Quotient Spaces

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 remaining space to still be a vector space, simply declaring all elements of \(W\) to be \(0\) is not enough.

Cosets

The main problem, as noted above, is that merely setting all elements of \(W\) to \(0\) gives no guarantee that the remaining space will be a vector space. For it to be a vector space, it must first be closed under operations: for any element \(v\) of \(V\) and any fixed \(0\neq w\in W\), we can always write \(v=(v-w)+w\), so if neither \(v\) nor \(v-w\) belongs to \(W\), then even after treating all elements of \(W\) as \(0\) we obtain

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

so 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 unchanged. 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 regard 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 \(v+W\) is the set of all vectors whose difference from \(v\) lies in \(W\), i.e. the vectors that we agreed to treat as equal to \(v\) in \(V/W\). 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 precisely when the difference of their representatives lies in \(W\).

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 \(v=v+0\in v'+W\), so there exists \(w\in W\) with \(v=v'+w\), 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 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 write the set of all cosets of \(W\) as \(V/W\) and call it the quotient space of \(V\) by \(W\). Addition and scalar multiplication on \(V/W\) are defined respectively by

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

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 representing vector \(v\), so we must check 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 equality we wanted. Hence the two operations on \(V/W\) are well defined.

Proposition 4 The set \(V/W\) equipped with the operations of 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 coset-wise from those on \(V\). For example, associativity of addition follows from

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

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

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

holds.

Proof

Let \(\dim W=k\), \(\dim V=n\), and choose a basis \(\{x_1,\ldots, x_k\}\) of \(W\). 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 \(x_i\in W\) for \(i\leq k\), so \(x_i+W=W\) is the zero vector of \(V/W\).

Next, to show linear independence, let scalars \(\alpha_{k+1},\ldots,\alpha_n\) satisfy

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

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

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

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 function \(p:V\rightarrow V/W\) defined by

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

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

Proposition 6 Let \(V\) be a \(\mathbb{K}\)-vector space, \(W\leq V\) a subspace, and \(L:V\rightarrow U\) 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

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

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 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 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 \(W=\ker L\) 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 \(W=\ker L\), Proposition 6 gives that 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 \(\im L\). 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 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.

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