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 by making all elements of \(W\) equal to \(0\) inside \(V\); however, simply declaring all elements of \(W\) to be \(0\) is not enough, because we want the remaining space to still be a vector space.

Cosets

The biggest problem, as pointed out above, is that merely setting all elements of \(W\) to \(0\) gives no guarantee that the remaining space will be a vector space. For instance, for this to be a vector space it must first be closed under addition, but 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 all elements of \(W\) as \(0\) we would have

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

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

This shows two things. First, to define \(V/W\) it is insufficient to simply set all elements of \(W\) to \(0\) and leave the rest untouched. Second, and more importantly, the simple calculation above actually gives a hint as to how \(V/W\) should be constructed. Namely, 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 \(v\).

From the definition, the coset \(v+W\) is the set of all vectors whose difference from \(v\) lies in \(W\), i.e. the vectors that we agreed to identify with \(v\) inside \(V/W\). This is an example of an equivalence class treated in 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 precisely when 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 two distinct cosets are always disjoint.

Definition of the Quotient Space

Now let us endow the cosets with a vector space structure.

Definition 3 For a \(\mathbb{K}\)-vector space \(V\) and its subspace \(W\leq V\), we write \(V/W\) for 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

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

respectively.

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

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

By assumption \(v-v_1\in W\) and \(v'-v_1'\in W\), so from the fact that \(W\) is closed under addition we obtain

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

and from the fact that \(W\) is closed under scalar multiplication we obtain

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

By Lemma 2 this is exactly the equality we wanted. Hence the two operations on \(V/W\) are well defined.

Proposition 4 With the operations of Definition 3, \(V/W\) 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 coset-wise from those on \(V\). For example, associativity of addition follows from the fact that for any \(v,v',v''\in V\) we have

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

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

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

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. In the case of \(V/W\), its dimension is determined directly 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 \(\dim W=k\) and \(\dim V=n\), and choose a basis \(\{x_1,\ldots, x_k\}\) of \(W\). Since this is a linearly independent subset of \(V\), 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 \(v\in V\), if \(v=\sum_{i=1}^n\alpha_ix_i\) then

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

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, let scalars \(\alpha_{k+1},\ldots,\alpha_n\) be such 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 there exist scalars \(\beta_1,\ldots,\beta_k\) with

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

which rearranges to

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

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 we have \(\dim(V/W)=n-k=\dim V-\dim W\).

First Isomorphism Theorem

One reason this post exists as a separate article is to give deeper 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 the formula

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

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

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

Thus \(p\) is a linear map, called the natural projection from \(V\) to \(V/W\). By definition \(p\) is surjective, and

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

Hence we see that every 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. It says that 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 there exists a unique linear map \(\bar L:V/W\rightarrow U\) defined by the formula

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

such that \(L=\bar L\circ p\).

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 \(L(v)=L(v')\). That is, 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 \(L':V/W\rightarrow U\) satisfies \(L=L'\circ p\), 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 \(L'=\bar L\).

Applying the above universal property to the case \(W=\ker L\), we obtain the following theorem, which is fundamental for 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 the formula

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

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 \(\im L\). Thus restricting the codomain to \(\im L\) gives a surjective map \(\bar L:V/\ker L\rightarrow\im L\). 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, hence an isomorphism. (§Isomorphisms, ⁋Lemma 2)

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

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

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 “folding away” \(\ker L\), \(L\) 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.
[Rom] S. Roman, Advanced linear algebra, 3rd ed., Graduate Texts in Mathematics 135, Springer, 2008.

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