 1}^{𝑟} 𝑏_{𝑗} 𝐯_{𝑗}) + 𝑈 .

By Lemma 4.6.14, $𝐯 - \sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗} \in 𝑈$ , so it equals $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖}$ for some scalars $𝑎_{𝑖}$ , and $𝐯 =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} +$ $\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗}$ .

Independence. Suppose $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} +$ $\sum_{𝑗} 𝑏_{𝑗} 𝐯_{𝑗} =$ $𝟎$ . Apply the linear map $𝜋$ . Since $𝜋 (𝐮_{𝑖}) = 𝟎$ (as $ker 𝜋 = 𝑈$ ), we get $\sum_{𝑗} 𝑏_{𝑗} (𝐯_{𝑗} + 𝑈) =$ $𝟎 +$ $𝑈$ , so all $𝑏_{𝑗} = 0$ because the cosets $𝐯_{𝑗} + 𝑈$ are independent. Then $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} = 𝟎$ , so all $𝑎_{𝑖} = 0$ .

For the last statement, $𝑈$ and $𝑉 / 𝑈$ have bases (Corollary 1.6.24), and every element of $𝑉 / 𝑈$ is a coset $𝐯 + 𝑈$ , so the basis of $𝑉 / 𝑈$ has the form $𝐯_{1} + 𝑈, \dots, 𝐯_{𝑟} + 𝑈$ . By the first part, $𝑉$ has a basis of length $𝑘 + 𝑟$ . □

Quotients and complements

A complement of $𝑈$ (Proposition 4.4.16) and the quotient $𝑉 / 𝑈$ have the same dimension. They are related by the quotient map.

Proposition 4.6.29 (Complements and the quotient). Let

𝑈

and

𝑊

be subspaces of

𝑉

, and let

𝜋 : 𝑉 \to 𝑉 / 𝑈

be the quotient map. Then

𝜋 ↾_{𝑊} : 𝑊 \to 𝑉 / 𝑈

is injective if and only if

𝑈 \cap 𝑊 = {𝟎}

, and surjective if and only if

𝑈 + 𝑊 = 𝑉

. In particular,

𝜋 ↾_{𝑊}

is an isomorphism if and only if

𝑉 = 𝑈 \oplus 𝑊

, so

𝑉 / 𝑈 ≅ 𝑊

for every complement

𝑊

𝑈

Proof.standard

Uses. Proposition 4.2.3 (Properties of restriction), Proposition 4.6.26 (The quotient map), Theorem 2.6.11 (Injectivity is decided at $𝟎$ ), Lemma 4.6.14 (Equal cosets), Theorem 4.4.7 (Criteria for direct sums)

Techniques. Restrict to a subspace (Technique 4.2.5)

Hint 1. The kernel of a restriction is the intersection of the subspace with the kernel (Proposition 4.2.3(c)).

Hint 2.

𝜋 ↾_{𝑊}

is surjective when every coset

𝐯 + 𝑈

contains a vector of

𝑊

. Translate this with Lemma 4.6.14.

Sketch.

ker 𝜋 ↾_{𝑊} =

𝑊 \cap

ker 𝜋 =

𝑊 \cap

𝑈

. Surjectivity says that every

𝐯

has some

𝐰 \in 𝑊

with

𝐯 - 𝐰 \in 𝑈

, i.e.

𝐯 \in 𝑈 + 𝑊

. Both together say

𝑉 = 𝑈 \oplus 𝑊

(Theorem 4.4.7).

By Proposition 4.2.3(c) and $ker 𝜋 = 𝑈$ , we have $ker 𝜋 ↾_{𝑊} = 𝑊 \cap 𝑈$ , so $𝜋 ↾_{𝑊}$ is injective if and only if $𝑈 \cap 𝑊 = {𝟎}$ (Theorem 2.6.11).

The image of $𝜋 ↾_{𝑊}$ consists of the cosets $𝐰 + 𝑈$ with $𝐰 \in 𝑊$ . If $𝑈 + 𝑊 = 𝑉$ , then every $𝐯$ is $𝐮 + 𝐰$ with $𝐮 \in 𝑈$ , $𝐰 \in 𝑊$ , and $𝐯 + 𝑈 = 𝐰 + 𝑈$ since $𝐯 - 𝐰 \in 𝑈$ ; so $𝜋 ↾_{𝑊}$ is surjective. Conversely, if it is surjective, then for every $𝐯 \in 𝑉$ there is $𝐰 \in 𝑊$ with $𝐯 + 𝑈 = 𝐰 + 𝑈$ , so $𝐯 - 𝐰 \in 𝑈$ (Lemma 4.6.14) and $𝐯 =$ $(𝐯 - 𝐰) +$ $𝐰 \in$ $𝑈 +$ $𝑊$ .

Hence $𝜋 ↾_{𝑊}$ is bijective if and only if $𝑈 \cap 𝑊 = {𝟎}$ and $𝑈 + 𝑊 = 𝑉$ , which by Theorem 4.4.7 says that $𝑉 = 𝑈 \oplus 𝑊$ . A bijective linear map is an isomorphism. □

Intuition (A complement is a choice; the quotient is not). The proposition says that a complement $𝑊$ meets every coset of $𝑈$ in exactly one vector. So a complement amounts to choosing one representative from each coset, in such a way that the chosen vectors form a subspace. In the plane, every line through $𝟎$ other than $𝑈$ meets each line parallel to $𝑈$ exactly once, and none of these lines is preferred. The quotient is the same space with the choice removed. This is why quotients, not complements, are the tool when a construction must not depend on choices, as in the first isomorphism theorem below, or when complements are hard to produce: in infinite dimensions their existence in general needs Zorn’s lemma (Section A.4), while $𝑉 / 𝑈$ always exists.

With Proposition 4.4.16(b) the proposition also gives a third proof of Theorem 4.6.27: $dim (𝑉 / 𝑈) =$ $dim 𝑊 =$ $dim 𝑉 -$ $dim 𝑈$ .

4.6.6Linear maps out of a quotient

Intuition (Maps that kill

𝑈

). A linear map out of

𝑉 / 𝑈

composed with

𝜋

is a linear map out of

𝑉

that sends all of

𝑈

𝟎

, because

𝜋

does. Conversely, let

𝑇 : 𝑉 \to 𝑊

be linear with

𝑇 (𝑈) = {𝟎}

. If

𝐯

and

𝐯^{'}

lie in the same coset, then

𝑇 (𝐯) -

𝑇 (𝐯^{'}) =

𝑇 (𝐯 - 𝐯^{'}) =

𝟎

, so

𝑇

is constant on each coset and makes sense as a function of the coset. So linear maps out of

𝑉 / 𝑈

are the same thing as linear maps out of

𝑉

that vanish on

𝑈

Theorem 4.6.30 (Universal property of the quotient). Let $𝑈$ be a subspace of $𝑉$ , let $𝜋 : 𝑉 \to 𝑉 / 𝑈$ be the quotient map, and let $𝑇 \in L (𝑉, 𝑊)$ .

(a)If $𝑈 \subseteq ker 𝑇$ , there is exactly one linear map $\tilde{𝑇} : 𝑉 / 𝑈 \to 𝑊$ with $\tilde{𝑇} \circ 𝜋 = 𝑇$ , namely
$\tilde{𝑇} (𝐯 + 𝑈) = 𝑇 (𝐯) .$
Its image is $im \tilde{𝑇} = im 𝑇$ , and its kernel is $ker \tilde{𝑇} =$ ${𝐯 + 𝑈 | 𝐯 \in ker 𝑇}$ .
(b)Conversely, for every $𝑆 \in L (𝑉 / 𝑈, 𝑊)$ the map $𝑆 \circ 𝜋 \in L (𝑉, 𝑊)$ vanishes on $𝑈$ .

The map $\tilde{𝑇}$ is the map induced by $𝑇$ . The equation $\tilde{𝑇} \circ 𝜋 = 𝑇$ is often drawn as a triangle:

\begin{matrix} 𝑉 & \overset{𝜋}{⟶} & 𝑉 / 𝑈 \\ 𝑇 ↘︎ & ↓ \tilde{𝑇} \\ 𝑊 \end{matrix}

Proof.standard

Uses. Proposition 4.6.26 (The quotient map), Lemma 4.6.14 (Equal cosets), Theorem 2.1.3 (Checking linearity), Definition 4.6.20 (Quotient space)

Techniques. Check that a definition does not depend on the representative (Technique 4.6.18), Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Uniqueness first: two maps

𝑆, 𝑆^{'}

with

𝑆 \circ 𝜋 = 𝑆^{'} \circ 𝜋

agree at every

𝜋 (𝐯)

. Why is that everything?

Hint 2. For existence, the formula

\tilde{𝑇} (𝐯 + 𝑈) = 𝑇 (𝐯)

is forced. Check that it is well defined: if

𝐯 - 𝐯^{'} \in 𝑈

, why is

𝑇 (𝐯) = 𝑇 (𝐯^{'})

Hint 3. Linearity: compute

\tilde{𝑇} (𝑎 (𝐯 + 𝑈) + (𝐰 + 𝑈))

by first combining the cosets. The image and kernel come from

\tilde{𝑇} (𝐯 + 𝑈) = 𝑇 (𝐯)

Sketch. Since

𝜋

is surjective,

𝑆 \circ 𝜋 = 𝑇

determines

𝑆

at every coset, which gives uniqueness and the formula. If

𝐯 - 𝐯^{'} \in 𝑈 \subseteq ker 𝑇

, then

𝑇 (𝐯) = 𝑇 (𝐯^{'})

, so the formula is well defined; it is linear because

𝑎 (𝐯 + 𝑈) +

(𝐰 + 𝑈) =

(𝑎 𝐯 + 𝐰) +

𝑈

and

𝑇

is linear. The values of

\tilde{𝑇}

are the values of

𝑇

, and

\tilde{𝑇} (𝐯 + 𝑈) = 𝟎

iff

𝐯 \in ker 𝑇

. (b)

𝑆 (𝜋 (𝐮)) =

𝑆 (𝟎) =

𝟎

for

𝐮 \in 𝑈

(a) Uniqueness. If $𝑆 \circ 𝜋 = 𝑇$ , then $𝑆 (𝐯 + 𝑈) =$ $𝑆 (𝜋 (𝐯)) =$ $𝑇 (𝐯)$ for every $𝐯 \in 𝑉$ . Every element of $𝑉 / 𝑈$ is $𝐯 + 𝑈$ for some $𝐯$ (Proposition 4.6.26), so $𝑆$ is determined by $𝑇$ : it must be given by the formula in the statement.

Well defined. Suppose $𝐯 + 𝑈 = 𝐯^{'} + 𝑈$ . Then $𝐯 - 𝐯^{'} \in 𝑈 \subseteq ker 𝑇$ (Lemma 4.6.14), so $𝑇 (𝐯) -$ $𝑇 (𝐯^{'}) =$ $𝑇 (𝐯 - 𝐯^{'}) =$ $𝟎$ . Hence the formula defines a function $\tilde{𝑇} : 𝑉 / 𝑈 \to 𝑊$ .

Linear. For $𝐯, 𝐰 \in 𝑉$ and $𝑎 \in 𝐹$ ,

\tilde{𝑇} (𝑎 (𝐯 + 𝑈) + (𝐰 + 𝑈)) = \tilde{𝑇} ((𝑎 𝐯 + 𝐰) + 𝑈) = 𝑇 (𝑎 𝐯 + 𝐰) = 𝑎 \tilde{𝑇} (𝐯 + 𝑈) + \tilde{𝑇} (𝐰 + 𝑈) .

By construction $\tilde{𝑇} (𝜋 (𝐯)) =$ $𝑇 (𝐯)$ , so $\tilde{𝑇} \circ 𝜋 = 𝑇$ .

Image and kernel. The values of $\tilde{𝑇}$ are the vectors $𝑇 (𝐯)$ , $𝐯 \in 𝑉$ , so $im \tilde{𝑇} = im 𝑇$ . And $\tilde{𝑇} (𝐯 + 𝑈) = 𝟎$ if and only if $𝑇 (𝐯) = 𝟎$ , so $ker \tilde{𝑇} =$ ${𝐯 + 𝑈 | 𝐯 \in ker 𝑇}$ . (This is well stated: whether $𝐯 \in ker 𝑇$ does not depend on the representative, since $ker 𝑇 \supseteq 𝑈$ .)

(b) $𝑆 \circ 𝜋$ is linear (Proposition 2.1.17), and for $𝐮 \in 𝑈$ , $(𝑆 \circ 𝜋) (𝐮) =$ $𝑆 (𝟎 + 𝑈) =$ $𝟎$ , since $𝐮 + 𝑈 = 𝟎 + 𝑈$ . □

So $𝑇 \mapsto \tilde{𝑇}$ and $𝑆 \mapsto 𝑆 \circ 𝜋$ are inverse bijections between the linear maps $𝑉 \to 𝑊$ that vanish on $𝑈$ and the linear maps $𝑉 / 𝑈 \to 𝑊$ . Note how the hypothesis $𝑈 \subseteq ker 𝑇$ entered: it is exactly what the well-definedness check needed. When $𝑈$ is not inside $ker 𝑇$ , there is no induced map at all. This is the quotient version of Exercise 2.6.43, with $𝜋$ as the surjection.

Technique 4.6.31 (Pass to a quotient). When a construction only depends on vectors up to elements of a subspace $𝑈$ , replace $𝑉$ by $𝑉 / 𝑈$ . By Theorem 4.6.30, a linear map that kills $𝑈$ is the same as a linear map on $𝑉 / 𝑈$ , and by Theorem 4.6.27 and Proposition 4.6.28, dimensions split as $dim 𝑉 =$ $dim 𝑈 +$ $dim (𝑉 / 𝑈)$ . Two typical uses:

Make a map injective. Dividing out exactly the kernel turns $𝑇$ into an injective map with the same image. This is the first isomorphism theorem (Theorem 4.6.34), and it is the standard way to show that some space is isomorphic to an image.
Split a problem in two. If an operator $𝑇$ maps $𝑈$ into itself, it acts on $𝑈$ and on $𝑉 / 𝑈$ (Proposition 4.6.32), and a statement about $𝑇$ often follows from the same statement for these two smaller operators, by induction on $dim 𝑉$ . Section 7.5 triangularizes operators this way.

The price is the well-definedness check (Technique 4.6.18) for every map defined on cosets.

Used in 17 places and 6 practice exercises

Proposition 4.6.32, Theorem 4.6.34, Corollary 4.6.35, Exercise 4.6.37, Exercise 4.6.42, Exercise 4.6.44, Exercise 4.6.45, Exercise 4.6.46, Exercise 4.6.47, Exercise 4.6.49, Exercise 4.6.50, Exercise 4.6.52, Exercise 4.6.53, Exercise 4.6.55, Exercise 4.8.14, Problem 4.18, Theorem 7.6.32; 6 practice exercises.

Operators on a quotient

Proposition 4.6.32 (Operators on a quotient). Let $𝑇 \in L (𝑉)$ and let $𝑈$ be a subspace of $𝑉$ with $𝑇 (𝐮) \in 𝑈$ for every $𝐮 \in 𝑈$ (that is, $𝑈$ is invariant under $𝑇$ ; see Section 7.2).

(a)There is exactly one $𝑇 / 𝑈 \in L (𝑉 / 𝑈)$ with $(𝑇 / 𝑈) (𝐯 + 𝑈) =$ $𝑇 (𝐯) +$ $𝑈$ for all $𝐯 \in 𝑉$ .
(b)Suppose $𝑉$ is finite-dimensional, let $𝒞︀ = 𝐮_{1}, \dots, 𝐮_{𝑘}$ be a basis of $𝑈$ , extend it to a basis $ℬ︀ =$ $𝐮_{1},$ $\dots,$ $𝐮_{𝑘},$ $𝐯_{1},$ $\dots,$ $𝐯_{𝑟}$ of $𝑉$ , and let $𝒟︀ =$ $𝐯_{1} +$ $𝑈,$ $\dots,$ $𝐯_{𝑟} +$ $𝑈$ (a basis of $𝑉 / 𝑈$ by Theorem 4.6.27). Regard $𝑇 ↾_{𝑈}$ as an operator on $𝑈$ . Then
${[𝑇]}_{ℬ︀ ℬ︀} = (\begin{matrix} {[𝑇 ↾_{𝑈}]}_{𝒞︀ 𝒞︀} & 𝐵 \\ 0 & {[𝑇 / 𝑈]}_{𝒟︀ 𝒟︀} \end{matrix})$
for some $𝐵 \in 𝐹^{𝑘 \times 𝑟}$ .

Proof.standard

Uses. Theorem 4.6.30 (Universal property of the quotient), Proposition 4.6.26 (The quotient map), Proposition 2.1.17 (Compositions of linear maps are linear), Definition 2.3.1 (Matrix of a linear map), Definition 2.4.33 (Block matrix)

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1. For (a), apply Theorem 4.6.30 to the composite

𝜋 \circ 𝑇 : 𝑉 \to 𝑉 / 𝑈

. What is its kernel?

Hint 2. For (b), read the matrix column by column. Where does

𝑇 (𝐮_{𝑗})

live? Expand

𝑇 (𝐯_{𝑗})

ℬ︀

and apply

𝜋

Sketch. (a)

𝜋 \circ 𝑇

is linear and kills

𝑈

, because

𝑇 (𝑈) \subseteq 𝑈 = ker 𝜋

, so it induces

𝑇 / 𝑈

, which is unique by Theorem 4.6.30. (b)

𝑇 (𝐮_{𝑗}) \in 𝑈

, so the first

𝑘

columns have zeros below row

𝑘

, and their top parts are the columns of

{[𝑇 ↾_{𝑈}]}_{𝒞︀ 𝒞︀}

. If

𝑇 (𝐯_{𝑗}) =

\sum_{𝑖} 𝑏_{𝑖 𝑗} 𝐮_{𝑖} +

\sum_{𝑖} 𝑐_{𝑖 𝑗} 𝐯_{𝑖}

, then

(𝑇 / 𝑈) (𝐯_{𝑗} + 𝑈) =

\sum_{𝑖} 𝑐_{𝑖 𝑗} (𝐯_{𝑖} + 𝑈)

, so the lower right block is

{[𝑇 / 𝑈]}_{𝒟︀ 𝒟︀}

(a) The composite $𝜋 \circ 𝑇 : 𝑉 \to 𝑉 / 𝑈$ is linear (Proposition 2.1.17). For $𝐮 \in 𝑈$ , $𝑇 (𝐮) \in 𝑈 = ker 𝜋$ , so $(𝜋 \circ 𝑇) (𝐮) = 𝟎$ . By Theorem 4.6.30(a), there is exactly one linear $𝑇 / 𝑈 : 𝑉 / 𝑈 \to 𝑉 / 𝑈$ with $(𝑇 / 𝑈) \circ 𝜋 = 𝜋 \circ 𝑇$ , that is, $(𝑇 / 𝑈) (𝐯 + 𝑈) =$ $𝑇 (𝐯) +$ $𝑈$ .

(b) Column $𝑗$ of ${[𝑇]}_{ℬ︀ ℬ︀}$ is the coordinate vector of the image of the $𝑗$ th basis vector (Definition 2.3.1). For $𝑗 \leq 𝑘$ , the vector $𝑇 (𝐮_{𝑗})$ lies in $𝑈$ , so it is a combination of $𝐮_{1}, \dots, 𝐮_{𝑘}$ alone: its last $𝑟$ coordinates are $0$ , and its first $𝑘$ coordinates form column $𝑗$ of ${[𝑇 ↾_{𝑈}]}_{𝒞︀ 𝒞︀}$ . For $𝑗 \leq 𝑟$ , write

𝑇 (𝐯_{𝑗}) = \sum_{𝑖 = 1}^{𝑘} 𝑏_{𝑖 𝑗} 𝐮_{𝑖} + \sum_{𝑖 = 1}^{𝑟} 𝑐_{𝑖 𝑗} 𝐯_{𝑖} .

Applying $𝜋$ , which kills the $𝐮_{𝑖}$ ,

(𝑇 / 𝑈) (𝐯_{𝑗} + 𝑈) = 𝑇 (𝐯_{𝑗}) + 𝑈 = \sum_{𝑖 = 1}^{𝑟} 𝑐_{𝑖 𝑗} (𝐯_{𝑖} + 𝑈),

so $(𝑐_{𝑖 𝑗})$ is ${[𝑇 / 𝑈]}_{𝒟︀ 𝒟︀}$ . Together these give the block form, with $𝐵 = (𝑏_{𝑖 𝑗})$ . □

Example 4.6.33 (A shift and its quotient). Let $𝑁 \in L (𝐹^{3})$ be the shift $𝑁 (𝐞_{1}) = 𝟎$ , $𝑁 (𝐞_{2}) = 𝐞_{1}$ , $𝑁 (𝐞_{3}) = 𝐞_{2}$ , and let $𝑈 = span (𝐞_{1})$ , which $𝑁$ maps into itself (to ${𝟎}$ ). On $𝐹^{3} / 𝑈$ , with basis $𝐞_{2} + 𝑈, 𝐞_{3} + 𝑈$ ,

(𝑁 / 𝑈) (𝐞_{2} + 𝑈) = 𝐞_{1} + 𝑈 = 𝟎 + 𝑈, (𝑁 / 𝑈) (𝐞_{3} + 𝑈) = 𝐞_{2} + 𝑈 .

So $𝑁 / 𝑈$ is the shift of a two-dimensional space, and in the standard basis

[𝑁] = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}), [𝑁 ↾_{𝑈}] = (\begin{matrix} 0 \end{matrix}), 𝐵 = (\begin{matrix} 1 & 0 \end{matrix}), [𝑁 / 𝑈] = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) .

Collapsing the last vector of the chain $𝐞_{3} \mapsto 𝐞_{2} \mapsto 𝐞_{1} \mapsto 𝟎$ leaves a chain one step shorter.

4.6.7The first isomorphism theorem

Intuition (Each slice becomes one vector). Return to the slices of Corollary 2.6.31. A linear map

𝑇

sends each coset of

ker 𝑇

to a single vector, and different cosets to different vectors. So

𝑇

is a bijection from the set of cosets of

ker 𝑇

onto

im 𝑇

. The quotient

𝑉 / ker 𝑇

makes that set of cosets into a vector space, and the bijection is linear.

Theorem 4.6.34 (First isomorphism theorem). Let $𝑇 \in L (𝑉, 𝑊)$ . The map

\tilde{𝑇} : 𝑉 / ker 𝑇 \to im 𝑇, \tilde{𝑇} (𝐯 + ker 𝑇) = 𝑇 (𝐯),

is well defined and is an isomorphism. In particular, $𝑉 / ker 𝑇 ≅ im 𝑇$ .

The first proof is an application of the universal property. The second proves well-definedness and injectivity at once, with a single chain of equivalences read in both directions.

Proof (by the universal property).routine

Uses. Theorem 4.6.30 (Universal property of the quotient), Proposition 2.7.7 (Bijective linear maps are invertible), Theorem 2.6.11 (Injectivity is decided at $𝟎$ )

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1. Apply Theorem 4.6.30 with

𝑈 = ker 𝑇

. What are the image and kernel of the induced map?

Hint 2. The kernel of

\tilde{𝑇}

is the set of cosets

𝐯 + ker 𝑇

with

𝐯 \in ker 𝑇

. How many different cosets is that?

Sketch. With

𝑈 = ker 𝑇

, Theorem 4.6.30 gives a linear

\tilde{𝑇}

with

im \tilde{𝑇} = im 𝑇

and

ker \tilde{𝑇} =

{𝐯 + ker 𝑇 | 𝐯 \in ker 𝑇} =

{ker 𝑇}

, the zero coset. So

\tilde{𝑇}

is injective, and as a map onto

im 𝑇

it is bijective.

Apply Theorem 4.6.30(a) with

𝑈 = ker 𝑇

; the hypothesis

𝑈 \subseteq ker 𝑇

holds trivially. It gives a well-defined linear map

\tilde{𝑇} : 𝑉 / ker 𝑇 \to 𝑊

with

\tilde{𝑇} (𝐯 + ker 𝑇) =

𝑇 (𝐯)

, image

im 𝑇

, and kernel

{𝐯 + ker 𝑇 | 𝐯 \in ker 𝑇}

. For

𝐯 \in ker 𝑇

we have

𝐯 + ker 𝑇 = ker 𝑇

, the zero vector of

𝑉 / ker 𝑇

(Corollary 4.6.15(c)). So

ker \tilde{𝑇}

is the zero subspace and

\tilde{𝑇}

is injective (Theorem 2.6.11). Regarded as a map into

im 𝑇

, it is also surjective, so it is a bijective linear map, hence an isomorphism (Proposition 2.7.7). □

Proof (by one chain of equivalences).routine

Uses. Lemma 4.6.14 (Equal cosets), Definition 2.6.5 (Kernel), Definition 4.6.20 (Quotient space), Proposition 2.7.7 (Bijective linear maps are invertible)

Techniques. Check that a definition does not depend on the representative (Technique 4.6.18)

Hint 1. For

𝐯, 𝐯^{'} \in 𝑉

, connect “

𝐯 +

ker 𝑇 =

𝐯^{'} +

ker 𝑇

” with “

𝑇 (𝐯) = 𝑇 (𝐯^{'})

” by a chain of equivalences.

Hint 2. Read the chain left to right: that is well-definedness. Read it right to left: that is injectivity.

Sketch.

𝐯 +

ker 𝑇 =

𝐯^{'} +

ker 𝑇

iff

𝐯 - 𝐯^{'} \in ker 𝑇

iff

𝑇 (𝐯) = 𝑇 (𝐯^{'})

. The forward implications make

\tilde{𝑇}

well defined, the backward ones make it injective. Linearity and surjectivity onto

im 𝑇

are immediate.

For $𝐯, 𝐯^{'} \in 𝑉$ , by Lemma 4.6.14 and linearity,

𝐯 + ker 𝑇 = 𝐯^{'} + ker 𝑇 \Leftrightarrow 𝐯 - 𝐯^{'} \in ker 𝑇 \Leftrightarrow 𝑇 (𝐯 - 𝐯^{'}) = 𝟎 \Leftrightarrow 𝑇 (𝐯) = 𝑇 (𝐯^{'}) .

Read from left to right, this says that the formula for $\tilde{𝑇}$ gives the same value for every representative, so $\tilde{𝑇}$ is well defined (Technique 4.6.18). Read from right to left, it says that $\tilde{𝑇}$ is injective. Linearity is the computation

\tilde{𝑇} (𝑎 (𝐯 + ker 𝑇) + (𝐰 + ker 𝑇)) = \tilde{𝑇} ((𝑎 𝐯 + 𝐰) + ker 𝑇) = 𝑇 (𝑎 𝐯 + 𝐰) = 𝑎 𝑇 (𝐯) + 𝑇 (𝐰),

and $\tilde{𝑇}$ is onto $im 𝑇$ because $𝑇 (𝐯) =$ $\tilde{𝑇} (𝐯 + ker 𝑇)$ . So $\tilde{𝑇}$ is a bijective linear map onto $im 𝑇$ , hence an isomorphism (Proposition 2.7.7). □

The second proof shows what is special about $𝑈 = ker 𝑇$ . For a subspace $𝑈$ strictly inside $ker 𝑇$ , the chain acquires a one-way step, “ $𝐯 - 𝐯^{'} \in 𝑈$ $\Rightarrow$ $𝐯 - 𝐯^{'} \in ker 𝑇$ ”. The induced map is still well defined, but it is no longer injective.

Corollary 4.6.35 (Rank–nullity, a third proof). Let $𝑉$ be finite-dimensional and $𝑇 \in L (𝑉, 𝑊)$ . Then $im 𝑇$ is finite-dimensional, and

dim 𝑉 = dim ker 𝑇 + dim im 𝑇 .

Proof.routine

Uses. Theorem 4.6.34 (First isomorphism theorem), Theorem 4.6.27 (Dimension of a quotient), Proposition 4.2.6 (Injective maps preserve dimension), Proposition 4.1.21 (Dimensions of subspaces)

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1.

im 𝑇

is isomorphic to a quotient of

𝑉

. What is the dimension of that quotient?

Hint 2. Use the first proof of Theorem 4.6.27. Using the second would be circular.

Sketch.

𝑉 / ker 𝑇

is finite-dimensional of dimension

dim 𝑉 - dim ker 𝑇

(Theorem 4.6.27, by extending a basis). The isomorphism

\tilde{𝑇}

carries it onto

im 𝑇

, so

im 𝑇

is finite-dimensional of the same dimension.

By Proposition 4.1.21,

ker 𝑇

is a finite-dimensional subspace of

𝑉

, and by Theorem 4.6.27 (first proof),

𝑉 / ker 𝑇

is finite-dimensional with

dim (𝑉 / ker 𝑇) =

dim 𝑉 -

dim ker 𝑇

. By Theorem 4.6.34,

\tilde{𝑇}

is an injective linear map on

𝑉 / ker 𝑇

with image

im 𝑇

. By Proposition 4.2.6(a),

im 𝑇

is finite-dimensional and

dim im 𝑇 =

dim (𝑉 / ker 𝑇) =

dim 𝑉 -

dim ker 𝑇

. □

The second proof of Theorem 4.6.27 used rank–nullity, so citing it here would be circular; the first used only Theorem 4.1.16. Unwound, this proof is close to the proof of Theorem 4.2.11 by extending a basis of the kernel: the basis $𝐯_{1} +$ $ker 𝑇,$ $\dots,$ $𝐯_{𝑟} +$ $ker 𝑇$ of $𝑉 / ker 𝑇$ is carried by $\tilde{𝑇}$ to $𝑇 (𝐯_{1}),$ $\dots,$ $𝑇 (𝐯_{𝑟})$ , the basis of $im 𝑇$ found there. What the quotient adds is that the space being counted, $𝑉 / ker 𝑇$ , no longer depends on the choice of the $𝐯_{𝑗}$ (see Remark 4.2.12).

Example 4.6.36 (The first isomorphism theorem at work).

(a)Evaluation. For $𝑐 \in 𝐹$ , the evaluation map ${ev}_{𝑐} : 𝐹 [𝑥] \to 𝐹$ , $𝑝 \mapsto 𝑝 (𝑐)$ (Example 2.1.9), is linear and surjective (it sends the constant polynomial $𝑎$ to $𝑎$ ). Its kernel is $𝑈_{𝑐} =$ ${𝑝 | 𝑝 (𝑐) = 0}$ . So $𝐹 [𝑥] / 𝑈_{𝑐} ≅ 𝐹$ , via $𝑝 + 𝑈_{𝑐} \mapsto 𝑝 (𝑐)$ : two polynomials lie in the same coset exactly when they agree at $𝑐$ . An infinite-dimensional space can have a one-dimensional quotient.
(b)Trace. The trace $tr : 𝐹^{𝑛 \times 𝑛} \to 𝐹$ is linear and surjective, since $tr (𝑈_{11}) = 1$ for the matrix unit $𝑈_{11}$ (Lemma 2.4.24). So $𝐹^{𝑛 \times 𝑛} / ker tr ≅ 𝐹$ , and by Theorem 4.6.27 the trace-zero matrices form a subspace of dimension $𝑛^{2} - 1$ . The coset of $𝐴$ is determined by $tr 𝐴$ .
(c)Fibers. For $𝐴 = (\begin{matrix} 1 & 2 \\ 1 & 2 \end{matrix})$ as in Figure 2.6.1, the cosets of $ker 𝑇_{𝐴}$ are the lines $𝑥_{1} + 2 𝑥_{2} = 𝑡$ , and $\tilde{𝑇_{𝐴}}$ sends the line $𝑥_{1} + 2 𝑥_{2} = 𝑡$ to $𝑡 (1, 1)$ . So $ℝ^{2} / ker 𝑇_{𝐴} ≅$ $span ((1, 1))$ .
(d)A quotient isomorphic to the whole space. The backward shift $𝐵 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ ,
$𝐵 (𝑎_{0} + 𝑎_{1} 𝑥 + \dots + 𝑎_{𝑛} 𝑥^{𝑛}) = 𝑎_{1} + 𝑎_{2} 𝑥 + \dots + 𝑎_{𝑛} 𝑥^{𝑛 - 1},$
is linear (it acts on coefficient sequences by deleting the first entry) and surjective, since $𝐵 (𝑥 𝑝) = 𝑝$ . Its kernel is the set of constant polynomials, $span (1)$ . So $𝐹 [𝑥] / span (1) ≅$ $𝐹 [𝑥]$ : dividing out a nonzero subspace gave back a space isomorphic to the original. In finite dimensions this is impossible, because $dim (𝑉 / 𝑈) =$ $dim 𝑉 -$ $dim 𝑈 <$ $dim 𝑉$ when $𝑈 \neq {𝟎}$ .

Exercise 4.6.37.standard Let $𝑈 =$ ${𝑝 \in 𝐹 [𝑥] | 𝑝 (1) = 𝑝 (2) = 0}$ .

(a)Show that $𝐹 [𝑥] / 𝑈 ≅ 𝐹^{2}$ , and that $𝑝 + 𝑈 = 𝑞 + 𝑈$ if and only if $𝑝 (1) = 𝑞 (1)$ and $𝑝 (2) = 𝑞 (2)$ .
(b)Show that $1 + 𝑈, 𝑥 + 𝑈$ is a basis of $𝐹 [𝑥] / 𝑈$ , and write $𝑥^{2} + 𝑈$ in this basis. What does your answer say over $𝔽_{2}$ ?

Solution.

Uses. Theorem 4.6.34 (First isomorphism theorem), Lemma 4.6.14 (Equal cosets), Proposition 1.2.18 (Degree and evaluation versus the operations)

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1.

𝑈

is the kernel of

𝐸 : 𝐹 [𝑥] \to 𝐹^{2}

𝐸 (𝑝) =

(𝑝 (1), 𝑝 (2))

. Is

𝐸

surjective? Try polynomials of degree at most

1

Hint 2. An isomorphism carries bases to bases, so compute

𝐸 (1)

and

𝐸 (𝑥)

. For

𝑥^{2} + 𝑈

, solve

𝐸 (𝑥^{2}) =

𝑎 𝐸 (1) +

𝑏 𝐸 (𝑥)

Sketch.

𝐸

is linear with kernel

𝑈

, and surjective:

𝑎 +

(𝑏 - 𝑎) (𝑥 - 1) \mapsto

(𝑎, 𝑏)

. The first isomorphism theorem gives

𝐹 [𝑥] / 𝑈 ≅ 𝐹^{2}

, and

𝑝 + 𝑈 = 𝑞 + 𝑈

iff

𝑝 - 𝑞 \in 𝑈

iff

𝐸 (𝑝) = 𝐸 (𝑞)

\tilde{𝐸}

sends

1 + 𝑈, 𝑥 + 𝑈

(1, 1), (1, 2)

, a basis. From

(1, 4) =

- 2 (1, 1) +

3 (1, 2)

𝑥^{2} +

𝑈 =

- 2 (1 + 𝑈) +

3 (𝑥 + 𝑈)

. Over

𝔽_{2}

this reads

𝑥^{2} + 𝑈 = 𝑥 + 𝑈

(a) Let $𝐸 : 𝐹 [𝑥] \to 𝐹^{2}$ , $𝐸 (𝑝) =$ $(𝑝 (1), 𝑝 (2))$ . It is linear by Proposition 1.2.18, and $ker 𝐸 = 𝑈$ . It is surjective: for $(𝑎, 𝑏) \in 𝐹^{2}$ , the polynomial $𝑝 =$ $𝑎 +$ $(𝑏 - 𝑎) (𝑥 - 1)$ has $𝑝 (1) = 𝑎$ and $𝑝 (2) =$ $𝑎 +$ $(𝑏 - 𝑎) =$ $𝑏$ . By Theorem 4.6.34, $\tilde{𝐸} : 𝐹 [𝑥] / 𝑈 \to 𝐹^{2}$ , $𝑝 +$ $𝑈 \mapsto$ $(𝑝 (1), 𝑝 (2))$ , is an isomorphism. Moreover $𝑝 + 𝑈 = 𝑞 + 𝑈$ if and only if $𝑝 - 𝑞 \in 𝑈$ (Lemma 4.6.14), that is, $𝑝 (1) - 𝑞 (1) = 0$ and $𝑝 (2) - 𝑞 (2) = 0$ .

(b) $\tilde{𝐸} (1 + 𝑈) = (1, 1)$ and $\tilde{𝐸} (𝑥 + 𝑈) = (1, 2)$ . These are independent, since $(1, 2) - (1, 1) = (0, 1)$ , so together with $(1, 1)$ they span $𝐹^{2}$ ; hence they form a basis of $𝐹^{2}$ , and their preimages $1 + 𝑈, 𝑥 + 𝑈$ form a basis of $𝐹 [𝑥] / 𝑈$ ( ${\tilde{𝐸}}^{- 1}$ is an isomorphism). Next, $\tilde{𝐸} (𝑥^{2} + 𝑈) = (1, 4)$ , and $𝑎 (1, 1) +$ $𝑏 (1, 2) =$ $(1, 4)$ gives $𝑎 + 𝑏 = 1$ , $𝑎 + 2 𝑏 = 4$ , so $𝑏 = 3$ and $𝑎 = - 2$ . Hence

𝑥^{2} + 𝑈 = - 2 (1 + 𝑈) + 3 (𝑥 + 𝑈) = (3 𝑥 - 2) + 𝑈 .

Check: $𝑥^{2} -$ $(3 𝑥 - 2) =$ $𝑥^{2} -$ $3 𝑥 +$ $2$ vanishes at $1$ and at $2$ . Over $𝔽_{2}$ we have $3 = 1$ and $2 = 0$ , so the answer reads $𝑥^{2} + 𝑈 = 𝑥 + 𝑈$ : there $2 = 0$ , so $𝑈$ is the set of polynomials vanishing at $0$ and $1$ , and $𝑥^{2} - 𝑥$ is one of them.

Exercises

Exercise 4.6.38.routine Let $𝑉 = {𝐹 [𝑥]}_{\leq 2}$ .

(a)Write down a basis of $𝑉 \times 𝑉$ and find $dim (𝑉 \times 𝑉)$ .
(b)Let $𝐷 =$ ${(𝑝, 𝑞) \in 𝑉 \times 𝑉 | 𝑝 (0) = 𝑞 (1)}$ . Show that $𝐷$ is a subspace of $𝑉 \times 𝑉$ , find $dim 𝐷$ , and find a basis of $𝐷$ .

Solution.

Uses. Theorem 4.6.5 (Dimension of a product), Proposition 4.6.7 (Maps into and out of a product), Theorem 4.2.11 (Rank–nullity), Theorem 4.1.10 (Independent lists are short)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. For (a), use Theorem 4.6.5 with the basis

1, 𝑥, 𝑥^{2}

of each factor.

Hint 2. For (b),

𝐷

is the kernel of

(𝑝, 𝑞) \mapsto

𝑝 (0) -

𝑞 (1)

, a linear map

𝑉 \times 𝑉 \to 𝐹

. Find its rank, then guess five independent elements of

𝐷

Sketch. (a)

(1, 𝟎),

(𝑥, 𝟎),

(𝑥^{2}, 𝟎),

(𝟎, 1),

(𝟎, 𝑥),

(𝟎, 𝑥^{2})

; dimension

6

. (b)

𝐷 = ker 𝜑

for the surjective

𝜑 (𝑝, 𝑞) =

𝑝 (0) -

𝑞 (1)

, so

dim 𝐷 = 5

. A basis:

(𝑥, 𝟎),

(𝑥^{2}, 𝟎),

(𝟎, 𝑥 - 1),

(𝟎, 𝑥^{2} - 1),

(1, 1)

(a) By Theorem 4.6.5, applied to the basis $1, 𝑥, 𝑥^{2}$ of $𝑉$ in each factor,

(1, 𝟎), (𝑥, 𝟎), (𝑥^{2}, 𝟎), (𝟎, 1), (𝟎, 𝑥), (𝟎, 𝑥^{2})

is a basis of $𝑉 \times 𝑉$ , which therefore has dimension $6$ .

(b) Let $𝜑 : 𝑉 \times 𝑉 \to 𝐹$ , $𝜑 (𝑝, 𝑞) =$ $𝑝 (0) -$ $𝑞 (1)$ . By Proposition 4.6.7(b), with $𝑆_{1} = {ev}_{0}$ and $𝑆_{2} = - {ev}_{1}$ , it is linear, and $𝐷 = ker 𝜑$ is a subspace. Since $𝜑 (1, 𝟎) = 1$ , the image is a nonzero subspace of $𝐹$ , hence all of $𝐹$ . By rank–nullity, $dim 𝐷 = 6 - 1 = 5$ . The five pairs

(𝑥, 𝟎), (𝑥^{2}, 𝟎), (𝟎, 𝑥 - 1), (𝟎, 𝑥^{2} - 1), (1, 1)

lie in $𝐷$ (for each, $𝑝 (0) = 𝑞 (1)$ ). They are independent: if $𝑎 (𝑥, 𝟎) +$ $𝑏 (𝑥^{2}, 𝟎) +$ $𝑐 (𝟎, 𝑥 - 1) +$ $𝑑 (𝟎, 𝑥^{2} - 1) +$ $𝑒 (1, 1) =$ $𝟎$ , the first component $𝑒 + 𝑎 𝑥 + 𝑏 𝑥^{2}$ is the zero polynomial, so $𝑎 = 𝑏 = 𝑒 = 0$ , and then the second component $𝑐 (𝑥 - 1) +$ $𝑑 (𝑥^{2} - 1) =$ $- (𝑐 + 𝑑) +$ $𝑐 𝑥 +$ $𝑑 𝑥^{2}$ vanishes, so $𝑐 = 𝑑 = 0$ . Five independent vectors in the five-dimensional space $𝐷$ form a basis (Theorem 4.1.10(b)).

Exercise 4.6.39.standard Let $𝑇 \in L (𝑉, 𝑊)$ . The graph of $𝑇$ is $𝐺_{𝑇} =$ ${(𝐯, 𝑇 (𝐯)) | 𝐯 \in 𝑉} \subseteq$ $𝑉 \times$ $𝑊$ .

(a)Show that $𝐺_{𝑇}$ is a subspace of $𝑉 \times 𝑊$ and that $𝐯 \mapsto (𝐯, 𝑇 (𝐯))$ is an isomorphism $𝑉 \to 𝐺_{𝑇}$ .
(b)Show that $𝑉 \times$ $𝑊 =$ $𝐺_{𝑇} \oplus$ $({𝟎} \times 𝑊)$ , and deduce that $(𝑉 \times 𝑊) / 𝐺_{𝑇} ≅ 𝑊$ .
(c)Let $𝐺$ be any subspace of $𝑉 \times 𝑊$ . Show that $𝐺 = 𝐺_{𝑆}$ for some $𝑆 \in L (𝑉, 𝑊)$ if and only if $𝐺 \cap ({𝟎} \times 𝑊) = {𝟎}$ and every $𝐯 \in 𝑉$ is the first component of some element of $𝐺$ .

Solution.

Uses. Proposition 4.6.7 (Maps into and out of a product), Proposition 2.6.8 (Kernels and images are subspaces), Theorem 4.4.7 (Criteria for direct sums), Proposition 4.6.29 (Complements and the quotient), Proposition 2.7.7 (Bijective linear maps are invertible)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. For (a), the map

𝐯 \mapsto (𝐯, 𝑇 (𝐯))

is linear by Proposition 4.6.7(a). Its first component makes it injective.

Hint 2. For (b), split

(𝐯, 𝐰)

(𝐯, 𝑇 (𝐯)) +

(𝟎, 𝐰 - 𝑇 (𝐯))

. Then use Proposition 4.6.29.

Hint 3. For (c), the two conditions say that each

𝐯

is the first component of exactly one element

(𝐯, 𝐰)

𝐺

. Define

𝑆 (𝐯) = 𝐰

and prove it is linear by uniqueness.

Sketch. (a)

𝜑 (𝐯) =

(𝐯, 𝑇 (𝐯))

is linear and injective with image

𝐺_{𝑇}

. (b)

(𝐯, 𝐰) =

(𝐯, 𝑇 𝐯) +

(𝟎, 𝐰 - 𝑇 𝐯)

, and

(𝐯, 𝑇 𝐯) = (𝟎, 𝐰)

forces

𝐯 = 𝟎

, then

𝐰 = 𝟎

. Then

(𝑉 \times 𝑊) / 𝐺_{𝑇} ≅

{𝟎} \times

𝑊 ≅

𝑊

. (c) The conditions give existence and uniqueness of

𝐰

with

(𝐯, 𝐰) \in 𝐺

; call it

𝑆 (𝐯)

. Since

𝐺

is a subspace,

(𝑎 𝐯 + 𝐯^{'}, 𝑎 𝑆 𝐯 + 𝑆 𝐯^{'}) \in

𝐺

, so

𝑆

is linear, and

𝐺 = 𝐺_{𝑆}

(a) Let $𝜑 : 𝑉 \to 𝑉 \times 𝑊$ , $𝜑 (𝐯) =$ $(𝐯, 𝑇 (𝐯))$ . It is linear by Proposition 4.6.7(a) (with the maps ${id}_{𝑉}$ and $𝑇$ ), its image is $𝐺_{𝑇}$ , which is therefore a subspace (Proposition 2.6.8), and it is injective because the first component of $𝜑 (𝐯)$ is $𝐯$ . So $𝜑 : 𝑉 \to 𝐺_{𝑇}$ is a bijective linear map, hence an isomorphism.

(b) Every $(𝐯, 𝐰) \in 𝑉 \times 𝑊$ equals $(𝐯, 𝑇 (𝐯)) +$ $(𝟎, 𝐰 - 𝑇 (𝐯))$ , a vector of $𝐺_{𝑇}$ plus a vector of ${𝟎} \times 𝑊$ . If $(𝐯, 𝑇 (𝐯)) =$ $(𝟎, 𝐰)$ lies in both subspaces, then $𝐯 = 𝟎$ , so $𝐰 = 𝑇 (𝟎) = 𝟎$ . So the sum is direct (Theorem 4.4.7) and $𝑉 \times$ $𝑊 =$ $𝐺_{𝑇} \oplus$ $({𝟎} \times 𝑊)$ . By Proposition 4.6.29, $(𝑉 \times 𝑊) / 𝐺_{𝑇} ≅$ ${𝟎} \times$ $𝑊$ , and $(𝟎, 𝐰) \mapsto 𝐰$ is an isomorphism ${𝟎} \times 𝑊 \to 𝑊$ .

(c) If $𝐺 = 𝐺_{𝑆}$ , then $(𝟎, 𝐰) \in 𝐺_{𝑆}$ forces $𝐰 = 𝑆 (𝟎) = 𝟎$ , and $𝐯$ is the first component of $(𝐯, 𝑆 (𝐯)) \in 𝐺_{𝑆}$ .

Conversely, assume the two conditions. For each $𝐯 \in 𝑉$ there is some $𝐰$ with $(𝐯, 𝐰) \in 𝐺$ , and only one: if $(𝐯, 𝐰),$ $(𝐯, 𝐰^{'}) \in$ $𝐺$ , then their difference $(𝟎, 𝐰 - 𝐰^{'})$ lies in $𝐺 \cap ({𝟎} \times 𝑊) = {𝟎}$ , so $𝐰 = 𝐰^{'}$ . Define $𝑆 (𝐯)$ to be this $𝐰$ . For $𝐯, 𝐯^{'} \in 𝑉$ and $𝑎 \in 𝐹$ , the vector $𝑎 (𝐯, 𝑆 (𝐯)) +$ $(𝐯^{'}, 𝑆 (𝐯^{'})) =$ $(𝑎 𝐯 + 𝐯^{'}, 𝑎 𝑆 (𝐯) + 𝑆 (𝐯^{'}))$ lies in $𝐺$ , so by uniqueness $𝑆 (𝑎 𝐯 + 𝐯^{'}) =$ $𝑎 𝑆 (𝐯) +$ $𝑆 (𝐯^{'})$ . Thus $𝑆$ is linear. By construction $𝐺_{𝑆} \subseteq 𝐺$ , and every $(𝐯, 𝐰) \in 𝐺$ has $𝐰 = 𝑆 (𝐯)$ , so $𝐺 = 𝐺_{𝑆}$ .

Exercise 4.6.40.routine For subsets $𝐴, 𝐵 \subseteq 𝑉$ and $𝑎 \in 𝐹$ , let $𝐴 +$ $𝐵 =$ ${𝐱 + 𝐲 | 𝐱 \in 𝐴, 𝐲 \in 𝐵}$ and $𝑎 𝐴 = {𝑎 𝐱 | 𝐱 \in 𝐴}$ . Let $𝑈$ be a subspace of $𝑉$ .

(a)Show that $(𝐯 + 𝑈) +$ $(𝐰 + 𝑈) =$ $(𝐯 + 𝐰) +$ $𝑈$ as sets.
(b)Show that $𝑎 (𝐯 + 𝑈) = 𝑎 𝐯 + 𝑈$ as sets when $𝑎 \neq 0$ .
(c)Compute $0 (𝐯 + 𝑈)$ , and explain why, when $𝑈 \neq {𝟎}$ , these set operations do not make the set of cosets into a vector space, although they agree with the operations of Definition 4.6.20 whenever the result is a coset.

Solution.

Uses. Definition 4.6.12 (Coset), Definition 4.6.20 (Quotient space)

Techniques. Double containment (Technique 1.3.14)

Hint 1. Prove each equality of sets by double containment. For

\supseteq

in (a), write

(𝐯 + 𝐰) +

𝐮 =

(𝐯 + 𝐮) +

(𝐰 + 𝟎)

Hint 2. For (b),

𝑎 𝐯 +

𝐮 =

𝑎 (𝐯 + 𝑎^{- 1} 𝐮)

. For (c), how many elements does a coset of

𝑈

have?

Sketch. (a)

(𝐯 + 𝐮) +

(𝐰 + 𝐮^{'}) =

(𝐯 + 𝐰) +

(𝐮 + 𝐮^{'})

, and conversely

(𝐯 + 𝐰) +

𝐮 =

(𝐯 + 𝐮) +

(𝐰 + 𝟎)

. (b) Similar, with

𝑎^{- 1}

. (c)

0 (𝐯 + 𝑈) = {𝟎}

, which is not a coset when

𝑈 \neq {𝟎}

, because every coset is in bijection with

𝑈

(a) An element of the left side is $(𝐯 + 𝐮) +$ $(𝐰 + 𝐮^{'}) =$ $(𝐯 + 𝐰) +$ $(𝐮 + 𝐮^{'})$ with $𝐮, 𝐮^{'} \in 𝑈$ , and $𝐮 + 𝐮^{'} \in 𝑈$ . Conversely, $(𝐯 + 𝐰) +$ $𝐮 =$ $(𝐯 + 𝐮) +$ $(𝐰 + 𝟎)$ is in the left side.

(b) An element of the left side is $𝑎 (𝐯 + 𝐮) =$ $𝑎 𝐯 +$ $𝑎 𝐮$ with $𝑎 𝐮 \in 𝑈$ . Conversely, $𝑎 𝐯 +$ $𝐮 =$ $𝑎 (𝐯 + 𝑎^{- 1} 𝐮)$ with $𝑎^{- 1} 𝐮 \in 𝑈$ .

(c) $0 (𝐯 + 𝑈) =$ ${0 𝐱 | 𝐱 \in 𝐯 + 𝑈} =$ ${𝟎}$ . A coset $𝐱 + 𝑈$ is in bijection with $𝑈$ via $𝐮 \mapsto 𝐱 + 𝐮$ , so when $𝑈 \neq {𝟎}$ it has at least two elements, and ${𝟎}$ is not a coset. So scalar multiplication by $0$ leads out of the set of cosets, and the set operations do not even define a scalar multiplication on it. By (a) and (b), the set operations agree with those of Definition 4.6.20 in all other cases. The quotient definition sets $0 (𝐯 + 𝑈) =$ $0 𝐯 +$ $𝑈 =$ $𝑈$ instead.

In particular, the sum in $𝑉 / 𝑈$ is the sum of the cosets as sets, which gives a second reason why addition is well defined.

Exercise 4.6.41.standard Let $𝑈$ be a subspace of $𝑉$ , let $𝐯_{1}, \dots, 𝐯_{𝑘} \in 𝑉$ , and let $𝑋 =$ $span (𝐯_{1}, \dots, 𝐯_{𝑘})$ .

(a)Show that $𝐯_{1} + 𝑈, \dots, 𝐯_{𝑘} + 𝑈$ is linearly independent in $𝑉 / 𝑈$ if and only if $𝐯_{1}, \dots, 𝐯_{𝑘}$ is linearly independent and $𝑋 \cap 𝑈 = {𝟎}$ .
(b)Show that $𝐯_{1} + 𝑈, \dots, 𝐯_{𝑘} + 𝑈$ spans $𝑉 / 𝑈$ if and only if $𝑋 + 𝑈 = 𝑉$ .
(c)Conclude that $𝐯_{1} + 𝑈, \dots, 𝐯_{𝑘} + 𝑈$ is a basis of $𝑉 / 𝑈$ if and only if $𝐯_{1}, \dots, 𝐯_{𝑘}$ is a basis of a complement of $𝑈$ .

Solution.

Uses. Proposition 4.6.26 (The quotient map), Proposition 2.1.16 (Linear maps preserve spans and linear relations), Lemma 4.6.14 (Equal cosets), Theorem 4.4.7 (Criteria for direct sums), Definition 1.6.1 (Basis)

Hint 1. A relation

\sum_{𝑗} 𝑐_{𝑗} (𝐯_{𝑗} + 𝑈) =

𝟎 +

𝑈

says exactly that

\sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗} \in 𝑈

Hint 2. For (b), the span of the cosets is

𝜋 (𝑋)

. When is

𝜋 (𝑋)

all of

𝑉 / 𝑈

Sketch. (a)

\sum_{𝑗} 𝑐_{𝑗} (𝐯_{𝑗} + 𝑈) =

𝑈

iff

\sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗} \in 𝑋 \cap 𝑈

. If

𝑋 \cap 𝑈 = {𝟎}

and the

𝐯_{𝑗}

are independent, then all

𝑐_{𝑗} = 0

; conversely, a relation among the

𝐯_{𝑗}

, or a nonzero vector of

𝑋 \cap 𝑈

, gives a nontrivial relation among the cosets. (b) The span of the cosets is

𝜋 (𝑋)

, and

𝜋 (𝑋) = 𝑉 / 𝑈

iff every

𝐯

is in

𝑋 + 𝑈

. (c) Combine:

𝐯_{1}, \dots, 𝐯_{𝑘}

is a basis of

𝑋

and

𝑉 = 𝑈 \oplus 𝑋

Throughout, $𝜋 : 𝑉 \to 𝑉 / 𝑈$ is the quotient map. For scalars $𝑐_{𝑗}$ , $\sum_{𝑗} 𝑐_{𝑗} (𝐯_{𝑗} + 𝑈) =$ $(\sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗}) +$ $𝑈$ , which is the zero coset if and only if $\sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗} \in 𝑈$ (Corollary 4.6.15(c)).

(a) Suppose the cosets are independent. If $\sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗} = 𝟎$ , then $\sum_{𝑗} 𝑐_{𝑗} (𝐯_{𝑗} + 𝑈) =$ $𝑈$ , so all $𝑐_{𝑗} = 0$ ; thus $𝐯_{1}, \dots, 𝐯_{𝑘}$ is independent. If $𝐱 =$ $\sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗} \in$ $𝑋 \cap$ $𝑈$ , then $\sum_{𝑗} 𝑐_{𝑗} (𝐯_{𝑗} + 𝑈) =$ $𝑈$ , so all $𝑐_{𝑗} = 0$ and $𝐱 = 𝟎$ . Conversely, assume both conditions and let $\sum_{𝑗} 𝑐_{𝑗} (𝐯_{𝑗} + 𝑈) =$ $𝑈$ . Then $\sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗} \in$ $𝑋 \cap$ $𝑈 =$ ${𝟎}$ , so $\sum_{𝑗} 𝑐_{𝑗} 𝐯_{𝑗} = 𝟎$ and all $𝑐_{𝑗} = 0$ by independence.

(b) The span of $𝜋 (𝐯_{1}),$ $\dots,$ $𝜋 (𝐯_{𝑘})$ is $im (𝜋 ↾_{𝑋}) =$ ${𝐱 + 𝑈 | 𝐱 \in 𝑋}$ (Proposition 2.1.16(a)). This is all of $𝑉 / 𝑈$ if and only if for every $𝐯 \in 𝑉$ there is $𝐱 \in 𝑋$ with $𝐯 - 𝐱 \in 𝑈$ (Lemma 4.6.14), that is, $𝐯 \in 𝑋 + 𝑈$ . So the cosets span $𝑉 / 𝑈$ if and only if $𝑋 + 𝑈 = 𝑉$ .

(c) By (a) and (b), the cosets form a basis of $𝑉 / 𝑈$ if and only if $𝐯_{1}, \dots, 𝐯_{𝑘}$ is independent (hence a basis of $𝑋$ ), $𝑋 \cap 𝑈 = {𝟎}$ , and $𝑋 + 𝑈 = 𝑉$ . The last two conditions say that $𝑋$ is a complement of $𝑈$ (Theorem 4.4.7). This matches Proposition 4.6.29: $𝜋 ↾_{𝑋}$ is an isomorphism exactly when $𝑋$ is a complement, and isomorphisms carry bases to bases.

Exercise 4.6.42.standard Let $𝑉 = 𝐹^{2 \times 2}$ and $𝑆 =$ ${𝐴 \in 𝑉 | 𝐴^{𝖳} = 𝐴}$ .

(a)Show that $𝐴 + 𝑆 \mapsto 𝑎_{12} - 𝑎_{21}$ is a well-defined isomorphism $𝑉 / 𝑆 \to 𝐹$ , where $𝐴 = (𝑎_{𝑖 𝑗})$ .
(b)Suppose $1 + 1 \neq 0$ in $𝐹$ . Show that $𝐾 = {𝐴 | 𝐴^{𝖳} = - 𝐴}$ is a complement of $𝑆$ .
(c)Show that over $𝔽_{2}$ the subspace $𝐾$ is not a complement of $𝑆$ , and find a complement.

Solution.

Uses. Theorem 4.6.34 (First isomorphism theorem), Proposition 4.6.29 (Complements and the quotient), Theorem 4.4.19 (Dimension of a sum), Lemma 2.4.24 (Matrix units)

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1. The function

𝜓 (𝐴) = 𝑎_{12} - 𝑎_{21}

is linear on

𝑉

. What is its kernel?

Hint 2. For (b), compute

𝑆 \cap 𝐾

and the dimensions of

𝑆

and

𝐾

. For (c), what is

- 𝐴

over

𝔽_{2}

Sketch. (a)

𝜓 (𝐴) = 𝑎_{12} - 𝑎_{21}

is linear and surjective with kernel

𝑆

; apply Theorem 4.6.34. (b)

𝐴 \in 𝑆 \cap 𝐾

gives

𝐴 = - 𝐴

, so

(1 + 1) 𝐴 = 0

and

𝐴 = 0

;

dim 𝑆 = 3

dim 𝐾 = 1

. (c) Over

𝔽_{2}

- 𝐴 = 𝐴

, so

𝐾 = 𝑆

span (𝑈_{12})

is a complement, since

𝜓 (𝑈_{12}) = 1

(a) Let $𝜓 : 𝑉 \to 𝐹$ , $𝜓 (𝐴) = 𝑎_{12} - 𝑎_{21}$ . It is linear, and surjective since $𝜓 (𝑈_{12}) = 1$ (Lemma 2.4.24). A $2 \times 2$ matrix is symmetric exactly when $𝑎_{12} = 𝑎_{21}$ , so $ker 𝜓 = 𝑆$ . By Theorem 4.6.34, $\tilde{𝜓} : 𝑉 / 𝑆 \to 𝐹$ , $𝐴 + 𝑆 \mapsto 𝑎_{12} - 𝑎_{21}$ , is well defined and an isomorphism.

(b) If $𝐴 \in 𝑆 \cap 𝐾$ , then $𝐴 = 𝐴^{𝖳} = - 𝐴$ , so $(1 + 1) 𝐴 = 0$ , and $𝐴 = 0$ after multiplying by ${(1 + 1)}^{- 1}$ . The space $𝑆$ has basis $𝑈_{11},$ $𝑈_{22},$ $𝑈_{12} +$ $𝑈_{21}$ , so $dim 𝑆 = 3$ . For $𝐴 \in 𝐾$ , the diagonal entries satisfy $𝑎_{𝑖 𝑖} = - 𝑎_{𝑖 𝑖}$ , so they vanish, and $𝑎_{21} = - 𝑎_{12}$ ; hence $𝐾 =$ $span (𝑈_{12} - 𝑈_{21})$ has dimension $1$ . By Theorem 4.4.19, $dim (𝑆 + 𝐾) =$ $3 +$ $1 -$ $0 =$ $4$ , so $𝑆 + 𝐾 = 𝑉$ and $𝑉 = 𝑆 \oplus 𝐾$ .

(c) Over $𝔽_{2}$ , $- 𝐴 = 𝐴$ for every $𝐴$ , so $𝐾 = 𝑆$ and $𝑆 \cap 𝐾 = 𝑆 \neq {0}$ . Let $𝐿 = span (𝑈_{12})$ . A nonzero multiple $𝑐 𝑈_{12}$ is not symmetric, so $𝐿 \cap 𝑆 = {0}$ , and $dim 𝑆 + dim 𝐿 = 4$ , so $𝑉 = 𝑆 \oplus 𝐿$ . Equivalently, by Proposition 4.6.29: $𝜓 (𝑈_{12}) = 1 \neq 0$ , so $𝜋 ↾_{𝐿}$ is an isomorphism onto the one-dimensional space $𝑉 / 𝑆$ .

The quotient $𝑉 / 𝑆 ≅ 𝐹$ is described by the same formula over every field; only the familiar complement of skew-symmetric matrices breaks down in characteristic $2$ .

Exercise 4.6.43.standard Let $𝑇 \in L (𝑉, 𝑊)$ , and let $𝑈$ be a subspace of $𝑉$ with $ker 𝑇 \subseteq 𝑈$ . A student defines $𝑆 : 𝑉 / 𝑈 \to 𝑊$ by $𝑆 (𝐯 + 𝑈) = 𝑇 (𝐯)$ , and argues:

“ $𝑆$ is well defined: if $𝑆 (𝐯 + 𝑈) =$ $𝑆 (𝐯^{'} + 𝑈)$ , then $𝑇 (𝐯) = 𝑇 (𝐯^{'})$ , so $𝐯 - 𝐯^{'} \in ker 𝑇 \subseteq 𝑈$ , hence $𝐯 + 𝑈 = 𝐯^{'} + 𝑈$ .”

Find the error, and give an example with $𝑉 = 𝑊 = ℝ^{2}$ in which the formula does not define a function on $𝑉 / 𝑈$ . What hypothesis would make the definition correct?

Solution.

Uses. Theorem 4.6.30 (Universal property of the quotient), Theorem 4.6.34 (First isomorphism theorem)

Techniques. Check that a definition does not depend on the representative (Technique 4.6.18)

Hint 1. Compare the student’s implication with the one in Technique 4.6.18. Which way does each go?

Hint 2. Take

𝑇 = id

, so that

ker 𝑇 = {𝟎}

is inside every

𝑈

Sketch. The student proved “equal outputs imply equal inputs”, which is injectivity. Well-definedness needs “equal inputs imply equal outputs”. With

𝑇 = {id}_{ℝ^{2}}

and

𝑈 = span (𝐞_{1})

𝐞_{1} + 𝑈 = 𝟎 + 𝑈

but

𝑇 (𝐞_{1}) \neq 𝑇 (𝟎)

. The right hypothesis is

𝑈 \subseteq ker 𝑇

The argument proves the wrong implication. It shows that if two cosets have the same value, then they are equal, which is injectivity of $𝑆$ , assuming $𝑆$ exists. Well-definedness is the converse: if $𝐯 + 𝑈 = 𝐯^{'} + 𝑈$ , then $𝑇 (𝐯) = 𝑇 (𝐯^{'})$ . That is, $𝐯 - 𝐯^{'} \in 𝑈$ must imply $𝐯 - 𝐯^{'} \in ker 𝑇$ , which needs $𝑈 \subseteq ker 𝑇$ , the opposite of the given containment.

Example: $𝑇 = {id}_{ℝ^{2}}$ , so $ker 𝑇 = {𝟎} \subseteq 𝑈$ for $𝑈 = span (𝐞_{1})$ . Then $𝐞_{1} + 𝑈 = 𝟎 + 𝑈$ , but the formula assigns $𝑇 (𝐞_{1}) = 𝐞_{1}$ to the first name and $𝑇 (𝟎) = 𝟎$ to the second.

By Theorem 4.6.30, the formula defines a (linear) map exactly when $𝑈 \subseteq ker 𝑇$ . Together with the student’s hypothesis $ker 𝑇 \subseteq 𝑈$ , that means $𝑈 = ker 𝑇$ , and then the student’s argument is a correct proof that $𝑆 = \tilde{𝑇}$ is injective, which is half of Theorem 4.6.34.

Exercise 4.6.44.standard Let $𝑈$ be a subspace of $𝑉$ , where $𝑉$ may be infinite-dimensional.

(a)Show that $dim (𝑉 / 𝑈) = 1$ if and only if $𝑈 = ker 𝑓$ for some nonzero linear map $𝑓 : 𝑉 \to 𝐹$ .
(b)Let $𝑓, 𝑔 : 𝑉 \to 𝐹$ be nonzero linear maps. Show that $ker 𝑓 = ker 𝑔$ if and only if $𝑔 = 𝑐 𝑓$ for some nonzero $𝑐 \in 𝐹$ .

Solution.

Uses. Theorem 4.6.34 (First isomorphism theorem), Theorem 4.6.30 (Universal property of the quotient), Proposition 4.6.26 (The quotient map), Proposition 1.6.19 (Coordinates respect linear combinations), Proposition 1.6.16 (The coordinate map is a bijection)

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1. A nonzero linear map

𝑓 : 𝑉 \to 𝐹

is surjective, since its image is a nonzero subspace of

𝐹

. Apply Theorem 4.6.34.

Hint 2. Conversely, if

dim (𝑉 / 𝑈) = 1

, compose

𝜋

with a coordinate map

𝑉 / 𝑈 \to 𝐹

Hint 3. For (b), both

𝑓

and

𝑔

induce isomorphisms

𝑉 / 𝑈 \to 𝐹

with

𝑈 = ker 𝑓 = ker 𝑔

. What are the linear maps

𝐹 \to 𝐹

Sketch. (a) If

𝑓 \neq 0

, then

im 𝑓 = 𝐹

and

𝑉 / ker 𝑓 ≅ 𝐹

. If

𝐯_{0} + 𝑈

is a basis of

𝑉 / 𝑈

, let

𝜑

be its coordinate map and

𝑓 = 𝜑 \circ 𝜋

; then

ker 𝑓 = ker 𝜋 = 𝑈

and

𝑓 (𝐯_{0}) = 1

. (b)

\tilde{𝑔} \circ {\tilde{𝑓}}^{- 1} : 𝐹 \to 𝐹

is multiplication by some

𝑐 \neq 0

, so

\tilde{𝑔} = 𝑐 \tilde{𝑓}

and

𝑔 = 𝑐 𝑓

(a) Suppose $𝑈 = ker 𝑓$ with $𝑓 \neq 0$ . Pick $𝐯$ with $𝑓 (𝐯) = 𝑎 \neq 0$ ; then $𝑓 ((𝑡 / 𝑎) 𝐯) = 𝑡$ for every $𝑡 \in 𝐹$ , so $𝑓$ is surjective. By Theorem 4.6.34, $𝑉 / 𝑈 ≅ 𝐹$ , which has dimension $1$ .

Conversely, suppose $dim (𝑉 / 𝑈) = 1$ , and let $𝐯_{0} + 𝑈$ be a basis of $𝑉 / 𝑈$ . Its coordinate map $𝜑 : 𝑉 / 𝑈 \to 𝐹$ , $𝑐 (𝐯_{0} + 𝑈) \mapsto 𝑐$ , is a linear bijection (Proposition 1.6.19, Proposition 1.6.16). Let $𝑓 = 𝜑 \circ 𝜋 : 𝑉 \to 𝐹$ . It is linear, and since $𝜑$ is injective, $ker 𝑓 = ker 𝜋 = 𝑈$ (Proposition 4.6.26). It is nonzero because $𝑓 (𝐯_{0}) =$ $𝜑 (𝐯_{0} + 𝑈) =$ $1$ .

(b) If $𝑔 = 𝑐 𝑓$ with $𝑐 \neq 0$ , then $𝑔 (𝐯) = 0$ if and only if $𝑓 (𝐯) = 0$ . Conversely, let $𝑈 = ker 𝑓 = ker 𝑔$ . As in (a), $𝑓$ and $𝑔$ are surjective, so Theorem 4.6.34 gives isomorphisms $\tilde{𝑓}, \tilde{𝑔} : 𝑉 / 𝑈 \to 𝐹$ with $𝑓 = \tilde{𝑓} \circ 𝜋$ and $𝑔 = \tilde{𝑔} \circ 𝜋$ . The map $ℎ = \tilde{𝑔} \circ {\tilde{𝑓}}^{- 1} : 𝐹 \to 𝐹$ is linear, so $ℎ (𝑡) = 𝑡 ℎ (1)$ for all $𝑡$ ; let $𝑐 = ℎ (1)$ , which is nonzero because $ℎ$ is bijective. Then $\tilde{𝑔} = ℎ \circ \tilde{𝑓} = 𝑐 \tilde{𝑓}$ , and $𝑔 =$ $\tilde{𝑔} \circ$ $𝜋 =$ $𝑐 \tilde{𝑓} \circ$ $𝜋 =$ $𝑐 𝑓$ .

Exercise 4.6.45.standard Let

𝐴 \in 𝐹^{𝑚 \times 𝑛}

and

𝐵 \in 𝐹^{𝑝 \times 𝑛}

. Show that

𝐵 = 𝐶 𝐴

for some

𝐶 \in 𝐹^{𝑝 \times 𝑚}

if and only if

ker 𝑇_{𝐴} \subseteq ker 𝑇_{𝐵}

Solution.

Uses. Theorem 4.6.30 (Universal property of the quotient), Theorem 4.6.34 (First isomorphism theorem), Proposition 4.1.19 (Extending linear maps and separating vectors), Proposition 2.3.17 (Every linear map $𝐹^{𝑛} \to 𝐹^{𝑚}$ is some $𝑇_{𝐴}$ )

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1. One direction is one line. For the other,

𝑇_{𝐵}

kills

ker 𝑇_{𝐴}

, so it factors through

𝐹^{𝑛} / ker 𝑇_{𝐴}

, and that quotient is isomorphic to

im 𝑇_{𝐴}

Hint 2. You get a linear map

𝑅_{0} : im 𝑇_{𝐴} \to 𝐹^{𝑝}

with

𝑅_{0} (𝐴 𝐱) = 𝐵 𝐱

. It is only defined on

im 𝑇_{𝐴}

. Extend it to

𝐹^{𝑚}

with Proposition 4.1.19(a).

Sketch. If

𝐵 = 𝐶 𝐴

and

𝐴 𝐱 = 𝟎

, then

𝐵 𝐱 = 𝟎

. Conversely, Theorem 4.6.30 gives

𝑆

with

𝑇_{𝐵} = 𝑆 \circ 𝜋

, and

𝑅_{0} =

𝑆 \circ

{\tilde{𝑇_{𝐴}}}^{- 1} : im 𝑇_{𝐴} \to

𝐹^{𝑝}

satisfies

𝑅_{0} (𝐴 𝐱) = 𝐵 𝐱

. Extend

𝑅_{0}

𝑅 \in L (𝐹^{𝑚}, 𝐹^{𝑝})

and let

𝐶

be its matrix.

If $𝐵 = 𝐶 𝐴$ and $𝐴 𝐱 = 𝟎$ , then $𝐵 𝐱 = 𝐶 (𝐴 𝐱) = 𝟎$ , so $ker 𝑇_{𝐴} \subseteq ker 𝑇_{𝐵}$ .

Conversely, suppose $ker 𝑇_{𝐴} \subseteq ker 𝑇_{𝐵}$ , and let $𝜋 : 𝐹^{𝑛} \to$ $𝐹^{𝑛} / ker 𝑇_{𝐴}$ be the quotient map. By Theorem 4.6.30(a), there is a linear $𝑆 : 𝐹^{𝑛} / ker 𝑇_{𝐴} \to$ $𝐹^{𝑝}$ with $𝑆 (𝐱 + ker 𝑇_{𝐴}) =$ $𝐵 𝐱$ . By Theorem 4.6.34, $\tilde{𝑇_{𝐴}} : 𝐹^{𝑛} / ker 𝑇_{𝐴} \to$ $im 𝑇_{𝐴}$ , $𝐱 + ker 𝑇_{𝐴} \mapsto 𝐴 𝐱$ , is an isomorphism. Let $𝑅_{0} =$ $𝑆 \circ$ ${\tilde{𝑇_{𝐴}}}^{- 1} : im 𝑇_{𝐴} \to$ $𝐹^{𝑝}$ ; it is linear, and

𝑅_{0} (𝐴 𝐱) = 𝑆 (𝐱 + ker 𝑇_{𝐴}) = 𝐵 𝐱 for all 𝐱 \in 𝐹^{𝑛} .

By Proposition 4.1.19(a), $𝑅_{0}$ extends to a linear map $𝑅 : 𝐹^{𝑚} \to 𝐹^{𝑝}$ . By Proposition 2.3.17, $𝑅 = 𝑇_{𝐶}$ for some $𝐶 \in 𝐹^{𝑝 \times 𝑚}$ . Then $𝐶 𝐴 𝐱 =$ $𝑅 (𝐴 𝐱) =$ $𝐵 𝐱$ for all $𝐱$ ; taking $𝐱 = 𝐞_{𝑗}$ shows that $𝐶 𝐴$ and $𝐵$ have the same columns, so $𝐶 𝐴 = 𝐵$ .

The extension step is where choices enter: for $𝑝 \geq 1$ , $𝐶$ is unique exactly when $im 𝑇_{𝐴} = 𝐹^{𝑚}$ , that is, $rank 𝐴 = 𝑚$ .

Exercise 4.6.46.hard Let $𝑇_{1} \in L (𝑉, 𝑊_{1})$ and $𝑇_{2} \in L (𝑉, 𝑊_{2})$ be surjective, and let $𝑇 =$ $(𝑇_{1}, 𝑇_{2}) : 𝑉 \to$ $𝑊_{1} \times$ $𝑊_{2}$ .

(a)Show that $ker 𝑇 =$ $ker 𝑇_{1} \cap$ $ker 𝑇_{2}$ .
(b)Show that $𝑇$ is surjective if and only if $ker 𝑇_{1} + ker 𝑇_{2} = 𝑉$ .
(c)Let $𝑈, 𝑊$ be subspaces of $𝑉$ . Show that $𝑉 \to 𝑉 / 𝑈 \times 𝑉 / 𝑊$ , $𝐯 \mapsto (𝐯 + 𝑈, 𝐯 + 𝑊)$ , has kernel $𝑈 \cap 𝑊$ , and is surjective if and only if $𝑈 + 𝑊 = 𝑉$ . Deduce that if $𝑈 + 𝑊 = 𝑉$ , then $𝑉 / (𝑈 \cap 𝑊) ≅$ $𝑉 / 𝑈 \times$ $𝑉 / 𝑊$ .

Solution.

Uses. Proposition 4.6.7 (Maps into and out of a product), Proposition 4.6.26 (The quotient map), Theorem 4.6.34 (First isomorphism theorem)

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1. For (b) “

\Leftarrow

”: given

(𝐰_{1}, 𝐰_{2})

, choose preimages

𝐱_{1}

under

𝑇_{1}

and

𝐱_{2}

under

𝑇_{2}

. You need one vector that works for both. Split

𝐱_{1} - 𝐱_{2}

along

ker 𝑇_{1} + ker 𝑇_{2}

Hint 2. For (b) “

\Rightarrow

”: to write

𝐯 \in 𝑉

as a sum, find

𝐱

with

𝑇 (𝐱) =

(𝑇_{1} (𝐯), 𝟎)

Hint 3. For (c), apply (a) and (b) to the two quotient maps.

Sketch. (a) Both components vanish. (b) If

𝐱_{1} - 𝐱_{2} = 𝐤_{1} + 𝐤_{2}

with

𝐤_{𝑖} \in ker 𝑇_{𝑖}

, then

𝐯 =

𝐱_{1} -

𝐤_{1} =

𝐱_{2} +

𝐤_{2}

maps to

(𝐰_{1}, 𝐰_{2})

. Conversely, if

𝑇 (𝐱) = (𝑇_{1} 𝐯, 𝟎)

, then

𝐯 =

(𝐯 - 𝐱) +

𝐱 \in

ker 𝑇_{1} +

ker 𝑇_{2}

. (c) The quotient maps are surjective with kernels

𝑈

and

𝑊

; then use Theorem 4.6.34.

$𝑇$ is linear by Proposition 4.6.7(a).

(a) $𝑇 (𝐯) = (𝟎, 𝟎)$ if and only if $𝑇_{1} (𝐯) = 𝟎$ and $𝑇_{2} (𝐯) = 𝟎$ .

(b) Suppose $ker 𝑇_{1} + ker 𝑇_{2} = 𝑉$ , and let $(𝐰_{1}, 𝐰_{2}) \in 𝑊_{1} \times 𝑊_{2}$ . Since $𝑇_{1}, 𝑇_{2}$ are surjective, there are $𝐱_{1}, 𝐱_{2} \in 𝑉$ with $𝑇_{1} (𝐱_{1}) = 𝐰_{1}$ and $𝑇_{2} (𝐱_{2}) = 𝐰_{2}$ . Write $𝐱_{1} - 𝐱_{2} = 𝐤_{1} + 𝐤_{2}$ with $𝐤_{𝑖} \in ker 𝑇_{𝑖}$ , and let

𝐯 = 𝐱_{1} - 𝐤_{1} = 𝐱_{2} + 𝐤_{2} .

Then $𝑇_{1} (𝐯) =$ $𝑇_{1} (𝐱_{1}) =$ $𝐰_{1}$ and $𝑇_{2} (𝐯) =$ $𝑇_{2} (𝐱_{2}) =$ $𝐰_{2}$ , so $𝑇 (𝐯) = (𝐰_{1}, 𝐰_{2})$ .

Conversely, suppose $𝑇$ is surjective, and let $𝐯 \in 𝑉$ . Choose $𝐱$ with $𝑇 (𝐱) =$ $(𝑇_{1} (𝐯), 𝟎)$ . Then $𝑇_{2} (𝐱) = 𝟎$ and $𝑇_{1} (𝐯 - 𝐱) = 𝟎$ , so $𝐯 =$ $(𝐯 - 𝐱) +$ $𝐱 \in$ $ker 𝑇_{1} +$ $ker 𝑇_{2}$ .

(c) The quotient maps $𝜋_{𝑈} : 𝑉 \to 𝑉 / 𝑈$ and $𝜋_{𝑊} : 𝑉 \to 𝑉 / 𝑊$ are surjective with kernels $𝑈$ and $𝑊$ (Proposition 4.6.26). The map in question is $(𝜋_{𝑈}, 𝜋_{𝑊})$ , so by (a) its kernel is $𝑈 \cap 𝑊$ , and by (b) it is surjective if and only if $𝑈 + 𝑊 = 𝑉$ . In that case Theorem 4.6.34 gives $𝑉 / (𝑈 \cap 𝑊) ≅$ $𝑉 / 𝑈 \times$ $𝑉 / 𝑊$ .

Part (c) is a linear-algebra version of the Chinese remainder theorem: when $𝑈 + 𝑊 = 𝑉$ , a pair of conditions “ $𝐯 \equiv 𝐚 (mod 𝑈)$ ” and “ $𝐯 \equiv 𝐛 (mod 𝑊)$ ” always has a solution, unique modulo $𝑈 \cap 𝑊$ .

Exercise 4.6.47.standard Let $𝑈_{1}, \dots, 𝑈_{𝑚}$ be subspaces of $𝑉$ such that each $𝑉 / 𝑈_{𝑖}$ is finite-dimensional. Here $𝑉$ may be infinite-dimensional.

(a)Show that $𝑉 / (𝑈_{1} \cap \dots \cap 𝑈_{𝑚})$ is finite-dimensional and
$dim (𝑉 / (𝑈_{1} \cap \dots \cap 𝑈_{𝑚})) \leq dim (𝑉 / 𝑈_{1}) + \dots + dim (𝑉 / 𝑈_{𝑚}) .$
(b)Deduce that if $𝑉$ is finite-dimensional, then $dim (𝑈_{1} \cap \dots \cap 𝑈_{𝑚}) \geq$ $dim 𝑈_{1} +$ $\dots +$ $dim 𝑈_{𝑚} -$ $(𝑚 - 1) dim 𝑉$ .
(c)Show that any two $4$ -dimensional subspaces and any $3$ -dimensional subspace of $𝐹^{5}$ have a nonzero vector in common.

Solution.

Uses. Proposition 4.6.7 (Maps into and out of a product), Theorem 4.6.34 (First isomorphism theorem), Theorem 4.6.5 (Dimension of a product), Proposition 4.1.21 (Dimensions of subspaces), Proposition 4.2.6 (Injective maps preserve dimension), Theorem 4.6.27 (Dimension of a quotient)

Techniques. Pass to a quotient (Technique 4.6.31), Dimension counting (Technique 4.2.24)

Hint 1. Find a linear map out of

𝑉

whose kernel is

𝑈_{1} \cap \dots \cap 𝑈_{𝑚}

and whose target is finite-dimensional.

Hint 2. Use

𝑇 : 𝑉 \to

𝑉 / 𝑈_{1} \times

\dots \times

𝑉 / 𝑈_{𝑚}

𝐯 \mapsto

(𝐯 + 𝑈_{1}, \dots, 𝐯 + 𝑈_{𝑚})

, and Theorem 4.6.34. The image need not be the whole product.

Sketch.

𝑇

has kernel

\cap_{𝑖} 𝑈_{𝑖}

, so

𝑉 / \cap_{𝑖} 𝑈_{𝑖} ≅ im 𝑇

, a subspace of a product of dimension

\sum_{𝑖} dim (𝑉 / 𝑈_{𝑖})

. (b) Substitute

dim (𝑉 / 𝑋) =

dim 𝑉 -

dim 𝑋

. (c)

4 + 4 + 3 - 2 \cdot 5 = 1

(a) Let $𝑇 : 𝑉 \to$ $𝑉 / 𝑈_{1} \times$ $\dots \times$ $𝑉 / 𝑈_{𝑚}$ , $𝑇 (𝐯) =$ $(𝐯 + 𝑈_{1}, \dots, 𝐯 + 𝑈_{𝑚})$ . It is linear by Proposition 4.6.7(a), and $𝑇 (𝐯) = 𝟎$ if and only if $𝐯 \in 𝑈_{𝑖}$ for every $𝑖$ , so $ker 𝑇 =$ $𝑈_{1} \cap$ $\dots \cap$ $𝑈_{𝑚} ≕ 𝑋$ . The product is finite-dimensional of dimension $\sum_{𝑖} dim (𝑉 / 𝑈_{𝑖})$ (Theorem 4.6.5), so its subspace $im 𝑇$ is finite-dimensional with $dim im 𝑇 \leq$ $\sum_{𝑖} dim (𝑉 / 𝑈_{𝑖})$ (Proposition 4.1.21). By Theorem 4.6.34, $\tilde{𝑇} : 𝑉 / 𝑋 \to im 𝑇$ is an isomorphism; its inverse is injective with finite-dimensional domain $im 𝑇$ , so by Proposition 4.2.6(a), $𝑉 / 𝑋$ is finite-dimensional with $dim (𝑉 / 𝑋) =$ $dim im 𝑇 \leq$ $\sum_{𝑖} dim (𝑉 / 𝑈_{𝑖})$ .

(b) By Theorem 4.6.27, (a) reads $dim 𝑉 -$ $dim 𝑋 \leq$ $\sum_{𝑖 = 1}^{𝑚} (dim 𝑉 - dim 𝑈_{𝑖}) =$ $𝑚 dim 𝑉 -$ $\sum_{𝑖} dim 𝑈_{𝑖}$ . Rearranging, $dim 𝑋 \geq$ $\sum_{𝑖} dim 𝑈_{𝑖} -$ $(𝑚 - 1) dim 𝑉$ .

(c) By (b), the intersection has dimension at least $4 + 4 + 3 - 2 \cdot 5 = 1$ , so it contains a nonzero vector.

Exercise 4.6.48.standard Let $𝑛 \geq 1$ , and let $𝑍 \subseteq 𝐹^{𝑛 \times 𝑛}$ be the set of matrices whose row sums and column sums are all $0$ . Let $𝟏 = (1, \dots, 1) \in 𝐹^{𝑛}$ . Use the map

Φ : 𝐹^{𝑛 \times 𝑛} \to 𝐹^{𝑛} \times 𝐹^{𝑛}, Φ (𝐴) = (𝐴 𝟏, 𝐴^{𝖳} 𝟏),

to show that $𝑍$ is a subspace of dimension ${(𝑛 - 1)}^{2}$ .

Solution.

Uses. Proposition 4.6.7 (Maps into and out of a product), Theorem 4.6.5 (Dimension of a product), Theorem 4.2.11 (Rank–nullity), Proposition 2.4.14 (Algebra of matrices)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1.

𝐴 𝟏

is the vector of row sums and

𝐴^{𝖳} 𝟏

the vector of column sums, so

𝑍 = ker Φ

. The image of

Φ

is not all of

𝐹^{𝑛} \times 𝐹^{𝑛}

: both vectors have the same sum of entries.

Hint 2. Let

𝐷 =

{(𝐫, 𝐜) | 𝑟_{1} + \dots + 𝑟_{𝑛} = 𝑐_{1} + \dots + 𝑐_{𝑛}}

, of dimension

2 𝑛 - 1

. To show

im Φ = 𝐷

, look for a preimage supported on the first row and first column.

Sketch.

𝑍 = ker Φ

im Φ \subseteq 𝐷

, since both sums equal the sum of all entries. For

(𝐫, 𝐜) \in 𝐷

with common sum

𝑠

𝐴 =

𝐫 𝐞_{1}^{𝖳} +

𝐞_{1} 𝐜^{𝖳} -

𝑠 𝐞_{1} 𝐞_{1}^{𝖳}

has

Φ (𝐴) = (𝐫, 𝐜)

. So

rank Φ = 2 𝑛 - 1

and

dim 𝑍 = 𝑛^{2} - 2 𝑛 + 1

$Φ$ is linear (Proposition 4.6.7(a)), and $𝐴 𝟏$ and $𝐴^{𝖳} 𝟏$ are the vectors of row sums and column sums of $𝐴$ . So $𝑍 = ker Φ$ , a subspace. Let $𝜎 (𝐱) = 𝑥_{1} + \dots + 𝑥_{𝑛}$ and

𝐷 = {(𝐫, 𝐜) \in 𝐹^{𝑛} \times 𝐹^{𝑛} | 𝜎 (𝐫) = 𝜎 (𝐜)} .

$𝐷$ is the kernel of $(𝐫, 𝐜) \mapsto$ $𝜎 (𝐫) -$ $𝜎 (𝐜)$ , a linear map onto $𝐹$ (it sends $(𝐞_{1}, 𝟎)$ to $1$ ), so $dim 𝐷 = 2 𝑛 - 1$ by Theorem 4.6.5 and rank–nullity.

Both $𝜎 (𝐴 𝟏)$ and $𝜎 (𝐴^{𝖳} 𝟏)$ equal the sum of all entries of $𝐴$ , so $im Φ \subseteq 𝐷$ . Conversely, let $(𝐫, 𝐜) \in 𝐷$ with $𝑠 = 𝜎 (𝐫) = 𝜎 (𝐜)$ , and let

𝐴 = 𝐫 𝐞_{1}^{𝖳} + 𝐞_{1} 𝐜^{𝖳} - 𝑠 𝐞_{1} 𝐞_{1}^{𝖳} .

Using $𝐞_{1}^{𝖳} 𝟏 = 1$ , $𝐜^{𝖳} 𝟏 = 𝑠$ , and $𝐫^{𝖳} 𝟏 = 𝑠$ ,

𝐴 𝟏 = 𝐫 + 𝑠 𝐞_{1} - 𝑠 𝐞_{1} = 𝐫, 𝐴^{𝖳} 𝟏 = 𝐞_{1} (𝐫^{𝖳} 𝟏) + 𝐜 - 𝑠 𝐞_{1} = 𝐜 .

So $im Φ = 𝐷$ , and by rank–nullity $dim 𝑍 =$ $𝑛^{2} -$ $(2 𝑛 - 1) =$ ${(𝑛 - 1)}^{2}$ .

Exercise 4.6.49.standard Let $𝑈$ and $𝑊$ be subspaces of $𝑉$ .

(a)Show that $𝐰 \mapsto 𝐰 + 𝑈$ is a surjective linear map $𝑊 \to (𝑈 + 𝑊) / 𝑈$ with kernel $𝑈 \cap 𝑊$ , and deduce the second isomorphism theorem: $𝑊 / (𝑈 \cap 𝑊) ≅$ $(𝑈 + 𝑊) / 𝑈$ .
(b)Use (a) and Theorem 4.6.27 to give another proof of Theorem 4.4.19.

Solution.

Uses. Proposition 4.6.26 (The quotient map), Proposition 4.2.3 (Properties of restriction), Theorem 4.6.34 (First isomorphism theorem), Theorem 4.6.27 (Dimension of a quotient)

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1. The map is the quotient map

𝑈 + 𝑊 \to (𝑈 + 𝑊) / 𝑈

restricted to

𝑊

. Why does every coset have a representative in

𝑊

Hint 2. For (b), take dimensions on both sides of (a).

Sketch.

(𝐮 + 𝐰) + 𝑈 = 𝐰 + 𝑈

, so the map is onto, and

𝐰 + 𝑈 = 𝑈

iff

𝐰 \in 𝑈

. By the first isomorphism theorem,

𝑊 / (𝑈 \cap 𝑊) ≅

(𝑈 + 𝑊) / 𝑈

. Then

dim 𝑊 -

dim (𝑈 \cap 𝑊) =

dim (𝑈 + 𝑊) -

dim 𝑈

(a) Let $𝜋 : 𝑈 +$ $𝑊 \to$ $(𝑈 + 𝑊) / 𝑈$ be the quotient map of the space $𝑈 + 𝑊$ by its subspace $𝑈$ , and $𝜑 = 𝜋 ↾_{𝑊}$ , which is linear. Every element of $(𝑈 + 𝑊) / 𝑈$ is $(𝐮 + 𝐰) +$ $𝑈 =$ $𝐰 +$ $𝑈 =$ $𝜑 (𝐰)$ for some $𝐮 \in 𝑈$ , $𝐰 \in 𝑊$ , so $𝜑$ is surjective. Its kernel is $𝑊 \cap ker 𝜋 = 𝑊 \cap 𝑈$ (Proposition 4.2.3(c)). By Theorem 4.6.34, $𝑊 / (𝑈 \cap 𝑊) ≅$ $im 𝜑 =$ $(𝑈 + 𝑊) / 𝑈$ .

(b) Let $𝑈$ and $𝑊$ be finite-dimensional. Then $𝑈 + 𝑊$ is finite-dimensional (it is spanned by bases of $𝑈$ and $𝑊$ ), and isomorphic finite-dimensional spaces have equal dimensions. By (a) and Theorem 4.6.27, applied in $𝑊$ and in $𝑈 + 𝑊$ ,

dim 𝑊 - dim (𝑈 \cap 𝑊) = dim (𝑈 + 𝑊) - dim 𝑈,

which rearranges to $dim (𝑈 + 𝑊) =$ $dim 𝑈 +$ $dim 𝑊 -$ $dim (𝑈 \cap 𝑊)$ .

Exercise 4.6.50.hard Let $𝑈$ be a subspace of $𝑉$ , and let $𝜋 : 𝑉 \to 𝑉 / 𝑈$ be the quotient map. For a subspace $𝑋$ of $𝑉$ with $𝑈 \subseteq 𝑋$ , write $𝑋 / 𝑈 =$ ${𝐱 + 𝑈 | 𝐱 \in 𝑋} \subseteq$ $𝑉 / 𝑈$ .

(a)Show that $𝑋 / 𝑈$ is a subspace of $𝑉 / 𝑈$ and that $𝜋^{- 1} (𝑋 / 𝑈) = 𝑋$ .
(b)Show that every subspace $𝑌$ of $𝑉 / 𝑈$ equals $𝑋 / 𝑈$ for exactly one subspace $𝑋$ of $𝑉$ containing $𝑈$ .
(c)(Third isomorphism theorem.) Show that $(𝑉 / 𝑈) / (𝑋 / 𝑈) ≅$ $𝑉 / 𝑋$ whenever $𝑈 \subseteq 𝑋 \subseteq 𝑉$ .
(d)If $𝑉$ is finite-dimensional, show that $dim (𝑋 / 𝑈) =$ $dim 𝑋 -$ $dim 𝑈$ .

Solution.

Uses. Proposition 4.6.26 (The quotient map), Proposition 2.6.8 (Kernels and images are subspaces), Theorem 4.6.30 (Universal property of the quotient), Theorem 4.6.34 (First isomorphism theorem), Lemma 4.6.14 (Equal cosets), Theorem 4.2.11 (Rank–nullity)

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1. (a)

𝑋 / 𝑈 = im (𝜋 ↾_{𝑋})

. For

𝜋^{- 1} (𝑋 / 𝑈) \subseteq 𝑋

: if

𝐯 + 𝑈 = 𝐱 + 𝑈

with

𝐱 \in 𝑋

, then

𝐯 - 𝐱 \in 𝑈 \subseteq 𝑋

Hint 2. (b) Given

𝑌

, take

𝑋 = 𝜋^{- 1} (𝑌)

. Check that it is a subspace containing

𝑈

and that

𝜋 (𝑋) = 𝑌

. Uniqueness comes from (a).

Hint 3. (c) Induce a map

𝑉 / 𝑈 \to 𝑉 / 𝑋

from the quotient map

𝑉 \to 𝑉 / 𝑋

with Theorem 4.6.30, and compute its kernel.

Sketch. (a)

𝑋 / 𝑈 = 𝜋 (𝑋)

is a subspace, and

𝐯 + 𝑈 \in 𝑋 / 𝑈

forces

𝐯 \in 𝑋 + 𝑈 = 𝑋

. (b)

𝑋 = 𝜋^{- 1} (𝑌)

is a subspace containing

ker 𝜋 = 𝑈

, with

𝜋 (𝑋) = 𝑌

since

𝜋

is onto; if also

𝑌 = 𝑋^{'} / 𝑈

, then

𝑋^{'} = 𝜋^{- 1} (𝑌) = 𝑋

by (a). (c)

𝜋_{𝑋}

kills

𝑈

, so it induces

𝜌 : 𝑉 / 𝑈 \to 𝑉 / 𝑋

𝐯 + 𝑈 \mapsto 𝐯 + 𝑋

, onto, with kernel

𝑋 / 𝑈

. (d)

𝜋 ↾_{𝑋}

has kernel

𝑈

and image

𝑋 / 𝑈

(a) $𝑋 / 𝑈 = im (𝜋 ↾_{𝑋})$ , the image of a linear map, so it is a subspace (Proposition 2.6.8). Clearly $𝑋 \subseteq 𝜋^{- 1} (𝑋 / 𝑈)$ . Conversely, if $𝜋 (𝐯) \in 𝑋 / 𝑈$ , then $𝐯 + 𝑈 = 𝐱 + 𝑈$ for some $𝐱 \in 𝑋$ , so $𝐯 - 𝐱 \in 𝑈 \subseteq 𝑋$ (Lemma 4.6.14), and $𝐯 = 𝐱 + (𝐯 - 𝐱) \in 𝑋$ .

(b) Let $𝑌$ be a subspace of $𝑉 / 𝑈$ and $𝑋 = 𝜋^{- 1} (𝑌)$ . It is a subspace of $𝑉$ : it contains $𝟎$ , and if $𝐯, 𝐯^{'} \in 𝑋$ and $𝑎 \in 𝐹$ , then $𝜋 (𝑎 𝐯 + 𝐯^{'}) =$ $𝑎 𝜋 (𝐯) +$ $𝜋 (𝐯^{'}) \in$ $𝑌$ . It contains $𝑈$ , because $𝜋 (𝑈) = {𝟎} \subseteq 𝑌$ . And $𝑋 / 𝑈 = 𝜋 (𝑋) = 𝑌$ : the containment $\subseteq$ is the definition of $𝑋$ , and every $𝐲 \in 𝑌$ is $𝜋 (𝐯)$ for some $𝐯$ (as $𝜋$ is surjective), which then lies in $𝑋$ . For uniqueness, if $𝑌 = 𝑋^{'} / 𝑈$ with $𝑈 \subseteq 𝑋^{'}$ , then $𝑋^{'} =$ $𝜋^{- 1} (𝑋^{'} / 𝑈) =$ $𝜋^{- 1} (𝑌) =$ $𝑋$ by (a).

(c) Let $𝜋_{𝑋} : 𝑉 \to 𝑉 / 𝑋$ be the quotient map. Its kernel $𝑋$ contains $𝑈$ , so by Theorem 4.6.30(a) it induces a linear $𝜌 : 𝑉 / 𝑈 \to 𝑉 / 𝑋$ , $𝜌 (𝐯 + 𝑈) = 𝐯 + 𝑋$ , with image $im 𝜋_{𝑋} = 𝑉 / 𝑋$ and kernel ${𝐯 + 𝑈 | 𝐯 \in 𝑋} =$ $𝑋 / 𝑈$ . By Theorem 4.6.34, $(𝑉 / 𝑈) / (𝑋 / 𝑈) ≅$ $𝑉 / 𝑋$ .

(d) $𝜋 ↾_{𝑋} : 𝑋 \to 𝑉 / 𝑈$ is linear with kernel $𝑋 \cap 𝑈 = 𝑈$ and image $𝑋 / 𝑈$ , so rank–nullity gives $dim 𝑋 =$ $dim 𝑈 +$ $dim (𝑋 / 𝑈)$ .

So the subspaces of $𝑉 / 𝑈$ correspond to the subspaces of $𝑉$ that contain $𝑈$ , with dimensions shifted down by $dim 𝑈$ . The set $𝑋 / 𝑈$ is also literally the quotient of the space $𝑋$ by its subspace $𝑈$ , which explains the notation.

Exercise 4.6.51.hard Let $𝑈$ and $𝑊$ be subspaces of $𝑉$ . For each statement, decide whether it holds when $𝑉$ is finite-dimensional, and whether it holds for all $𝑉$ .

(a)If $𝑈 \neq {𝟎}$ , then $𝑉 / 𝑈 ≇ 𝑉$ .
(b)If $𝑉 / 𝑈 ≅ 𝑉 / 𝑊$ , then $𝑈 ≅ 𝑊$ .
(c)If $𝑈 ≅ 𝑊$ , then $𝑉 / 𝑈 ≅ 𝑉 / 𝑊$ .

Solution.

Uses. Theorem 4.6.27 (Dimension of a quotient), Theorem 4.5.6 (Classification by dimension), Example 4.6.36 (The first isomorphism theorem at work), Example 4.6.23 (Quotient spaces), Theorem 4.6.34 (First isomorphism theorem), Example 2.1.9 (Maps of polynomials)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. In finite dimensions, everything is decided by dimensions (Theorem 4.5.6 and Theorem 4.6.27).

Hint 2. In general, look in

𝐹 [𝑥]

. The backward shift of Example 4.6.36(d) gives

𝐹 [𝑥] / span (1) ≅

𝐹 [𝑥]

, and multiplication by

𝑥

maps

𝐹 [𝑥]

isomorphically onto the polynomials with constant term

0

Sketch. Finite dimensions: all three are true, by comparing

dim 𝑉 - dim 𝑈

and

dim 𝑉 - dim 𝑊

. In general all three are false in

𝑉 = 𝐹 [𝑥]

: (a)

𝑈 = span (1)

; (b)

𝑈 = span (1)

𝑊 = {𝟎}

; (c)

𝑈 = 𝐹 [𝑥]

𝑊 =

im 𝑀 =

{𝑝 | 𝑝 (0) = 0}

, where

𝑉 / 𝑈 = {𝟎}

but

𝑉 / 𝑊 ≅ 𝐹

Finite-dimensional $𝑉$ . All three statements are true. By Theorem 4.6.27, $dim (𝑉 / 𝑈) =$ $dim 𝑉 -$ $dim 𝑈$ and $dim (𝑉 / 𝑊) =$ $dim 𝑉 -$ $dim 𝑊$ , and finite-dimensional spaces are isomorphic exactly when their dimensions agree (Theorem 4.5.6).

(a) $dim (𝑉 / 𝑈) =$ $dim 𝑉 -$ $dim 𝑈 <$ $dim 𝑉$ when $𝑈 \neq {𝟎}$ .

(b) $dim 𝑉 -$ $dim 𝑈 =$ $dim 𝑉 -$ $dim 𝑊$ gives $dim 𝑈 = dim 𝑊$ .

(c) $dim 𝑈 = dim 𝑊$ gives $dim (𝑉 / 𝑈) =$ $dim (𝑉 / 𝑊)$ .

General $𝑉$ . All three fail for $𝑉 = 𝐹 [𝑥]$ .

(a) By Example 4.6.36(d), $𝐹 [𝑥] / span (1) ≅$ $𝐹 [𝑥]$ , although $span (1) \neq {𝟎}$ .

(b) Take $𝑈 = span (1)$ and $𝑊 = {𝟎}$ . Then $𝑉 / 𝑈 ≅ 𝑉 ≅ 𝑉 / 𝑊$ (Example 4.6.23(c)), but $𝑈$ is one-dimensional and $𝑊$ is zero, so $𝑈 ≇ 𝑊$ .

(c) Take $𝑈 = 𝐹 [𝑥]$ and $𝑊 = {𝑝 | 𝑝 (0) = 0}$ . Multiplication by $𝑥$ , $𝑀 (𝑝) = 𝑥 𝑝$ , is linear and injective (Example 2.1.9), and its image is $𝑊$ (a polynomial with constant term $0$ is $𝑥$ times the polynomial with its coefficients shifted down). So $𝑈 = 𝐹 [𝑥] ≅ 𝑊$ . But $𝑉 / 𝑈 = {𝟎}$ , while $𝑉 / 𝑊 ≅ 𝐹$ by Theorem 4.6.34 applied to ${ev}_{0}$ , which is surjective with kernel $𝑊$ .

The finite-dimensional arguments all rest on subtracting dimensions, which is exactly what is unavailable in infinite dimensions.

Exercise 4.6.52.hard Let $𝑉$ be finite-dimensional, $𝑇 \in L (𝑉)$ , and $𝑈$ a subspace with $𝑇 (𝑈) \subseteq 𝑈$ . Write $𝑇_{𝑈} \in L (𝑈)$ for $𝑇 ↾_{𝑈}$ regarded as an operator on $𝑈$ , and $𝑇 / 𝑈$ for the induced operator on $𝑉 / 𝑈$ (Proposition 4.6.32).

(a)Show that $rank 𝑇 \geq$ $rank 𝑇_{𝑈} +$ $rank (𝑇 / 𝑈)$ .
(b)Show that equality holds if and only if $im 𝑇 \cap 𝑈 = 𝑇 (𝑈)$ .
(c)Give an example where the inequality is strict. What does (a) say about block upper triangular matrices?

Solution.

Uses. Proposition 4.6.32 (Operators on a quotient), Proposition 4.2.3 (Properties of restriction), Theorem 4.2.11 (Rank–nullity), Proposition 4.1.21 (Dimensions of subspaces)

Techniques. Restrict to a subspace (Technique 4.2.5), Pass to a quotient (Technique 4.6.31)

Hint 1. Apply rank–nullity to

𝜋 ↾_{im 𝑇}

, where

𝜋 : 𝑉 \to 𝑉 / 𝑈

Hint 2. Its kernel is

im 𝑇 \cap 𝑈

, which contains

𝑇 (𝑈) = im 𝑇_{𝑈}

. Its image is

𝜋 (im 𝑇)

. Compare it with

im (𝑇 / 𝑈)

Sketch.

𝜋 ↾_{im 𝑇}

has kernel

im 𝑇 \cap 𝑈 \supseteq 𝑇 (𝑈)

and image

{𝑇 𝐯 + 𝑈} =

im (𝑇 / 𝑈)

. So

rank 𝑇 =

dim (im 𝑇 \cap 𝑈) +

rank (𝑇 / 𝑈) \geq

rank 𝑇_{𝑈} +

rank (𝑇 / 𝑈)

, with equality iff

im 𝑇 \cap 𝑈 = 𝑇 (𝑈)

. Example: the

2 \times 2

shift with

𝑈 = span (𝐞_{1})

(a) Let $𝑋 = im 𝑇$ and $𝜋 : 𝑉 \to 𝑉 / 𝑈$ . The restriction $𝜋 ↾_{𝑋}$ is linear with kernel $𝑋 \cap 𝑈$ (Proposition 4.2.3(c)) and image ${𝑇 (𝐯) + 𝑈 | 𝐯 \in 𝑉} =$ ${(𝑇 / 𝑈) (𝐯 + 𝑈)} =$ $im (𝑇 / 𝑈)$ . By rank–nullity,

rank 𝑇 = dim 𝑋 = dim (𝑋 \cap 𝑈) + rank (𝑇 / 𝑈) .

Since $𝑇 (𝑈) \subseteq 𝑈$ and $𝑇 (𝑈) \subseteq 𝑋$ , we have $im 𝑇_{𝑈} =$ $𝑇 (𝑈) \subseteq$ $𝑋 \cap$ $𝑈$ , so $dim (𝑋 \cap 𝑈) \geq$ $rank 𝑇_{𝑈}$ (Proposition 4.1.21), and (a) follows.

(b) By the computation in (a), equality holds if and only if $dim (𝑋 \cap 𝑈) =$ $dim 𝑇 (𝑈)$ , which, as $𝑇 (𝑈) \subseteq 𝑋 \cap 𝑈$ , holds if and only if $𝑇 (𝑈) = 𝑋 \cap 𝑈$ (Proposition 4.1.21(b)).

(c) Let $𝑁 \in L (𝐹^{2})$ with $𝑁 (𝐞_{1}) = 𝟎$ , $𝑁 (𝐞_{2}) = 𝐞_{1}$ , and $𝑈 = span (𝐞_{1})$ . Then $𝑁_{𝑈} = 0$ , and $(𝑁 / 𝑈) (𝐞_{2} + 𝑈) =$ $𝐞_{1} +$ $𝑈 =$ $𝟎 +$ $𝑈$ , so $𝑁 / 𝑈 = 0$ . Both ranks are $0$ , but $rank 𝑁 = 1$ . Here $im 𝑁 \cap 𝑈 = 𝑈$ , while $𝑁 (𝑈) = {𝟎}$ .

In matrix terms (Proposition 4.6.32(b)): if $𝑀 = (\begin{matrix} 𝐴 & 𝐵 \\ 0 & 𝐶 \end{matrix})$ with square diagonal blocks, then $rank 𝑀 \geq$ $rank 𝐴 +$ $rank 𝐶$ , and the example $(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ shows that the off-diagonal block can add rank.

Exercise 4.6.53.hard Reprove Frobenius’s rank inequality (Exercise 4.3.31(a)) with quotients: for $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ , $𝐵 \in 𝐹^{𝑛 \times 𝑝}$ , and $𝐶 \in 𝐹^{𝑝 \times 𝑞}$ ,

rank (𝐴 𝐵) + rank (𝐵 𝐶) \leq rank 𝐵 + rank (𝐴 𝐵 𝐶) .

Use the map $im 𝑇_{𝐵} / im 𝑇_{𝐵 𝐶} \to$ $im 𝑇_{𝐴 𝐵} / im 𝑇_{𝐴 𝐵 𝐶}$ induced by $𝑇_{𝐴}$ .

Solution.

Uses. Theorem 4.6.30 (Universal property of the quotient), Theorem 4.6.27 (Dimension of a quotient), Theorem 4.2.7 (The image of $𝑇_{𝐴}$ ), Theorem 4.2.11 (Rank–nullity)

Techniques. Pass to a quotient (Technique 4.6.31)

Hint 1. Check the containments

im 𝑇_{𝐵 𝐶} \subseteq im 𝑇_{𝐵}

and

im 𝑇_{𝐴 𝐵 𝐶} \subseteq

im 𝑇_{𝐴 𝐵}

, and that

𝑇_{𝐴}

maps

im 𝑇_{𝐵}

into

im 𝑇_{𝐴 𝐵}

Hint 2. Define

𝜑 : im 𝑇_{𝐵} \to

im 𝑇_{𝐴 𝐵} / im 𝑇_{𝐴 𝐵 𝐶}

𝜑 (𝐱) =

𝐴 𝐱 +

im 𝑇_{𝐴 𝐵 𝐶}

. Why does it vanish on

im 𝑇_{𝐵 𝐶}

? Why is the induced map surjective?

Hint 3. A surjective linear map cannot increase dimension. Translate both quotient dimensions with Theorem 4.6.27.

Sketch. Let

𝑋 =

im 𝑇_{𝐵} \supseteq

𝑌 =

im 𝑇_{𝐵 𝐶}

and

𝑋^{'} =

im 𝑇_{𝐴 𝐵} \supseteq

𝑌^{'} =

im 𝑇_{𝐴 𝐵 𝐶}

. The map

𝐱 \mapsto 𝐴 𝐱 + 𝑌^{'}

𝑋

kills

𝑌

(as

𝐴 𝐵 𝐶 𝐳 \in 𝑌^{'}

), so it induces

𝑋 / 𝑌 \to 𝑋^{'} / 𝑌^{'}

, which is onto since

𝐴 𝐵 𝐳 + 𝑌^{'}

is the image of

𝐵 𝐳 + 𝑌

. Hence

rank (𝐴 𝐵) -

rank (𝐴 𝐵 𝐶) \leq

rank 𝐵 -

rank (𝐵 𝐶)

Let $𝑋 = im 𝑇_{𝐵}$ , $𝑌 = im 𝑇_{𝐵 𝐶}$ , $𝑋^{'} = im 𝑇_{𝐴 𝐵}$ , and $𝑌^{'} = im 𝑇_{𝐴 𝐵 𝐶}$ . Since $𝐵 𝐶 𝐳 = 𝐵 (𝐶 𝐳)$ and $𝐴 𝐵 𝐶 𝐳 =$ $𝐴 𝐵 (𝐶 𝐳)$ , we have $𝑌 \subseteq 𝑋$ and $𝑌^{'} \subseteq 𝑋^{'}$ , and $𝐴 𝑋 \subseteq 𝑋^{'}$ since $𝐴 (𝐵 𝐳) = (𝐴 𝐵) 𝐳$ .

Define $𝜑 : 𝑋 \to 𝑋^{'} / 𝑌^{'}$ by $𝜑 (𝐱) = 𝐴 𝐱 + 𝑌^{'}$ . It is linear, being the composite of $𝑇_{𝐴} ↾_{𝑋} : 𝑋 \to 𝑋^{'}$ with the quotient map $𝑋^{'} \to 𝑋^{'} / 𝑌^{'}$ . For $𝐱 = 𝐵 𝐶 𝐳 \in 𝑌$ , $𝐴 𝐱 = 𝐴 𝐵 𝐶 𝐳 \in 𝑌^{'}$ , so $𝜑 (𝐱) = 𝟎$ . By Theorem 4.6.30, $𝜑$ induces a linear map $\tilde{𝜑} : 𝑋 / 𝑌 \to 𝑋^{'} / 𝑌^{'}$ with $\tilde{𝜑} (𝐱 + 𝑌) =$ $𝐴 𝐱 +$ $𝑌^{'}$ . It is surjective: every element of $𝑋^{'} / 𝑌^{'}$ is $𝐴 𝐵 𝐳 +$ $𝑌^{'} =$ $\tilde{𝜑} (𝐵 𝐳 + 𝑌)$ .

By rank–nullity, $dim (𝑋^{'} / 𝑌^{'}) =$ $dim im \tilde{𝜑} \leq$ $dim (𝑋 / 𝑌)$ . By Theorem 4.6.27 and Theorem 4.2.7,

rank (𝐴 𝐵) - rank (𝐴 𝐵 𝐶) \leq rank 𝐵 - rank (𝐵 𝐶),

which rearranges to the inequality.

The quotient proof makes the meaning visible: $𝑇_{𝐴}$ maps “what $𝐵$ reaches beyond $𝐵 𝐶$ ” onto “what $𝐴 𝐵$ reaches beyond $𝐴 𝐵 𝐶$ ”, and a linear map cannot make a space bigger.

Exercise 4.6.54.hard Let $𝐹$ be a field with $𝑞$ elements, let $𝑉$ be an $𝑛$ -dimensional space over $𝐹$ , and let $𝑈$ be a subspace of dimension $𝑘$ . Fix one complement $𝑊_{0}$ of $𝑈$ .

(a)For $𝑆 \in L (𝑊_{0}, 𝑈)$ , show that $𝑊_{𝑆} =$ ${𝐰 + 𝑆 (𝐰) | 𝐰 \in 𝑊_{0}}$ is a complement of $𝑈$ .
(b)Show that every complement of $𝑈$ is $𝑊_{𝑆}$ for exactly one $𝑆 \in L (𝑊_{0}, 𝑈)$ .
(c)Deduce that $𝑈$ has exactly $𝑞^{𝑘 (𝑛 - 𝑘)}$ complements. How many lines in $𝔽_{3}^{2}$ are complements of a given line?

Solution.

Uses. Theorem 4.4.7 (Criteria for direct sums), Proposition 4.4.16 (Complements exist), Proposition 4.6.29 (Complements and the quotient), Theorem 4.5.8 (Dimension of $L (𝑉, 𝑊)$ ), Proposition 4.1.34 (Size of a vector space over a finite field), Proposition 4.2.6 (Injective maps preserve dimension)

Techniques. Uniqueness via two candidates (Technique 1.1.8), Dimension counting (Technique 4.2.24)

Hint 1. A complement is a choice of one vector in each coset of

𝑈

, made linearly (Proposition 4.6.29). Starting from

𝑊_{0}

, each

𝐰 \in 𝑊_{0}

can be moved within its coset by a vector of

𝑈

, namely

𝑆 (𝐰)

Hint 2. For (b), given a complement

𝑊

, decompose each

𝐰 \in 𝑊_{0}

along

𝑉 = 𝑈 \oplus 𝑊

𝐰 = 𝐮 + 𝐱

, and set

𝑆 (𝐰) = - 𝐮

Hint 3. For (c),

L (𝑊_{0}, 𝑈)

has dimension

(𝑛 - 𝑘) 𝑘

(Theorem 4.5.8); count its elements.

Sketch. (a)

𝑊_{𝑆}

is the image of

𝐰 \mapsto 𝐰 + 𝑆 𝐰

. If

𝐰 + 𝑆 𝐰 \in 𝑈

, then

𝐰 \in 𝑈 \cap 𝑊_{0} = {𝟎}

; and

𝑈 + 𝑊_{𝑆} \supseteq 𝑈 + 𝑊_{0} = 𝑉

. (b) If

𝑉 = 𝑈 \oplus 𝑊

, let

- 𝑆 (𝐰)

be the

𝑈

-component of

𝐰 \in 𝑊_{0}

;

𝑆

is linear by uniqueness of components,

𝑊_{𝑆} \subseteq 𝑊

, and dimensions agree. If

𝑊_{𝑆} = 𝑊_{𝑆^{'}}

, comparing

𝐰 +

𝑆 𝐰 =

𝐰^{'} +

𝑆^{'} 𝐰^{'}

𝑊_{0} \oplus 𝑈

gives

𝑆 = 𝑆^{'}

. (c)

| L (𝑊_{0}, 𝑈) | =

𝑞^{𝑘 (𝑛 - 𝑘)}

. In

𝔽_{3}^{2}

3

By Proposition 4.4.16(b), $dim 𝑊_{0} = 𝑛 - 𝑘$ .

(a) $𝑊_{𝑆}$ is the image of the linear map $𝐰 \mapsto 𝐰 + 𝑆 (𝐰)$ , $𝑊_{0} \to 𝑉$ , so it is a subspace. If $𝐰 + 𝑆 (𝐰) \in 𝑈$ , then $𝐰 \in 𝑈$ (as $𝑆 (𝐰) \in 𝑈$ ), so $𝐰 \in 𝑈 \cap 𝑊_{0} = {𝟎}$ and the vector is $𝟎$ . So $𝑈 \cap 𝑊_{𝑆} = {𝟎}$ . Every $𝐰 \in 𝑊_{0}$ equals $(𝐰 + 𝑆 (𝐰)) -$ $𝑆 (𝐰) \in$ $𝑊_{𝑆} +$ $𝑈$ , so $𝑈 + 𝑊_{𝑆} \supseteq 𝑈 + 𝑊_{0} = 𝑉$ . Hence $𝑉 = 𝑈 \oplus 𝑊_{𝑆}$ (Theorem 4.4.7).

(b) Let $𝑊$ be a complement of $𝑈$ . Each $𝐰 \in 𝑊_{0}$ has a unique decomposition $𝐰 = 𝐮 + 𝐱$ with $𝐮 \in 𝑈$ , $𝐱 \in 𝑊$ ; set $𝑆 (𝐰) = - 𝐮$ . If $𝐰 = 𝐮 + 𝐱$ and $𝐰^{'} = 𝐮^{'} + 𝐱^{'}$ , then $𝑎 𝐰 +$ $𝐰^{'} =$ $(𝑎 𝐮 + 𝐮^{'}) +$ $(𝑎 𝐱 + 𝐱^{'})$ is the decomposition of $𝑎 𝐰 + 𝐰^{'}$ , so $𝑆$ is linear, with values in $𝑈$ . Now $𝐰 + 𝑆 (𝐰) = 𝐱 \in 𝑊$ , so $𝑊_{𝑆} \subseteq 𝑊$ . Both have dimension $𝑛 - 𝑘$ : $𝑊$ by Proposition 4.4.16(b), and $𝑊_{𝑆}$ because $𝐰 \mapsto 𝐰 + 𝑆 (𝐰)$ is injective (if $𝐰 + 𝑆 (𝐰) = 𝟎$ , then $𝐰 =$ $- 𝑆 (𝐰) \in$ $𝑈 \cap$ $𝑊_{0} =$ ${𝟎}$ ) and Proposition 4.2.6(a). So $𝑊_{𝑆} = 𝑊$ .

For uniqueness, suppose $𝑊_{𝑆} = 𝑊_{𝑆^{'}}$ . For $𝐰 \in 𝑊_{0}$ , the vector $𝐰 + 𝑆 (𝐰)$ lies in $𝑊_{𝑆^{'}}$ , so $𝐰 +$ $𝑆 (𝐰) =$ $𝐰^{'} +$ $𝑆^{'} (𝐰^{'})$ for some $𝐰^{'} \in 𝑊_{0}$ . Then $𝐰 -$ $𝐰^{'} =$ $𝑆^{'} (𝐰^{'}) -$ $𝑆 (𝐰) \in$ $𝑊_{0} \cap$ $𝑈 =$ ${𝟎}$ , so $𝐰^{'} = 𝐰$ and $𝑆 (𝐰) = 𝑆^{'} (𝐰)$ . Hence $𝑆 = 𝑆^{'}$ .

(c) By (a) and (b), $𝑆 \mapsto 𝑊_{𝑆}$ is a bijection from $L (𝑊_{0}, 𝑈)$ onto the set of complements of $𝑈$ . The space $L (𝑊_{0}, 𝑈)$ has dimension $(𝑛 - 𝑘) 𝑘$ (Theorem 4.5.8), so it has $𝑞^{𝑘 (𝑛 - 𝑘)}$ elements (Proposition 4.1.34). For a line $𝑈$ in $𝔽_{3}^{2}$ , $𝑛 = 2$ and $𝑘 = 1$ , so there are $3^{1} = 3$ complements. Indeed, $𝔽_{3}^{2}$ has $(9 - 1) / 2 = 4$ lines, and every line other than $𝑈$ meets $𝑈$ only in $𝟎$ .

Exercise 4.6.55.challenge Let $𝑈$ be a subspace of $𝐹 [𝑥]$ with $𝑈 \neq {𝟎}$ and $𝑥 𝑝 \in 𝑈$ for every $𝑝 \in 𝑈$ . Let $𝑑$ be the smallest degree of a nonzero element of $𝑈$ .

(a)Show that $𝑈$ contains a polynomial $𝑝 =$ $𝑥^{𝑑} +$ $𝑐_{𝑑 - 1} 𝑥^{𝑑 - 1} +$ $\dots +$ $𝑐_{1} 𝑥 +$ $𝑐_{0}$ , and that ${𝐹 [𝑥]}_{\leq 𝑑 - 1}$ is a complement of $𝑈$ (read ${𝐹 [𝑥]}_{\leq - 1}$ as ${𝟎}$ ). Conclude that $dim (𝐹 [𝑥] / 𝑈) = 𝑑$ .
(b)Suppose $𝑑 \geq 1$ . Show that multiplication by $𝑥$ induces an operator $𝑋$ on $𝐹 [𝑥] / 𝑈$ with $𝑋 (𝑞 + 𝑈) = 𝑥 𝑞 + 𝑈$ , and find the matrix of $𝑋$ in the basis $1 +$ $𝑈,$ $𝑥 +$ $𝑈,$ $\dots,$ $𝑥^{𝑑 - 1} +$ $𝑈$ .

Solution.

Uses. Proposition 4.6.29 (Complements and the quotient), Proposition 4.6.32 (Operators on a quotient), Example 2.1.9 (Maps of polynomials), Proposition 1.2.18 (Degree and evaluation versus the operations), Definition 2.3.1 (Matrix of a linear map)

Techniques. Pass to a quotient (Technique 4.6.31), Extremal choice and minimal counterexamples (Technique 1.5.9)

Hint 1. Scale a nonzero element of degree

𝑑

to make it monic. By minimality of

𝑑

, which polynomials of degree less than

𝑑

lie in

𝑈

Hint 2. To show

{𝐹 [𝑥]}_{\leq 𝑑 - 1} +

𝑈 =

𝐹 [𝑥]

, induct on the degree of

𝑞

: subtract a multiple of

𝑥^{𝑒 - 𝑑} 𝑝 \in 𝑈

to kill the leading term of

𝑞

Hint 3. For (b), apply Proposition 4.6.32 to

𝑀 (𝑞) = 𝑥 𝑞

. The last basis vector goes to

𝑥^{𝑑} + 𝑈

; rewrite

𝑥^{𝑑}

modulo

𝑝

Sketch. (a) A nonzero element of degree

𝑑

divided by its leading coefficient is

𝑝

. Nonzero elements of

𝑈

have degree

\geq 𝑑

, so

𝑈 \cap

{𝐹 [𝑥]}_{\leq 𝑑 - 1} =

{𝟎}

. If

deg 𝑞 = 𝑒 \geq 𝑑

with leading coefficient

𝑏

, then

𝑥^{𝑒 - 𝑑} 𝑝 \in 𝑈

and

deg (𝑞 - 𝑏 𝑥^{𝑒 - 𝑑} 𝑝) <

𝑒

, so by induction

𝑞 \in {𝐹 [𝑥]}_{\leq 𝑑 - 1} + 𝑈

. By Proposition 4.6.29,

𝐹 [𝑥] / 𝑈 ≅

{𝐹 [𝑥]}_{\leq 𝑑 - 1}

. (b)

𝑀 (𝑈) \subseteq 𝑈

gives

𝑋 = 𝑀 / 𝑈

, and

𝑋 (𝑥^{𝑗} + 𝑈) =

𝑥^{𝑗 + 1} +

𝑈

for

𝑗 < 𝑑 - 1

𝑋 (𝑥^{𝑑 - 1} + 𝑈) =

𝑥^{𝑑} +

𝑈 =

- (𝑐_{0} + \dots + 𝑐_{𝑑 - 1} 𝑥^{𝑑 - 1}) +

𝑈

: the companion matrix of

𝑝

(a) Let $𝑝_{0} \in 𝑈$ be nonzero of degree $𝑑$ with leading coefficient $𝑎 \neq 0$ . Then $𝑝 = 𝑎^{- 1} 𝑝_{0} \in 𝑈$ has degree $𝑑$ and leading coefficient $1$ , so it has the stated form. By the hypothesis applied $𝑗$ times, $𝑥^{𝑗} 𝑝 \in 𝑈$ for every $𝑗 \geq 0$ , and $𝑥^{𝑗} 𝑝$ has degree $𝑑 + 𝑗$ and leading coefficient $1$ .

Intersection. A nonzero element of $𝑈$ has degree at least $𝑑$ , by the choice of $𝑑$ , so $𝑈 \cap$ ${𝐹 [𝑥]}_{\leq 𝑑 - 1} =$ ${𝟎}$ .

Sum. We show by strong induction on $𝑒$ that every $𝑞$ of degree $𝑒$ lies in ${𝐹 [𝑥]}_{\leq 𝑑 - 1} + 𝑈$ ; the zero polynomial lies there trivially. If $𝑒 \leq 𝑑 - 1$ , then $𝑞 \in {𝐹 [𝑥]}_{\leq 𝑑 - 1}$ . If $𝑒 \geq 𝑑$ , let $𝑏$ be the leading coefficient of $𝑞$ . Then $𝑞 - 𝑏 𝑥^{𝑒 - 𝑑} 𝑝$ is zero or has degree less than $𝑒$ , because the coefficients of $𝑥^{𝑒}$ cancel (Proposition 1.2.18). By induction it lies in ${𝐹 [𝑥]}_{\leq 𝑑 - 1} + 𝑈$ , and since $𝑏 𝑥^{𝑒 - 𝑑} 𝑝 \in 𝑈$ , so does $𝑞$ .

Hence $𝐹 [𝑥] =$ ${𝐹 [𝑥]}_{\leq 𝑑 - 1} \oplus$ $𝑈$ , and by Proposition 4.6.29, $𝑞 \mapsto 𝑞 + 𝑈$ restricts to an isomorphism ${𝐹 [𝑥]}_{\leq 𝑑 - 1} \to$ $𝐹 [𝑥] / 𝑈$ . So $dim (𝐹 [𝑥] / 𝑈) = 𝑑$ , and $1 +$ $𝑈,$ $𝑥 +$ $𝑈,$ $\dots,$ $𝑥^{𝑑 - 1} +$ $𝑈$ is a basis of $𝐹 [𝑥] / 𝑈$ (the image of the basis $1, 𝑥, \dots, 𝑥^{𝑑 - 1}$ ). If $𝑑 = 0$ , then $𝑝 = 1$ and $𝑈 = 𝐹 [𝑥]$ .

(b) Multiplication by $𝑥$ , $𝑀 (𝑞) = 𝑥 𝑞$ , is linear (Example 2.1.9), and $𝑀 (𝑈) \subseteq 𝑈$ by hypothesis. By Proposition 4.6.32(a), $𝑋 = 𝑀 / 𝑈$ is an operator on $𝐹 [𝑥] / 𝑈$ with $𝑋 (𝑞 + 𝑈) = 𝑥 𝑞 + 𝑈$ . On the basis,

𝑋 (𝑥^{𝑗} + 𝑈) = 𝑥^{𝑗 + 1} + 𝑈 (0 \leq 𝑗 \leq 𝑑 - 2),

and, since $𝑝 \in 𝑈$ ,

𝑋 (𝑥^{𝑑 - 1} + 𝑈) = 𝑥^{𝑑} + 𝑈 = (𝑥^{𝑑} - 𝑝) + 𝑈 = - 𝑐_{0} (1 + 𝑈) - 𝑐_{1} (𝑥 + 𝑈) - \dots - 𝑐_{𝑑 - 1} (𝑥^{𝑑 - 1} + 𝑈) .

So the matrix of $𝑋$ has $1$ s just below the diagonal, last column $(- 𝑐_{0}, - 𝑐_{1}, \dots, - 𝑐_{𝑑 - 1})$ , and zeros elsewhere:

(\begin{matrix} 0 & 0 & \dots & 0 & - 𝑐_{0} \\ 1 & 0 & \dots & 0 & - 𝑐_{1} \\ 0 & 1 & \dots & 0 & - 𝑐_{2} \\ ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & 1 & - 𝑐_{𝑑 - 1} \end{matrix}) .

This is the companion matrix of $𝑝$ , which reappears in Section 7.3.

One can show that $𝑈$ is exactly the set of polynomial multiples of $𝑝$ (polynomial multiplication and division come in Section 7.1). So the quotients of $𝐹 [𝑥]$ by such subspaces are finite-dimensional, and multiplication by $𝑥$ on them is the basic example of an operator with a cyclic vector.

More practice

The practice bank has 16 exercises on this section’s topics.