Quotient Spaces

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In this post, we define the quotient space \(V/W\) for a \(\mathbb{K}\)-vector space \(V\) and its subspace \(W\). Intuitively, this is the space obtained from \(V\) by making every element of \(W\) equal to \(0\); however, since we want the resulting space to remain a vector space, simply declaring every element of \(W\) to be \(0\) is insufficient.

Cosets

The main obstacle, as noted above, is that merely setting all elements of \(W\) to \(0\) does not guarantee that the remaining space will be a vector space. For it to be a vector space, it must first be closed under operations. Any element \(v\) of \(V\) can always be written as \(v=(v-w)+w\) for a fixed \(w\in W\), so if \(v\) and \(v-w\) do not belong to \(W\), then even after treating every element of \(W\) as \(0\),

\[v-(v-w)=w=0\]

the difference of \(v\) and \(v-w\) becomes \(0\) even though they are not equal.

This reveals two things. First, defining \(V/W\) by simply setting all elements of \(W\) to \(0\) and leaving the rest unchanged is inadequate. Second, and more importantly, the simple computation above actually hints at how to construct \(V/W\): if the difference of two vectors \(v,v'\) lies in \(W\), then we must treat them as the same element inside \(V/W\).

Definition 1 Let a \(\mathbb{K}\)-vector space \(V\) and its subspace \(W\leq V\) be given. For any \(v\in V\), the set

\[v+W=\{v+w\mid w\in W\}\]

is called the coset of \(W\) containing \(v\).

By definition, the coset \(v+W\) consists of all vectors whose difference from \(v\) lies in \(W\)—precisely the vectors that we agreed to treat as equal to \(v\) inside \(V/W\). This is an example of an equivalence class from set theory ([Set Theory] §Equivalence Relations, ⁋Definition 4), but all we need is the following fact, claimed in the introduction, that two cosets are equal if and only if the difference of their representatives lies in \(W\).

Lemma 2 For a \(\mathbb{K}\)-vector space \(V\) and its subspace \(W\leq V\), and for any two vectors \(v,v'\in V\), the following equivalence holds:

\[v+W=v'+W\iff v-v'\in W\]

Proof

First, if \(v-v'\in W\), then for any representative \(v+w\in v+W\) we have \(v+w=v'+\bigl((v-v')+w\bigr)\in v'+W\), and the converse holds in the same way, so \(v+W=v'+W\).

Conversely, if \(v+W=v'+W\), then since \(v=v+0\in v'+W\) there exists \(w\in W\) such that \(v=v'+w\), and therefore \(v-v'=w\in W\).

In particular, \(v+W=W\) is equivalent to \(v\in W\), and from this we see that any two distinct cosets are always disjoint.

Definition of the Quotient Space

Now we endow the cosets with a vector space structure.

Definition 3 For a \(\mathbb{K}\)-vector space \(V\) and its subspace \(W\leq V\), we denote by \(V/W\) the set of all cosets of \(W\) and call it the quotient space of \(V\) by \(W\). Addition and scalar multiplication on \(V/W\) are defined by the formulas

\[(v+W)+(v'+W)=(v+v')+W,\qquad \alpha(v+W)=(\alpha v)+W\]

In this case, the linear map \(p: V\rightarrow V/W\) sending any \(v\in V\) to \(v+W\) is called the natural projection.

In the above definition, addition and scalar multiplication are described via a representative vector \(v\) of the coset, so we must verify that they are well defined independent of the choice of representative. That is, if \(v+W=v_1+W\) and \(v'+W=v_1'+W\), then

\[(v+v')+W=(v_1+v_1')+W,\qquad (\alpha v)+W=(\alpha v_1)+W\]

must hold. By assumption \(v-v_1\in W\) and \(v'-v_1'\in W\), so since \(W\) is closed under addition,

\[(v+v')-(v_1+v_1')=(v-v_1)+(v'-v_1')\in W\]

and since \(W\) is closed under scalar multiplication,

\[(\alpha v)-(\alpha v_1)=\alpha(v-v_1)\in W\]

By Lemma 2 this is exactly the condition we need. Therefore the two operations on \(V/W\) are well defined.

Proposition 4 The set \(V/W\) equipped with the operations from Definition 3 is a \(\mathbb{K}\)-vector space. The additive identity is \(0+W=W\), and the additive inverse of \(v+W\) is \((-v)+W\).

Proof

All vector space axioms follow immediately from the fact that the operations on \(V/W\) are induced from those on \(V\) coset-wise. For example, associativity of addition follows from the fact that for any \(v,v',v''\in V\),

\[\bigl((v+W)+(v'+W)\bigr)+(v''+W)=\bigl((v+v')+v''\bigr)+W=\bigl(v+(v'+v'')\bigr)+W=(v+W)+\bigl((v'+W)+(v''+W)\bigr)\]

which is a direct consequence of the associativity of addition in \(V\). Commutativity, distributivity, and the scalar multiplication axioms are verified in the same way. Meanwhile, for any \(v\in V\),

\[(v+W)+(0+W)=(v+0)+W=v+W,\qquad (v+W)+((-v)+W)=(v-v)+W=0+W\]

so \(0+W\) is the additive identity and \((-v)+W\) is the additive inverse of \(v+W\).

Dimension of the Quotient Space

The only invariant of a vector space is its dimension. For \(V/W\), the dimension is determined immediately from the dimensions of \(V\) and \(W\).

Theorem 5 For a finite-dimensional \(\mathbb{K}\)-vector space \(V\) and its subspace \(W\leq V\), the formula

\[\dim(V/W)=\dim V-\dim W\]

holds.

Proof

Let \(\dim W=k\), \(\dim V=n\), and choose a basis \(\{x_1,\ldots, x_k\}\) of \(W\). This is a linearly independent subset of \(V\), so by §Dimension of Vector Spaces, ⁋Proposition 5 we can extend it to a basis \(\{x_1,\ldots, x_k, x_{k+1},\ldots, x_n\}\) of \(V\). We show that the cosets

\[x_{k+1}+W,\quad\ldots,\quad x_n+W\]

form a basis of \(V/W\).

First, they span \(V/W\). For any \(v\in V\), writing \(v=\sum_{i=1}^n\alpha_ix_i\) gives

\[v+W=\sum_{i=1}^n\alpha_i(x_i+W)=\sum_{i=k+1}^n\alpha_i(x_i+W)\]

where the last equality holds because for \(i\leq k\) we have \(x_i\in W\), so \(x_i+W=W\) is the zero vector of \(V/W\).

Next, to show linear independence, suppose for scalars \(\alpha_{k+1},\ldots,\alpha_n\) that

\[\sum_{i=k+1}^n\alpha_i(x_i+W)=0+W\]

Then as seen before, \(\sum_{i=k+1}^n\alpha_ix_i+W=W=0+W\), so by Lemma 2 we have \(\sum_{i=k+1}^n\alpha_ix_i\in W\). Hence we can express this vector in terms of the basis of \(W\), so for suitable scalars \(\beta_1,\ldots,\beta_k\),

\[\sum_{i=k+1}^n\alpha_ix_i=\sum_{i=1}^k\beta_ix_i\]

and rearranging gives

\[-\sum_{i=1}^k\beta_ix_i+\sum_{i=k+1}^n\alpha_ix_i=0\]

But the left-hand side is a linear combination of \(\{x_1,\ldots, x_n\}\), and since these form a basis of \(V\) they are linearly independent, so all coefficients must be \(0\); in particular \(\alpha_{k+1}=\cdots=\alpha_n=0\).

Therefore \(\{x_{k+1}+W,\ldots, x_n+W\}\) is a basis of \(V/W\), and since it has \(n-k\) elements, \(\dim(V/W)=n-k=\dim V-\dim W\).

First Isomorphism Theorem

One reason this post exists as a separate article is to give a more essential meaning to §Isomorphisms, ⁋Theorem 7 (Rank-nullity theorem). In this final section we resolve this.

For any \(\mathbb{K}\)-vector space \(V\) and subspace \(W\leq V\), consider the function \(p:V\rightarrow V/W\) defined by

\[p(v)=v+W\]

Then the operations in Definition 3 are defined precisely so that \(p\) satisfies the two equations

\[p(\alpha v)=(\alpha v)+W=\alpha(v+W)=\alpha p(v),\qquad p(v+v')=(v+v')+W=(v+W)+(v'+W)=p(v)+p(v')\]

That is, \(p\) is a linear map, which we call the natural projection from \(V\) to \(V/W\). By definition \(p\) is surjective, and

\[\ker p=\{v\in V\mid v+W=W\}=W\]

From this we see that any subspace can be realized as the kernel of a suitable linear map.

The most important property of the natural projection is the following universal property: any linear map sending \(W\) to \(0\) factors uniquely through \(V/W\).

Proposition 6 Let a \(\mathbb{K}\)-vector space \(V\) and a subspace \(W\leq V\) be given, and let \(L:V\rightarrow U\) be a linear map to another \(\mathbb{K}\)-vector space \(U\) satisfying \(W\subseteq\ker L\). Then the linear map \(\bar L:V/W\rightarrow U\) defined by

\[\bar L(v+W)=L(v)\]

exists uniquely and satisfies \(L=\bar L\circ p\).

Proof

First we show that \(\bar L\) is well defined. If \(v+W=v'+W\), then \(v-v'\in W\subseteq\ker L\), so

\[L(v)-L(v')=L(v-v')=0\]

and therefore \(L(v)=L(v')\). Thus the value of \(\bar L(v+W)\) is independent of the choice of representative. That \(\bar L\) is linear follows from

\[\bar L\bigl(\alpha(v+W)+(v'+W)\bigr)=\bar L\bigl((\alpha v+v')+W\bigr)=L(\alpha v+v')=\alpha L(v)+L(v')=\alpha\bar L(v+W)+\bar L(v'+W)\]

Also, for any \(v\in V\) we have \((\bar L\circ p)(v)=\bar L(v+W)=L(v)\), so \(L=\bar L\circ p\). Finally, if a linear map \(L':V/W\rightarrow U\) satisfying \(L=L'\circ p\) is given, then since \(p\) is surjective, for any \(v+W\in V/W\) we have \(L'(v+W)=L'(p(v))=L(v)=\bar L(v+W)\), and therefore \(L'=\bar L\).

In particular, applying the above universal property to the case \(W=\ker L\) yields the following theorem, which is central to classifying vector spaces.

Theorem 7 (First Isomorphism Theorem) For two \(\mathbb{K}\)-vector spaces \(V,U\) and a linear map \(L:V\rightarrow U\), the linear map \(\bar L:V/\ker L\rightarrow \im L\) defined by

\[\bar L(v+\ker L)=L(v)\]

is an isomorphism. That is, \(V/\ker L\cong\im L\).

Proof

Setting \(W=\ker L\), by Proposition 6 the linear map \(\bar L:V/\ker L\rightarrow U\) defined by \(\bar L(v+\ker L)=L(v)\) is well defined, and its image equals \(\im L\). Therefore, restricting the codomain to \(\im L\), the map \(\bar L:V/\ker L\rightarrow\im L\) is surjective. On the other hand, if \(\bar L(v+\ker L)=0\), then \(L(v)=0\), i.e. \(v\in\ker L\), so \(v+\ker L=\ker L\) is the zero vector of \(V/\ker L\). Hence \(\ker\bar L=\{0\}\), and therefore \(\bar L\) is injective. (§Linear Maps, ⁋Proposition 8) That is, \(\bar L\) is a bijective linear map, and thus an isomorphism. (§Isomorphisms, ⁋Lemma 2)

Combining Theorem 7 (First Isomorphism Theorem) and Theorem 5 above, we recover the rank-nullity theorem. Indeed, for finite-dimensional \(V\),

\[\rank L=\dim\im L=\dim(V/\ker L)=\dim V-\dim\ker L=\dim V-\nullity L\]

holds, which is exactly the formula of §Isomorphisms, ⁋Theorem 7 (Rank-nullity theorem). That is, the rank-nullity theorem is nothing more than the statement that after “collapsing” \(\ker L\), \(L\) becomes injective, expressed in the language of dimensions.

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References

[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.

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