4.4 Products and Quotient Spaces

So far every new vector space in this book has been found inside an old one, as a subspace. This section builds new spaces in two other ways. The first glues spaces together side by side, making a product; it is the “external” direct sum promised in Chapter 1. The second, the quotient space $V/U$ , is subtler and more important: it collapses a subspace $U$ to a single point and keeps only what $V$ knows beyond $U$ . Quotients let us state precisely what a linear map remembers and what it forgets, in the three Isomorphism Theorems.

4.4.1. Products of vector spaces

In Chapter 1 we wrote $V = U \oplus W$ for subspaces $U$ and $W$ of one space $V$ (Definition 1.7.2). The two pieces already lived in a common space, and “direct” was a property of how they sat there. But often we want to combine two spaces that have nothing to do with each other, such as $\mathbb{R}^2$ and $\mathbb{R}[x]_{\le 1}$ , into one space in which each keeps its identity. There is no common space to add inside, so we build one.

The model is $F^2$ . A vector of $F^2$ is a pair $(a, b)$ of scalars, and we add and scale pairs entry by entry. Nothing in that recipe needs the entries to be scalars: the entries could be vectors, taken from any vector spaces over $F$ .

The product of vector spaces is the space of lists with one entry from each space, added and scaled entry by entry.

Definition 4.4.1 (Product of vector spaces)

Let $V_1, \dots, V_k$ be vector spaces over the same field $F$ . Their product (or external direct sum) is the Cartesian product (Definition 0.3.10) $V_1 \times \dots \times V_k = \{ (v_1, \dots, v_k) : v_i \in V_i \text{ for each } i \},$ with addition and scalar multiplication defined entrywise: for $a \in F$ , $\begin{aligned} (v_1, \dots, v_k) + (w_1, \dots, w_k) &\mathrel{:=} (v_1 + w_1, \dots, v_k + w_k), \\ a(v_1, \dots, v_k) &\mathrel{:=} (av_1, \dots, av_k). \end{aligned}$

In words: the $i$ -th entry of a sum is the sum of the $i$ -th entries, computed in $V_i$ , and similarly for scalar multiples. Two elements are equal exactly when all their entries are equal. Every operation happens inside one $V_i$ at a time, which is why all the $V_i$ must share the field $F$ : the same scalar $a$ has to act on every entry.

It is a vector space. Both operations land in $V_1 \times \dots \times V_k$ , since $v_i + w_i \in V_i$ and $av_i \in V_i$ . Each axiom of Definition 1.1.1 is an equation between lists, so it holds when it holds in each entry, and in each entry it is the same axiom in $V_i$ . For example, (VS6) reads $\begin{aligned} (a + b)(v_1, \dots, v_k) &= ((a + b)v_1, \dots, (a + b)v_k) \\ &= (av_1 + bv_1, \dots, av_k + bv_k) \\ &= a(v_1, \dots, v_k) + b(v_1, \dots, v_k), \end{aligned}$ using (VS6) in each $V_i$ . The zero vector is $(0, \dots, 0)$ , with the zero of $V_i$ in the $i$ -th entry, and the additive inverse of $(v_1, \dots, v_k)$ is $(-v_1, \dots, -v_k)$ . The other axioms are checked in exactly the same way.

Examples.

The standard example. $F \times \dots \times F$ ( $n$ factors) is $F^n$ with its usual operations.
Unrelated spaces. An element of $\mathbb{R}^2 \times \mathbb{R}[x]_{\le 1}$ is a pair $((a, b), c + dx)$ , and $((1, 0), x) + ((2, 5), 3 - x) = ((3, 5), 3)$ .
Augmented data. A linear system $\mathbf{A}\mathbf{x} = \mathbf{b}$ with $m$ equations in $n$ unknowns is a single element $(\mathbf{A}, \mathbf{b})$ of $M_{m \times n}(F) \times F^m$ .
Degenerate cases. For $k = 1$ the product is $V_1$ itself, with entries written in parentheses. And $V \times \{0\}$ consists of the pairs $(v, 0)$ ; the map $v \mapsto (v, 0)$ is a linear bijection $V \to V \times \{0\}$ , so a zero factor adds nothing.

Nonexample by minimal change. Take $V_1 = \mathbb{C}$ as a vector space over $\mathbb{C}$ and $V_2 = \mathbb{R}$ over $\mathbb{R}$ . The set $\mathbb{C} \times \mathbb{R}$ is fine, but the entrywise rule $i \cdot (z, t) = (iz, it)$ fails: $it \notin \mathbb{R}$ for $t \ne 0$ . The clause that fails is “over the same field”. Restricting the scalars of $\mathbb{C}$ to $\mathbb{R}$ (Definition 1.9.1) repairs it, and $\mathbb{C} \times \mathbb{R}$ over $\mathbb{R}$ is a vector space of dimension $3$ .

Bases of the factors combine into a basis of the product. For $v \in V_i$ , write $\iota_i(v)$ for the list with $v$ in the $i$ -th entry and $0$ in every other entry. The map $\iota_i \colon V_i \to V_1 \times \dots \times V_k$ is linear, because the operations are entrywise.

Theorem 4.4.1 (Dimension of a Product)

Let $V_1, \dots, V_k$ be finite-dimensional vector spaces over $F$ , and let $(b_{i1}, \dots, b_{in_i})$ be a basis of $V_i$ for each $i$ . Then the list of all vectors $\iota_i(b_{ij})$ , for $i = 1, \dots, k$ and $j = 1, \dots, n_i$ , is a basis of $V_1 \times \dots \times V_k$ . In particular, $\dim(V_1 \times \dots \times V_k) = \dim V_1 + \dots + \dim V_k .$

Proof. Spanning. Let $(v_1, \dots, v_k)$ be in the product. Since the operations are entrywise, $(v_1, \dots, v_k) = \iota_1(v_1) + \dots + \iota_k(v_k)$ . Each $v_i$ is a combination $\sum_j c_{ij}b_{ij}$ , and $\iota_i$ is linear, so $\iota_i(v_i) = \sum_j c_{ij}\,\iota_i(b_{ij})$ (Theorem 2.1.4). Hence $(v_1, \dots, v_k)$ is a combination of the vectors $\iota_i(b_{ij})$ .

Independence. Let $\sum_{i=1}^{k}\sum_{j=1}^{n_i} c_{ij}\,\iota_i(b_{ij}) = 0$ . The $i$ -th entry of the left side is $\sum_j c_{ij}b_{ij}$ , because every $\iota_{i'}(b_{i'j})$ with $i' \ne i$ has $0$ there. So $\sum_j c_{ij}b_{ij} = 0$ in $V_i$ for each $i$ , and the independence of the basis of $V_i$ gives $c_{ij} = 0$ for all $j$ .

The list has $n_1 + \dots + n_k$ vectors, which gives the dimension (Definition 1.6.1).

For instance, $\dim(\mathbb{R}^2 \times \mathbb{R}[x]_{\le 1}) = 2 + 2 = 4$ , so this product is isomorphic to $\mathbb{R}^4$ by Theorem 2.5.4.

How does the external product relate to the internal direct sum of Chapter 1? If $U_1, \dots, U_k$ are subspaces of $V$ , there is a natural linear map from the product to $V$ : add up the entries. The sum is direct exactly when this map loses no information.

Theorem 4.4.2 (Internal and External Direct Sums)

(a) Let $U_1, \dots, U_k$ be subspaces of a vector space $V$ , and let $S \colon U_1 \times \dots \times U_k \to V, \qquad S(u_1, \dots, u_k) = u_1 + \dots + u_k .$ Then $S$ is linear with $\operatorname{im} S = U_1 + \dots + U_k$ . It is injective if and only if the sum $U_1 + \dots + U_k$ is direct. In that case $S$ is an isomorphism from $U_1 \times \dots \times U_k$ onto $U_1 \oplus \dots \oplus U_k$ .
(b) Conversely, let $V_1, \dots, V_k$ be vector spaces over $F$ . Then the images $\iota_i(V_i)$ are subspaces of $V_1 \times \dots \times V_k$ , each isomorphic to $V_i$ via $\iota_i$ , and $V_1 \times \dots \times V_k = \iota_1(V_1) \oplus \dots \oplus \iota_k(V_k).$

Proof.

(a) For lists $(u_i)$ , $(u_i')$ , and $a \in F$ , regrouping the finite sums gives $S((u_i) + (u_i')) = \sum_i (u_i + u_i') = S((u_i)) + S((u_i'))$ and $S(a(u_i)) = \sum_i au_i = aS((u_i))$ , so $S$ is linear. Its image is $\{ u_1 + \dots + u_k : u_i \in U_i \}$ , which is $U_1 + \dots + U_k$ by Definition 1.7.1. Its kernel consists of the lists with $u_1 + \dots + u_k = 0$ . By Theorem 2.2.3, $S$ is injective if and only if the only such list is $(0, \dots, 0)$ , which is condition (b) of Theorem 1.7.5, equivalent to the sum being direct. In that case $S$ , viewed as a map onto its image, is a linear bijection, hence an isomorphism (Definition 2.5.2).
(b) Each $\iota_i$ is linear and injective (its kernel is $\{0\}$ , since $\iota_i(v)$ has $v$ as its $i$ -th entry), so $\iota_i(V_i)$ is a subspace (Theorem 2.2.2) and $\iota_i$ is a bijection onto it. Every list satisfies $(v_1, \dots, v_k) = \iota_1(v_1) + \dots + \iota_k(v_k)$ , so the product is the sum of the subspaces $\iota_i(V_i)$ . If $\iota_1(v_1) + \dots + \iota_k(v_k) = 0$ , then $(v_1, \dots, v_k) = (0, \dots, 0)$ , so every piece $\iota_i(v_i)$ is $0$ . By Theorem 1.7.5 ((b) $\Rightarrow$ (a)) the sum is direct.

So internal and external direct sums are the same thing seen from two sides. An internal direct sum is a product that happens to live inside $V$ ; a product is an internal direct sum of copies of its factors. Many books write $\oplus$ for both. This book keeps $\times$ for the external construction, so that $U \oplus W$ always refers to subspaces of one space.

Quick check. Use Theorem 4.4.2 (a) and Rank–Nullity to explain why, for finite-dimensional subspaces $U, W$ of $V$ , the sum $U + W$ is direct if and only if $\dim(U + W) = \dim U + \dim W$ .

Solution. The addition map $S \colon U \times W \to V$ has image $U + W$ , and $\dim(U \times W) = \dim U + \dim W$ by Theorem 4.4.1. By Theorem 2.3.1, $\dim U + \dim W = \dim(U + W) + \dim \operatorname{ker} S$ . So $\dim(U + W) = \dim U + \dim W$ exactly when $\operatorname{ker} S = \{0\}$ , that is, when $S$ is injective, which by the theorem means the sum is direct. This recovers Theorem 1.7.4 (a) $\Leftrightarrow$ (c).

4.4.2. Cosets

The second construction starts from a picture you have already seen. In Chapter 1 we compared the line $U = \{ x + 2y = 0 \}$ through the origin with the parallel line $L = \{ x + 2y = 3 \}$ , which is not a subspace. In Chapter 3 we found that the solutions of a consistent system $\mathbf{A}\mathbf{x} = \mathbf{b}$ form a set $\mathbf{p} + \operatorname{nul}(\mathbf{A})$ , a subspace shifted by one particular solution (Theorem 3.1.3). Both are translates of a subspace. They deserve a name.

Definition 4.4.2 (Coset)

Let $U$ be a subspace of a vector space $V$ , and let $v \in V$ . The coset of $U$ through $v$ is the subset $v + U \mathrel{:=} \{ v + u : u \in U \} \subseteq V .$ Any $w \in v + U$ is called a representative of the coset.

In $\mathbb{R}^2$ , with $U = \{ x + 2y = 0 \} = \operatorname{span}((2, -1))$ , the coset through $(3, 0)$ is the line $\{ x + 2y = 3 \}$ : a point $(3, 0) + t(2, -1) = (3 + 2t, -t)$ satisfies $x + 2y = 3$ , and every solution of $x + 2y = 3$ is of this form with $t = -y$ . In general $(a, b) + U = \{ x + 2y = a + 2b \}$ , so the cosets of $U$ are exactly the lines parallel to $U$ .

$Diagram. Labels: x, y, mathbf 0, cvec v = (1, 1), cvec v' = (3, 0), U = \x + 2y = 0\, cvec v + U = \x + 2y = 3\, \x + 2y = -2\, +\,(2, -1)$

The picture shows three cosets of $U$ : $U$ itself (solid), the line through $\mathbf{v} = (1, 1)$ (dashed), and another parallel line (dotted). The points $\mathbf{v} = (1, 1)$ and $\mathbf{v}' = (3, 0)$ lie on the same dashed line, because $\mathbf{v}' - \mathbf{v} = (2, -1) \in U$ ; so they are two representatives of one coset. That observation is the whole theory of cosets.

Lemma 4.4.3 (When two cosets are equal)

Let $U$ be a subspace of $V$ . For $v, w \in V$ , write $v \sim w$ if $v - w \in U$ .

(a) $\sim$ is an equivalence relation on $V$ , and the equivalence class of $v$ is the coset $v + U$ .
(b) $v + U = w + U$ if and only if $v - w \in U$ .
(c) Two cosets of $U$ are either equal or disjoint, and every vector of $V$ lies in exactly one coset.

Proof.

(a) We check Definition 0.5.2. (E1) $v - v = 0 \in U$ . (E2) If $v - w \in U$ , then $w - v = (-1)(v - w) \in U$ by Theorem 1.1.7 and closure under scalar multiplication. (E3) If $v - w \in U$ and $w - x \in U$ , then $v - x = (v - w) + (w - x) \in U$ by closure under addition. For the class, $w \sim v$ means $w - v = u$ for some $u \in U$ , that is, $w = v + u \in v + U$ ; so the class $[v]$ of Definition 0.5.3 is $v + U$ .
(b) By (a) and Lemma 0.5.1 (a), $v + U = w + U$ if and only if $v \sim w$ , that is, $v - w \in U$ .
(c) This is Lemma 0.5.1 (b), together with $v \in v + U$ (take $u = 0$ ).

More examples.

Vertical lines. In $\mathbb{R}^3$ , let $U = \operatorname{span}(\mathbf{e}_3)$ , the $z$ -axis. The coset $(a, b, c) + U$ is the vertical line through $(a, b, 0)$ , and $(a, b, c) + U = (a', b', c') + U$ exactly when $a = a'$ and $b = b'$ . A coset is determined by the shadow of any of its points on the $xy$ -plane.
Polynomials with a prescribed value. In $F[x]_{\le n}$ , let $U = \{ p : p(0) = 0 \}$ , a subspace (the kernel of evaluation at $0$ ). Then $p + U = q + U$ exactly when $(p - q)(0) = 0$ , that is, $p(0) = q(0)$ . So $p + U$ is the set of all polynomials with the same constant term as $p$ .
Degenerate cases. If $U = \{0\}$ , each coset $v + U = \{v\}$ is a single vector. If $U = V$ , there is only one coset, $v + V = V$ , because $v - w \in V$ always.

Nonexample by minimal change. Try to use the line $L = \{ x + 2y = 3 \}$ , which is not a subspace, in place of $U$ . The rule “ $\mathbf{v} \sim \mathbf{w}$ if $\mathbf{v} - \mathbf{w} \in L$ ” is not even reflexive: $\mathbf{v} - \mathbf{v} = \mathbf{0} \notin L$ . What fails is the hypothesis that $U$ is a subspace, which (E1), (E2), and (E3) each used.

Warning. A coset has many names, and it is usually not a subspace. In the example above, $(1, 1) + U$ and $(3, 0) + U$ are the same set, although $(1, 1) \ne (3, 0)$ . Never conclude $v = w$ from $v + U = w + U$ ; conclude only $v - w \in U$ . And $v + U$ is a subspace only when it contains $0$ , that is, when $v \in U$ and so $v + U = U$ . The dashed line above is not a subspace.

4.4.3. The quotient space

Now we want to treat each coset as a single point. The motivation is a kind of deliberate forgetting. Suppose we only care about vectors of $\mathbb{R}^2$ “up to” adding something from $U$ , for instance because we only ever measure the quantity $x + 2y$ . Then $(1, 1)$ and $(3, 0)$ carry the same information, and the space of all the information we keep is the set of lines parallel to $U$ . We would like this set to be a vector space, so that we can add and scale lines.

The recipe is the one used for $\mathbb{Z}/n\mathbb{Z}$ in Chapter 0 (Example (Addition modulo $n$ is well defined)): define the operations through representatives, then check that the choice of representatives does not matter.

The quotient space $V/U$ is $V$ with everything in $U$ declared to be zero.

Definition 4.4.3 (Quotient space)

Let $U$ be a subspace of a vector space $V$ over $F$ . The quotient space $V/U$ (read “ $V$ modulo $U$ ”) is the set of all cosets of $U$ , $V/U \mathrel{:=} \{ v + U : v \in V \},$ with the operations $\begin{aligned} (v + U) + (w + U) &\mathrel{:=} (v + w) + U, \\ a(v + U) &\mathrel{:=} av + U \qquad (v, w \in V,\ a \in F). \end{aligned}$ The quotient map is $\pi \colon V \to V/U$ , $\pi(v) = v + U$ .

In words: an element of $V/U$ is a whole coset, a subset of $V$ . By Lemma 4.4.3 (a), $V/U$ is the quotient set of $V$ by the relation $v \sim w \Leftrightarrow v - w \in U$ (Definition 0.5.5), and $\pi$ is its quotient map. To add two cosets, pick a representative of each, add them in $V$ , and take the coset of the result. To scale a coset, scale a representative. The operations are written with representatives, so before anything else we must check that they are well defined.

The danger is real. By Lemma 4.4.3, $(1, 1) + U$ and $(3, 0) + U$ are one element of $\mathbb{R}^2/U$ . If the sum $((1, 1) + U) + (\mathbf{w} + U)$ depended on which of the two names we used, the “definition” would assign two values to one pair of cosets, and it would not define a function at all (Definition 0.4.1). The theorem below rules this out, and then shows that the operations make $V/U$ a vector space.

Theorem 4.4.4 (The Quotient Space Is a Vector Space)

Let $U$ be a subspace of a vector space $V$ over $F$ .

(a) The addition and scalar multiplication in Definition 4.4.3 are well defined: the right-hand sides depend only on the cosets $v + U$ and $w + U$ , not on the representatives $v$ and $w$ .
(b) With these operations $V/U$ is a vector space over $F$ . Its zero vector is $0 + U = U$ , and the additive inverse of $v + U$ is $(-v) + U$ .
(c) The quotient map $\pi \colon V \to V/U$ is linear and surjective, and $\operatorname{ker} \pi = U$ .

Idea. Chapter 0’s Theorem 0.5.3 turns a function on $V$ that is constant on classes into a function on $V/U$ . It is stated for functions of one variable, so we use it one variable at a time. ① For scalar multiplication by a fixed $a$ : if $v - v' \in U$ , then $av - av' = a(v - v') \in U$ . ② For addition, first freeze the second summand $w$ and vary the first; then show that the resulting function on cosets does not change when $w$ is replaced by an equivalent $w'$ . Each check is one line and uses exactly one closure property of $U$ . Once the operations exist, every axiom for cosets is the same axiom in $V$ , passed through $\pi$ .

Proof. Throughout, $v \sim v'$ means $v - v' \in U$ , and by Lemma 4.4.3 (b), $v \sim v'$ exactly when $v + U = v' + U$ .

(a) Scalar multiplication. Fix $a \in F$ , and let $g_a \colon V \to V/U$ , $g_a(v) = av + U$ . If $v \sim v'$ , then $av - av' = a(v - v') \in U$ by (VS5) and closure of $U$ under scalar multiplication, so $g_a(v) = g_a(v')$ . By Theorem 0.5.3 (a), there is exactly one function $\bar g_a \colon V/U \to V/U$ with $\bar g_a(v + U) = av + U$ for every $v \in V$ . We define $a(v + U) \mathrel{:=} \bar g_a(v + U)$ .
Addition, first variable. Fix $w \in V$ , and let $f_{w} \colon V \to V/U$ , $f_{w}(v) = (v + w) + U$ . If $v \sim v'$ , then $(v + w) - (v' + w) = v - v' \in U$ , so $f_{w}(v) = f_{w}(v')$ . By Theorem 0.5.3 (a), there is exactly one function $\bar f_{w} \colon V/U \to V/U$ with $\bar f_{w}(v + U) = (v + w) + U$ for every $v$ .

Addition, second variable. Let $\mathcal{F}$ be the set of all functions $V/U \to V/U$ , and let $h \colon V \to \mathcal{F}$ , $h(w) = \bar f_{w}$ . Suppose $w \sim w'$ . For every $v \in V$ , $(v + w) - (v + w') = w - w' \in U$ , so $\bar f_{w}(v + U) = (v + w) + U = (v + w') + U = \bar f_{w'}(v + U).$ Since every element of $V/U$ has the form $v + U$ , the functions $\bar f_{w}$ and $\bar f_{w'}$ are equal, that is, $h(w) = h(w')$ . By Theorem 0.5.3 (a) again, there is exactly one function $\bar h \colon V/U \to \mathcal{F}$ with $\bar h(w + U) = \bar f_{w}$ . We define $X + Y \mathrel{:=} \bar h(Y)(X)$ for $X, Y \in V/U$ . Then $(v + U) + (w + U) = \bar h(w + U)(v + U) = \bar f_{w}(v + U) = (v + w) + U,$ which is the formula of Definition 4.4.3, and its value depends only on the two cosets.
(b) Every element of $V/U$ is $\pi(v) = v + U$ for some $v$ , and by (a) we have $\pi(v) + \pi(w) = \pi(v + w)$ and $a\pi(v) = \pi(av)$ . Let $X = \pi(x)$ , $Y = \pi(y)$ , $Z = \pi(z)$ be in $V/U$ , and $a, b \in F$ . Each axiom follows by moving $\pi$ outside, using the axiom in $V$ , and moving $\pi$ back:
- (VS1) $X + Y = \pi(x + y) = \pi(y + x) = Y + X$ .
- (VS2) $(X + Y) + Z = \pi((x + y) + z) = \pi(x + (y + z)) = X + (Y + Z)$ .
- (VS3) $X + \pi(0) = \pi(x + 0) = \pi(x) = X$ , so $\pi(0) = 0 + U = U$ is a zero vector.
- (VS4) $X + \pi(-x) = \pi(x + (-x)) = \pi(0)$ , so $\pi(-x) = (-x) + U$ is an additive inverse of $X$ .
- (VS5) $a(X + Y) = \pi(a(x + y)) = \pi(ax + ay) = aX + aY$ .
- (VS6) $(a + b)X = \pi((a + b)x) = \pi(ax + bx) = aX + bX$ .
- (VS7) $a(bX) = \pi(a(bx)) = \pi((ab)x) = (ab)X$ .
- (VS8) $1X = \pi(1x) = \pi(x) = X$ .
Hence $V/U$ is a vector space over $F$ .
(c) The identities $\pi(v + w) = \pi(v) + \pi(w)$ and $\pi(av) = a\pi(v)$ from (b) say that $\pi$ is linear. It is surjective by the definition of $V/U$ . Finally, $v \in \operatorname{ker} \pi$ means $v + U = 0 + U$ , which by Lemma 4.4.3 (b) means $v = v - 0 \in U$ . So $\operatorname{ker} \pi = U$ .

Part (c) has a pleasant reading: every subspace is a kernel. Given $U \subseteq V$ , the quotient map is a linear map whose kernel is exactly $U$ .

Examples.

Parallel lines. In $\mathbb{R}^2/U$ with $U = \{ x + 2y = 0 \}$ , write $\ell_c$ for the line $\{ x + 2y = c \}$ . Since $(a, b) \in \ell_{a + 2b}$ , the definition gives $\ell_c + \ell_d = \ell_{c + d}$ and $t\,\ell_c = \ell_{tc}$ : to add two parallel lines, add their “heights” $c$ and $d$ . The zero vector is $\ell_0 = U$ .
Constant terms. In $F[x]_{\le n}/U$ with $U = \{ p : p(0) = 0 \}$ , the coset $p + U$ is determined by $p(0)$ , and $(p + U) + (q + U) = (p + q) + U$ is determined by $p(0) + q(0)$ . All that survives of a polynomial is its constant term.
Degenerate cases. $V/\{0\}$ has the cosets $\{v\}$ , added exactly as in $V$ : nothing is forgotten. $V/V$ has the single element $V$ , so it is the zero space: everything is forgotten.

Nonexample by minimal change. Replace the line $U$ by the union of the two axes, $S = \operatorname{span}(\mathbf{e}_1) \cup \operatorname{span}(\mathbf{e}_2) \subseteq \mathbb{R}^2$ . It still contains $\mathbf{0}$ and is still closed under scalar multiplication, but it is not closed under addition. Then “ $\mathbf{v} - \mathbf{w} \in S$ ” is not transitive: $(1, 1) - (0, 1) = (1, 0) \in S$ and $(0, 1) - (0, 0) = (0, 1) \in S$ , but $(1, 1) - (0, 0) = (1, 1) \notin S$ . Concretely, the “cosets” $(1, 1) + S$ and $(0, 0) + S = S$ both contain $(0, 1) = (1, 1) + (-1, 0)$ , yet they are different sets, since $\mathbf{0} \in S$ but $(1, 1) + \mathbf{s} = \mathbf{0}$ would need $\mathbf{s} = (-1, -1) \notin S$ . The sets $\mathbf{v} + S$ do not split $\mathbb{R}^2$ into disjoint pieces, so there is no set “ $\mathbb{R}^2/S$ ” of the kind the definition needs. The clause that fails is “ $U$ is a subspace”, specifically closure under addition.

Why this definition. Each part of the hypothesis that $U$ is a subspace has a job: $0 \in U$ makes $\sim$ reflexive; closure under scalar multiplication makes $\sim$ symmetric and scaling of cosets well defined; closure under addition makes $\sim$ transitive and addition of cosets well defined. The nonexample shows what goes wrong when one of them is missing. The notation $V/U$ and the phrase “modulo $U$ ” come from $\mathbb{Z}/n\mathbb{Z}$ , where integers are identified when their difference lies in $n\mathbb{Z}$ .

Warning. $V/U$ is not a subspace of $V$ . Its elements are subsets of $V$ , not vectors of $V$ , and its addition is a new operation. For example, $\mathbb{R}^3/\operatorname{span}(\mathbf{e}_3)$ is not the $xy$ -plane: its elements are vertical lines. The $xy$ -plane is a complement of the $z$ -axis, one of many (Chapter 1), whereas $V/U$ involves no choice at all. The two are isomorphic, as we show below, but they are different objects.

In finite dimension the size of a quotient is exactly what one would guess from the picture: we lose $\dim U$ dimensions.

Theorem 4.4.5 (Dimension of a Quotient)

Let $V$ be finite-dimensional and $U$ a subspace of $V$ . Then $V/U$ is finite-dimensional and $\dim(V/U) = \dim V - \dim U .$ More precisely, if $(u_1, \dots, u_m, w_1, \dots, w_s)$ is a basis of $V$ such that $(u_1, \dots, u_m)$ is a basis of $U$ , then $(w_1 + U, \dots, w_s + U)$ is a basis of $V/U$ .

Proof. By Theorem 4.4.4 (c), $\pi \colon V \to V/U$ is linear, surjective, and has kernel $U$ . By the Rank–Nullity Theorem (Theorem 2.3.1), $\operatorname{im} \pi = V/U$ is finite-dimensional and $\dim V = \dim U + \dim(V/U)$ , which is the formula.

For the basis, let $v + U \in V/U$ and write $v = \sum_i a_iu_i + \sum_j b_jw_j$ . Since $\pi$ is linear and $\pi(u_i) = U$ is the zero vector, Theorem 2.1.4 gives $v + U = \sum_j b_j(w_j + U)$ . So the list $(w_1 + U, \dots, w_s + U)$ spans $V/U$ . Its length is $s = \dim V - \dim U = \dim(V/U)$ , so it is a basis by Theorem 1.6.4 (b).

Such a basis always exists: take a basis of $U$ and extend it (Theorem 1.6.5). So to find a basis of a quotient, extend a basis of $U$ to one of $V$ and keep the cosets of the new vectors.

Example (A basis of a quotient of

\mathbb{R}^3

)

Let $U = \operatorname{span}((1, 1, 1)) \subseteq \mathbb{R}^3$ . Find a basis of $\mathbb{R}^3/U$ , and write $(2, 5, -1) + U$ in it.

Solution. Extend $(1, 1, 1)$ by $\mathbf{e}_1$ and $\mathbf{e}_2$ . The list $((1, 1, 1), \mathbf{e}_1, \mathbf{e}_2)$ is independent: in $a(1, 1, 1) + b\mathbf{e}_1 + c\mathbf{e}_2 = (a + b, a + c, a) = \mathbf{0}$ , the third entry gives $a = 0$ , and then $b = c = 0$ . It has length $3$ , so it is a basis of $\mathbb{R}^3$ (Theorem 1.6.4). By Theorem 4.4.5, $(\mathbf{e}_1 + U, \mathbf{e}_2 + U)$ is a basis of $\mathbb{R}^3/U$ , which has dimension $3 - 1 = 2$ .

To write $(2, 5, -1) + U$ , solve $(2, 5, -1) = a(1, 1, 1) + b\mathbf{e}_1 + c\mathbf{e}_2$ : the third entry gives $a = -1$ , then $b = 3$ and $c = 6$ . Hence $(2, 5, -1) + U = 3(\mathbf{e}_1 + U) + 6(\mathbf{e}_2 + U),$ and indeed $(2, 5, -1) - (3, 6, 0) = (-1, -1, -1) \in U$ . The coordinates $(3, 6)$ are $(x - z, y - z)$ evaluated at $(2, 5, -1)$ : they are exactly the quantities that do not change when a multiple of $(1, 1, 1)$ is added.

The warning above said that a quotient is not a complement. The next result says how they are related: any complement is a concrete model of the quotient.

Proposition 4.4.6 (A complement is a model of the quotient)

Let $U$ be a subspace of $V$ , and let $W$ be a subspace with $V = U \oplus W$ . Then the restriction of the quotient map, $W \to V/U$ , $w \mapsto w + U$ , is an isomorphism.

Proof. The restriction of the linear map $\pi$ to $W$ is linear. Injective: if $w + U = U$ with $w \in W$ , then $w \in U$ by Lemma 4.4.3 (b), so $w \in U \cap W = \{0\}$ (Theorem 1.7.4); by Theorem 2.2.3 the restriction is injective. Surjective: given $v + U$ , write $v = u + w$ with $u \in U$ , $w \in W$ ; then $v - w = u \in U$ , so $v + U = w + U$ . A linear bijection is an isomorphism (Definition 2.5.2).

So every complement of $U$ is isomorphic to $V/U$ , which gives a second proof that all complements have the same dimension. The quotient is the choice-free object that all the complements imitate.

Quick check. In $\mathbb{R}^2/U$ with $U = \operatorname{span}((1, 1))$ , is $(2, 3) + U = (0, 1) + U$ ? Is $(1, 0) + U = (0, 1) + U$ ? What is $\dim(\mathbb{R}^2/U)$ ?

Solution. By Lemma 4.4.3 (b): $(2, 3) - (0, 1) = (2, 2) \in U$ , so the first two cosets are equal. $(1, 0) - (0, 1) = (1, -1) \notin U$ , so the second pair are different cosets. By Theorem 4.4.5, $\dim(\mathbb{R}^2/U) = 2 - 1 = 1$ .

4.4.4. The First Isomorphism Theorem

A linear map $T \colon V \to W$ forgets exactly the information in its kernel: $Tv = Tv'$ if and only if $T(v - v') = 0$ , that is, $v - v' \in \operatorname{ker} T$ . In the language of this section, $T$ takes the same value at two vectors exactly when they lie in the same coset of $\operatorname{ker} T$ . So $T$ is really a function of the coset, and on cosets it forgets nothing.

Theorem 4.4.7 (First Isomorphism Theorem)

Let $T \colon V \to W$ be a linear map, and let $K = \operatorname{ker} T$ . Then the rule $\bar T \colon V/K \to W, \qquad \bar T(v + K) = Tv$ defines a linear map, which is injective with $\operatorname{im} \bar T = \operatorname{im} T$ , and satisfies $\bar T \circ \pi = T$ . Hence $V/\operatorname{ker} T \cong \operatorname{im} T .$

Proof. Well defined. $K$ is a subspace by Theorem 2.2.1, so $V/K$ is a vector space. If $v + K = v' + K$ , then $v - v' \in K$ by Lemma 4.4.3 (b), so $Tv - Tv' = T(v - v') = 0$ . Hence $T$ is constant on each coset of $K$ , and by Theorem 0.5.3 (a) there is exactly one function $\bar T \colon V/K \to W$ with $\bar T(v + K) = Tv$ , that is, $\bar T \circ \pi = T$ .

Linear. For $v, w \in V$ and $a \in F$ , by Definition 4.4.3 and the linearity of $T$ , $\begin{aligned} \bar T\big((v + K) + (w + K)\big) &= \bar T\big((v + w) + K\big) = T(v + w) \\ &= Tv + Tw = \bar T(v + K) + \bar T(w + K), \end{aligned}$ and similarly $\bar T(a(v + K)) = \bar T(av + K) = T(av) = a\bar T(v + K)$ .

Injective. If $\bar T(v + K) = 0$ , then $Tv = 0$ , so $v \in K$ and $v + K = K$ , the zero of $V/K$ . By Theorem 2.2.3, $\bar T$ is injective.

Image. The values of $\bar T$ are the vectors $Tv$ , $v \in V$ , so $\operatorname{im} \bar T = \operatorname{im} T$ . Regarded as a map $V/K \to \operatorname{im} T$ , $\bar T$ is a linear bijection, hence an isomorphism (Definition 2.5.2).

So every linear map factors as “collapse the kernel, then embed”: $T = \bar T \circ \pi$ , with $\pi$ surjective and $\bar T$ injective. When $V$ is finite-dimensional, taking dimensions with Theorem 4.4.5 gives $\dim V - \dim \operatorname{ker} T = \dim \operatorname{im} T$ , which is Rank–Nullity again. The theorem is stronger than the count, because it names the isomorphism, and it holds in infinite dimension too.

The practical use runs the other way: to identify a quotient $V/U$ , find a surjective linear map out of $V$ whose kernel is exactly $U$ .

Example (Identifying quotients)

Use the First Isomorphism Theorem to show:

(a) $\mathbb{R}^3/L \cong \mathbb{R}^2$ , where $L = \operatorname{span}((1, 1, 1))$ ;
(b) $\mathbb{R}[x]_{\le n}/U \cong \mathbb{R}$ , where $U = \{ p : p(0) = 0 \}$ .

Solution.

(a) Let $T \colon \mathbb{R}^3 \to \mathbb{R}^2$ , $T(x, y, z) = (x - z,\ y - z)$ , which is linear because each entry is a linear expression. It is surjective, since $T(a, b, 0) = (a, b)$ . Its kernel is $\{ x = z,\ y = z \} = \{ (t, t, t) \} = L$ . By Theorem 4.4.7, $(x, y, z) + L \mapsto (x - z, y - z)$ is an isomorphism $\mathbb{R}^3/L \to \mathbb{R}^2$ . This explains the coordinates found in Example (A basis of a quotient of $\mathbb{R}^3$ ). The check that it is well defined is visible: adding $(t, t, t)$ to $(x, y, z)$ does not change $x - z$ or $y - z$ .
(b) Let $\operatorname{ev}_0 \colon \mathbb{R}[x]_{\le n} \to \mathbb{R}$ , $p \mapsto p(0)$ . It is linear by Theorem 0.8.5, surjective because the constant polynomial $c$ maps to $c$ , and its kernel is $U$ by definition. By Theorem 4.4.7, $p + U \mapsto p(0)$ is an isomorphism $\mathbb{R}[x]_{\le n}/U \to \mathbb{R}$ . In particular $\dim(\mathbb{R}[x]_{\le n}/U) = 1$ by Theorem 2.5.4, so $\dim U = (n + 1) - 1 = n$ by Theorem 4.4.5.

4.4.5. The Second and Third Isomorphism Theorems

Two more identifications come from the same method: write down a surjective map, compute its kernel, and apply the First Isomorphism Theorem. The first compares a sum of subspaces with an intersection.

Theorem 4.4.8 (Second Isomorphism Theorem)

Let $U$ and $W$ be subspaces of a vector space $V$ . Then $U \cap W$ is a subspace of $U$ , $W$ is a subspace of $U + W$ , and $U/(U \cap W) \to (U + W)/W, \qquad u + (U \cap W) \mapsto u + W$ is a well-defined isomorphism. Hence $(U + W)/W \cong U/(U \cap W)$ .

Idea. Picture two planes $U$ and $W$ in $\mathbb{R}^3$ meeting in a line. Modulo $W$ , a vector of $U + W$ is represented by its $U$ -part alone, since the $W$ -part is forgotten. The only vectors of $U$ that become zero are those already in $W$ . So the map “ $u \mapsto u + W$ ” from $U$ is surjective, with kernel $U \cap W$ .

Proof. $U \cap W$ is a subspace (Theorem 1.2.2) contained in $U$ , and $W$ is a subspace contained in $U + W$ (Theorem 1.7.1), so both quotients exist. Let $\pi_W \colon U + W \to (U + W)/W$ be the quotient map, and let $S \colon U \to (U + W)/W$ , $Su = u + W$ , be its restriction to $U$ ; $S$ is linear because $\pi_W$ is (Theorem 4.4.4).

$S$ is surjective. An element of $(U + W)/W$ is $(u + w) + W$ with $u \in U$ , $w \in W$ . Since $(u + w) - u = w \in W$ , Lemma 4.4.3 (b) gives $(u + w) + W = u + W = Su$ .

$\operatorname{ker} S = U \cap W$ . For $u \in U$ , $Su = W$ means $u \in W$ (Lemma 4.4.3 (b)), that is, $u \in U \cap W$ .

By Theorem 4.4.7 applied to $S$ , the rule $u + (U \cap W) \mapsto Su = u + W$ is a well-defined injective linear map with image $\operatorname{im} S = (U + W)/W$ , hence an isomorphism. (Directly: if $u - u' \in U \cap W$ , then $u - u' \in W$ , so $u + W = u' + W$ , and the value does not depend on the representative.)

If $U$ and $W$ are finite-dimensional, then so is $U + W$ (it is spanned by the union of bases, Proposition 1.7.2), and Theorem 4.4.5 turns the theorem into $\dim(U + W) - \dim W = \dim U - \dim(U \cap W)$ . That is the dimension formula for sums (Theorem 1.7.3), now as a consequence of an isomorphism rather than of a basis computation. An exercise below asks for the details.

The third theorem says that quotients can be taken in stages.

Theorem 4.4.9 (Third Isomorphism Theorem)

Let $U \subseteq W \subseteq V$ be subspaces. Then $W/U = \{ w + U : w \in W \}$ is a subspace of $V/U$ , and $(V/U)/(W/U) \to V/W, \qquad (v + U) + W/U \mapsto v + W$ is a well-defined isomorphism. Hence $(V/U)/(W/U) \cong V/W$ .

Proof. Let $S \colon V/U \to V/W$ , $S(v + U) = v + W$ .

Well defined. If $v + U = v' + U$ , then $v - v' \in U \subseteq W$ , so $v + W = v' + W$ by Lemma 4.4.3 (b). By Theorem 0.5.3 (a), applied to $V \to V/W$ , $v \mapsto v + W$ , the rule defines a function on $V/U$ .

Linear and surjective. $\begin{aligned} S((v + U) + (w + U)) &= S((v + w) + U) = (v + w) + W \\ &= S(v + U) + S(w + U), \end{aligned}$ and similarly for scalars, by Definition 4.4.3 in both quotients. Every $v + W$ equals $S(v + U)$ .

Kernel. $S(v + U) = W$ means $v \in W$ . If $v \in W$ , then $v + U \in W/U$ . Conversely, if $v + U = w + U$ with $w \in W$ , then $v - w \in U \subseteq W$ , so $v \in W$ . Hence $\operatorname{ker} S = W/U$ . In particular $W/U$ is a subspace of $V/U$ , by Theorem 2.2.1.

By Theorem 4.4.7 applied to $S$ , $(V/U)/(W/U) \cong \operatorname{im} S = V/W$ , via $(v + U) + W/U \mapsto S(v + U) = v + W$ .

The notation $W/U$ is used with two meanings, and they agree: the quotient space of $W$ by $U$ , and the set of cosets $w + U$ with $w \in W$ , sitting inside $V/U$ . The coset $w + U$ is the same subset of $V$ in both readings, and the operations agree because both are computed with representatives. For example, with $V = \mathbb{R}^3$ , $U$ the $z$ -axis, and $W$ the $xz$ -plane, $V/W \cong \mathbb{R}$ through the $y$ -coordinate, and the theorem says that first forgetting $z$ and then forgetting $x$ leaves the same thing as forgetting both at once.

4.4.6. The universal property of the quotient

The proofs of the three theorems repeated one move: a linear map that vanishes on $U$ passes to $V/U$ . We record that move once, in its general form. It is the standard way to define linear maps out of a quotient.

Theorem 4.4.10 (Universal Property of the Quotient)

Let $U$ be a subspace of $V$ , $\pi \colon V \to V/U$ the quotient map, and $T \colon V \to W$ a linear map.

(a) If $U \subseteq \operatorname{ker} T$ , then there is exactly one linear map $\bar T \colon V/U \to W$ with $\bar T \circ \pi = T$ , namely $\bar T(v + U) = Tv$ . Its image is $\operatorname{im} T$ , and it is injective if and only if $U = \operatorname{ker} T$ .
(b) Conversely, if some function $\bar T \colon V/U \to W$ satisfies $\bar T \circ \pi = T$ , then $U \subseteq \operatorname{ker} T$ .

Proof.

(a) If $v - v' \in U \subseteq \operatorname{ker} T$ , then $Tv = Tv'$ . By Theorem 0.5.3 (a) there is exactly one function $\bar T$ with $\bar T(v + U) = Tv$ , and it is linear by the computation in the proof of Theorem 4.4.7, which used only this formula. Any linear $\bar T$ with $\bar T \circ \pi = T$ satisfies this formula, so it is this one. The image is $\{ Tv \} = \operatorname{im} T$ . The kernel of $\bar T$ is $\{ v + U : v \in \operatorname{ker} T \}$ ; it is $\{ U \}$ , the zero subspace, exactly when every $v \in \operatorname{ker} T$ lies in $U$ , that is, when $\operatorname{ker} T \subseteq U$ , which together with $U \subseteq \operatorname{ker} T$ means $U = \operatorname{ker} T$ . Apply Theorem 2.2.3.
(b) Let $u \in U$ . Then $\pi(u) = U = \pi(0)$ , so $Tu = \bar T(\pi(u)) = \bar T(\pi(0)) = T(0) = 0$ by Theorem 2.1.1. Hence $U \subseteq \operatorname{ker} T$ .

In words: linear maps out of $V/U$ are the same as linear maps out of $V$ that kill $U$ . The First Isomorphism Theorem is the case $U = \operatorname{ker} T$ . For example, the map $T \colon \mathbb{R}^3 \to \mathbb{R}$ , $T(x, y, z) = x$ , kills the $z$ -axis $U$ , so it induces $\bar T \colon \mathbb{R}^3/U \to \mathbb{R}$ , $(x, y, z) + U \mapsto x$ . Since $\operatorname{ker} T$ is the $yz$ -plane, strictly larger than $U$ , this $\bar T$ is not injective: it also kills the coset $\mathbf{e}_2 + U$ .

Quotients will return in Chapter 9, where an operator $T$ with $T(U) \subseteq U$ induces an operator on $V/U$ , and this is the engine of induction arguments on the dimension.

4.4.7. Exercises

A. Check your understanding

Exercise (A1)

(a) Define the quotient space $V/U$ and its two operations.
(b) State the condition, in terms of $v - w$ , for $v + U = w + U$ .
(c) True or false: $V/U$ is a subspace of $V$ . Justify your answer.
(d) State the First Isomorphism Theorem.
(e) If $U$ is a $2$ -dimensional subspace of $\mathbb{R}^5$ , what is $\dim(\mathbb{R}^5/U)$ ? What is $\dim(\mathbb{R}^5 \times U)$ ?
(f) What must be checked before the formula $(v + U) + (w + U) = (v + w) + U$ can be used as a definition, and which property of $U$ makes the check work?

Solution.

(a) For a subspace $U$ of $V$ , $V/U = \{ v + U : v \in V \}$ , with $(v + U) + (w + U) = (v + w) + U$ and $a(v + U) = av + U$ (Definition 4.4.3).
(b) $v + U = w + U$ if and only if $v - w \in U$ (Lemma 4.4.3).
(c) False. The elements of $V/U$ are cosets, which are subsets of $V$ , not vectors of $V$ .
(d) If $T \colon V \to W$ is linear, then $v + \operatorname{ker} T \mapsto Tv$ is a well-defined isomorphism $V/\operatorname{ker} T \to \operatorname{im} T$ (Theorem 4.4.7).
(e) $5 - 2 = 3$ by Theorem 4.4.5, and $5 + 2 = 7$ by Theorem 4.4.1.
(f) That the right-hand side does not depend on the representatives: if $v - v' \in U$ and $w - w' \in U$ , then $(v + w) + U = (v' + w') + U$ . This works because $(v + w) - (v' + w') = (v - v') + (w - w')$ and $U$ is closed under addition.

B. Practice

Exercise (B1: A basis of a quotient of

\mathbb{R}^4

)

Let $U = \operatorname{span}((1, 0, 1, 0), (0, 1, 0, 1)) \subseteq \mathbb{R}^4$ .

(a) Find $\dim(\mathbb{R}^4/U)$ and a basis of $\mathbb{R}^4/U$ .
(b) Write $(3, 1, 4, 1) + U$ in your basis.
(c) Decide whether $(1, 2, 3, 4) + U = (3, 4, 1, 2) + U$ .

Solution.

(a) The two spanning vectors are independent (look at the first two entries), so $\dim U = 2$ and $\dim(\mathbb{R}^4/U) = 4 - 2 = 2$ by Theorem 4.4.5. Extend by $\mathbf{e}_1, \mathbf{e}_2$ . If $a(1, 0, 1, 0) + b(0, 1, 0, 1) + c\mathbf{e}_1 + d\mathbf{e}_2 = (a + c, b + d, a, b) = \mathbf{0}$ , then $a = b = 0$ from the last two entries and $c = d = 0$ from the first two. So the four vectors are independent, hence a basis of $\mathbb{R}^4$ (Theorem 1.6.4), and by Theorem 4.4.5 $(\mathbf{e}_1 + U, \mathbf{e}_2 + U)$ is a basis of $\mathbb{R}^4/U$ .
(b) Solve $(3, 1, 4, 1) = (a + c, b + d, a, b)$ : $a = 4$ , $b = 1$ , $c = -1$ , $d = 0$ . Since the $U$ -part becomes zero in the quotient, $(3, 1, 4, 1) + U = -1(\mathbf{e}_1 + U) + 0(\mathbf{e}_2 + U)$ . Check: $(3, 1, 4, 1) + \mathbf{e}_1 = (4, 1, 4, 1) = 4(1, 0, 1, 0) + (0, 1, 0, 1) \in U$ .
(c) The difference is $(1, 2, 3, 4) - (3, 4, 1, 2) = (-2, -2, 2, 2)$ . A vector $(a, b, a, b)$ of $U$ has equal first and third entries, but $-2 \ne 2$ . So the difference is not in $U$ , and the cosets are different by Lemma 4.4.3.

Exercise (B2: A map that factors through a quotient)

Let $T \colon \mathbb{R}[x]_{\le 2} \to \mathbb{R}^2$ , $T(p) = (p(1), p'(1))$ , and $U = \operatorname{span}((x - 1)^2)$ .

(a) Show that $U \subseteq \operatorname{ker} T$ , so that $T$ induces a linear map $\bar T \colon \mathbb{R}[x]_{\le 2}/U \to \mathbb{R}^2$ with $\bar T(p + U) = T(p)$ .
(b) Prove that $\bar T$ is an isomorphism.
(c) Compute $\bar T(x^2 + U)$ , and find the coset $\bar T^{-1}(0, 1)$ .

Solution.

(a) $T$ is linear, since evaluation and differentiation are. For $q = (x - 1)^2$ , $q(1) = 0$ and $q'(1) = 2(1 - 1) = 0$ , so $T(q) = \mathbf{0}$ , and hence $T(cq) = \mathbf{0}$ for all $c$ . So $U \subseteq \operatorname{ker} T$ , and Theorem 4.4.10 (a) gives the linear map $\bar T$ .
(b) $T(1) = (1, 0)$ and $T(x - 1) = (0, 1)$ , so $T$ is surjective and $\operatorname{rank} T = 2$ . By Theorem 2.3.1, $\dim \operatorname{ker} T = 3 - 2 = 1 = \dim U$ , so $\operatorname{ker} T = U$ by Theorem 1.6.8. By Theorem 4.4.10 (a), $\bar T$ is injective with image $\operatorname{im} T = \mathbb{R}^2$ . Hence $\bar T$ is a linear bijection, an isomorphism.
(c) $\bar T(x^2 + U) = T(x^2) = (1, 2)$ . Since $T(x - 1) = (0, 1)$ , $\bar T^{-1}(0, 1) = (x - 1) + U$ , the set of all $x - 1 + c(x - 1)^2$ .

Exercise (B3: Matrices modulo scalars)

Let $n \ge 1$ and $S = \{ c\mathbf{I}_n : c \in F \} \subseteq M_n(F)$ .

(a) Show that $S$ is a subspace and find $\dim(M_n(F)/S)$ .
(b) For $n = 2$ , find a basis of $M_2(F)/S$ .
(c) Show that $\mathbf{A} + S = \mathbf{B} + S$ if and only if $\mathbf{A} - \mathbf{B}$ is a scalar matrix, and decide whether $\begin{pmatrix} 3 & 1 \\ 0 & 5 \end{pmatrix} + S = \begin{pmatrix} 1 & 1 \\ 0 & 3 \end{pmatrix} + S$ .

Solution.

(a) $S = \operatorname{span}(\mathbf{I}_n)$ is a subspace by Theorem 1.3.1, of dimension $1$ since $\mathbf{I}_n \ne \mathbf{O}$ . By Theorem 4.4.5, $\dim(M_n(F)/S) = n^2 - 1$ .
(b) Extend $(\mathbf{I}_2)$ by $\mathbf{E}_{11}, \mathbf{E}_{12}, \mathbf{E}_{21}$ . If $a\mathbf{I}_2 + b\mathbf{E}_{11} + c\mathbf{E}_{12} + d\mathbf{E}_{21} = \begin{pmatrix} a + b & c \\ d & a \end{pmatrix} = \mathbf{O}$ , then $a = c = d = 0$ and then $b = 0$ . So the four matrices are independent, hence a basis of $M_2(F)$ (Theorem 1.6.4), and $(\mathbf{E}_{11} + S, \mathbf{E}_{12} + S, \mathbf{E}_{21} + S)$ is a basis of $M_2(F)/S$ by Theorem 4.4.5.
(c) This is Lemma 4.4.3 (b) for the subspace $S$ . Here the difference is $\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = 2\mathbf{I}_2 \in S$ , so the two cosets are equal.

C. Going deeper

Exercise (C1: The dimension formula from isomorphisms)

Let $U$ and $W$ be finite-dimensional subspaces of a vector space $V$ .

(a) Using Theorem 4.4.8 and Theorem 4.4.5, prove that $\dim(U + W) = \dim U + \dim W - \dim(U \cap W)$ .
(b) Give a second proof: apply Rank–Nullity to the addition map $S \colon U \times W \to V$ , $S(u, w) = u + w$ , after showing that $u \mapsto (u, -u)$ is an isomorphism from $U \cap W$ onto $\operatorname{ker} S$ .

Solution.

(a) $U + W$ is finite-dimensional, being spanned by a basis of $U$ followed by a basis of $W$ (Proposition 1.7.2), and $U \cap W \subseteq U$ is finite-dimensional (Theorem 1.6.7). By Theorem 4.4.8, $(U + W)/W \cong U/(U \cap W)$ , and isomorphic finite-dimensional spaces have equal dimension (Theorem 2.5.4). Applying Theorem 4.4.5 to both sides, $\dim(U + W) - \dim W = \dim U - \dim(U \cap W),$ which rearranges to the formula.
(b) The map $S$ is linear with image $U + W$ by Theorem 4.4.2 (a). Let $J \colon U \cap W \to U \times W$ , $Ju = (u, -u)$ ; it is well defined because $u \in U$ and $-u \in W$ , and linear because the operations are entrywise. It is injective, since $(u, -u) = (0, 0)$ forces $u = 0$ . Its image lies in $\operatorname{ker} S$ , since $u + (-u) = 0$ . Conversely, if $(u, w) \in \operatorname{ker} S$ , then $w = -u$ ; so $u = -w \in W$ , hence $u \in U \cap W$ and $(u, w) = Ju$ . Thus $J$ is an isomorphism onto $\operatorname{ker} S$ , and $\dim \operatorname{ker} S = \dim(U \cap W)$ . By Theorem 4.4.1 and Theorem 2.3.1, $\dim U + \dim W = \dim(U \times W) = \dim(U + W) + \dim(U \cap W).$

Exercise (C2: The operator induced on a quotient)

Let $V$ be finite-dimensional, $T \in \mathcal{L}(V)$ , and $U$ a subspace with $Tu \in U$ for every $u \in U$ .

(a) Prove that $\bar T(v + U) \mathrel{:=} Tv + U$ is a well-defined linear operator on $V/U$ .
(b) Let $(u_1, \dots, u_m)$ be a basis of $U$ , extended to a basis $\mathcal{B} = (u_1, \dots, u_m, w_1, \dots, w_s)$ of $V$ . Show that $[T]_{\mathcal{B}}^{\mathcal{B}} = \begin{pmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{O} & D \end{pmatrix},$ where $\mathbf{A} \in M_m(F)$ is the matrix of the restriction $T|_U \colon U \to U$ in $(u_1, \dots, u_m)$ , $D \in M_s(F)$ is the matrix of $\bar T$ in $(w_1 + U, \dots, w_s + U)$ , and $\mathbf{O}$ is the $s \times m$ zero matrix.
(c) Check (b) when $T$ is differentiation on $\mathbb{R}[x]_{\le 2}$ , with $U = \mathbb{R}[x]_{\le 1}$ and $\mathcal{B} = (1, x, x^2)$ .

Hint: for (a), apply Theorem 4.4.10 to $\pi \circ T$ .

Solution.

(a) The map $\pi \circ T \colon V \to V/U$ , $v \mapsto Tv + U$ , is linear as a composition of linear maps (Theorem 2.4.2). For $u \in U$ , $Tu \in U$ , so $(\pi \circ T)(u) = U$ , the zero of $V/U$ ; thus $U \subseteq \operatorname{ker}(\pi \circ T)$ . By Theorem 4.4.10 (a) there is exactly one linear map $\bar T \colon V/U \to V/U$ with $\bar T(v + U) = Tv + U$ .
(b) For $j \le m$ , $Tu_j \in U$ , so it is a combination of $u_1, \dots, u_m$ only, and its $\mathcal{B}$ -coordinates are $0$ in the last $s$ places. These first $m$ columns, restricted to their first $m$ entries, are the columns of the matrix $\mathbf{A}$ of $T|_U$ (which maps $U$ to $U$ by hypothesis and is linear). For the remaining columns, write $Tw_k = \sum_i b_{ik}u_i + \sum_l d_{lk}w_l$ . Passing to $V/U$ , where $u_i + U = U$ is zero, gives $\bar T(w_k + U) = \sum_l d_{lk}(w_l + U)$ . So the $s \times s$ matrix $(d_{lk})$ is the matrix $D$ of $\bar T$ in the basis $(w_1 + U, \dots, w_s + U)$ of Theorem 4.4.5, and the bottom-right block of $[T]_{\mathcal{B}}^{\mathcal{B}}$ is $D$ .
(c) $T(1) = 0$ , $T(x) = 1$ , $T(x^2) = 2x$ , so $[T]_{\mathcal{B}}^{\mathcal{B}} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}$ . Differentiation maps $U = \mathbb{R}[x]_{\le 1}$ into itself. The block $\mathbf{A} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ is the matrix of $T$ on $U$ in $(1, x)$ , the lower-left block is $\begin{pmatrix} 0 & 0 \end{pmatrix}$ , and the block $D$ is the $1 \times 1$ matrix $\mathbf{O}_1$ : indeed $\bar T(x^2 + U) = 2x + U = U$ , the zero coset, since $2x \in U$ .

Exercise (C3: Counting cosets over a finite field)

Let $F = \mathbb{F}_q$ be a finite field with $q$ elements, $V$ a vector space over $F$ of dimension $n$ , and $U$ a subspace of dimension $m$ .

(a) Show that $\lvert V \rvert = q^n$ .
(b) Show that every coset $v + U$ has exactly $\lvert U \rvert$ elements, and deduce $\lvert V/U \rvert = q^{n - m}$ by counting.
(c) Check that (b) agrees with Theorem 4.4.5.

Solution.

(a) Choose a basis $(v_1, \dots, v_n)$ . By Theorem 1.5.1, $(a_1, \dots, a_n) \mapsto a_1v_1 + \dots + a_nv_n$ is a bijection $F^n \to V$ , and $F^n$ has $q^n$ elements.
(b) The map $U \to v + U$ , $u \mapsto v + u$ , is surjective by Definition 4.4.2 and injective because $v + u = v + u'$ implies $u = u'$ . So $\lvert v + U \rvert = \lvert U \rvert = q^m$ by (a) applied to $U$ . By Lemma 4.4.3 (c), the cosets partition $V$ into $\lvert V/U \rvert$ disjoint sets of size $q^m$ , so $q^n = \lvert V/U \rvert \cdot q^m$ , and $\lvert V/U \rvert = q^{n - m}$ .
(c) By Theorem 4.4.5, $V/U$ is a vector space over $\mathbb{F}_q$ of dimension $n - m$ , so by (a) it has $q^{n - m}$ elements, in agreement with (b). For example, $\mathbb{F}_2^3$ modulo a line has $8/2 = 4$ cosets.