Quotient Spaces

In this post, for a \(\mathbb{K}\)-vector space \(V\) and its subspace \(W\), we define its quotient space \(V/W\). Intuitively, this is the space obtained inside \(V\) by making all elements of \(W\) equal to \(0\); however, since we want the space remaining even after making all elements of \(W\) equal to \(0\) like this to still be a vector space, simply declaring all elements of \(W\) to be \(0\) cannot achieve this.

Cosets

The main 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 operations, but any element of \(V\), \(v\), can always be written, for a fixed \(0\neq w\in W\), in the form \(v=(v-w)+w\); thus if \(v\) and \(v-w\) do not belong to \(W\), then even after treating all elements of \(W\) as \(0\), we have

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

so that even though \(v\) and \(v-w\) are not equal, their difference becomes \(0\).

This shows two things. First, to define \(V/W\), it is insufficient to simply set all elements of \(W\) to \(0\) and leave the remaining elements as they are. Second, and more importantly, the simple calculation above actually gives a hint as to how to construct \(V/W\): namely, if the difference of two vectors \(v,v'\) lies in \(W\), then we must treat them inside \(V/W\) as the same element.

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

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

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

In the definition, the coset \(v+W\) is the set of all vectors whose difference from \(v\) belongs to \(W\), that is, the set of all vectors that we agreed in the introduction to treat inside \(V/W\) as equal to \(v\). 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 above, that determines whether two cosets are equal by the difference of their representatives.

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

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

holds.

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 \(v=v+0\in v'+W\), we have \(v=v'+w\) for some \(w\in W\), 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 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 the set of all cosets of \(W\) by \(V/W\) and call it the quotient space by \(W\) of \(V\). Addition and scalar multiplication on \(V/W\) are defined respectively by

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

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

In the definition above, addition and scalar multiplication are described via a vector \(v\) representing 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

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

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

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

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

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

as well. By Lemma 2, this is precisely the equality we wanted. Therefore, the two operations on \(V/W\) are well defined.

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

Proof

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

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

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

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

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

Dimension of the Quotient Space

The only invariant of a vector space is its dimension. If \(V\) is finite-dimensional, then in the case of \(V/W\), its 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 of \(W\), say \(\{x_1,\ldots, x_k\}\). Since this is a linearly independent subset of \(V\), by §Dimension of Vector Spaces, ⁋Proposition 5 we can extend it to obtain a basis of \(V\), \(\{x_1,\ldots, x_k, x_{k+1},\ldots, x_n\}\). 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 \(i\leq k\), \(x_i\in W\) and so \(x_i+W=W\) is the zero vector of \(V/W\).

Next, to show that they are linearly independent, suppose that for scalars \(\alpha_{k+1},\ldots,\alpha_n\),

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

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

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

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

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 formula

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

which defines a function \(p:V\rightarrow V/W\). Then the operations of Definition 3 are defined precisely so that \(p\) satisfies the two identities

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

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

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

holds. 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. This says that any linear map sending \(W\) to \(0\) factors uniquely through \(V/W\).

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

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

there exists a unique linear map \(\bar{L}:V/W\rightarrow U\) 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 hence \(L(v)=L(v')\). 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 \(L=L'\circ p\) holds for a linear map \(L':V/W\rightarrow U\), 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 \(L:V\rightarrow U\), defined by the equation

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

the linear map \(\bar{L}:V/\ker L\rightarrow \im L\) is an isomorphism. That is, \(V/\ker L\cong\im L\).

Proof

Setting \(W=\ker L\), by Proposition 6, the linear map defined by \(\bar{L}(v+\ker L)=L(v)\), namely \(\bar{L}:V/\ker L\rightarrow U\), is well defined, and its image equals \(\im L\). Restricting the codomain to \(\im L\), we obtain that \(\bar{L}:V/\ker L\rightarrow\im L\) 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) Thus \(\bar{L}\) is a bijective linear map, hence 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). In other words, the rank-nullity theorem is nothing more than the statement that \(L\) becomes injective after “folding away” \(\ker L\), 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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