Study on Quotient Spaces

me@iany.me (Ian Yang) — Tue, 18 Nov 2025 21:23:25 +0800

I’m reading Linear Algebra Done Right by Axler and found the section on quotient spaces difficult to understand, so I researched and took these notes.

Definitions

3.95 notion: $v + U$

Suppose $v \in V$ and $U \subseteq V$ . Then $v + U$ is the subset of $V$ defined by

 $v + U = \{v + u : u \in U\}.$

Also called a translate. Attention that a translate is a set.

3.97 definition: translate

Suppose $v \in V$ and $U \subseteq V$ , the set $v + U$ is said to be a translate of $U$ .

Quotient space is a set of all translates (set of sets):

3.99 definition: quotient space, $V/U$

Suppose $U$ is a subspace of $V$ . Then the quotient space $V/U$ is the set of all translates of $U$ . Thus

 $V/U = \{v + U : v \in V\}.$

Quotient space is a set of sets. There are duplicates for each $v \in V$ because for some $v_1, v_2 \in V$ , $v_1 + U$ and $v_2 + U$ can be identical set.

A quotient space $V/U$ is formed by “collapsing” a subspace $U$ to zero within a larger vector space $V$ . This construction is based on an equivalence relation where two vectors $x, y \in V$ are considered equivalent if their difference lies in $U$ —that is, $x \sim y$ if and only if $x - y \in U$ . wikipedia

Lemmas

3.101 two translates of a subspace are equal or disjoint

Suppose $U$ is a subspace of $V$ and $v, w \in V$ . Then

 $v - w \in U \iff v + U = w + U \iff (v + U) \cap (w + U) \neq \emptyset$

If two translates are not disjoint (the union set is not empty), they must be equal. So they are equal or disjoint.

All distinct translates of a subspace are disjoint. Given any $v \in V$ , it belongs to only one translate.

Since the quotient space $V/U$ is a set of translates of a subspace, it is like a disjoint partition of values in $V$ . By using the definition of quotient map

3.104 definition: quotient map, $\pi$

Suppose $U$ is a subspace of $V$ . The quotient map $\pi : V \to V/U$ is the linear map defined by

 $\pi(v) = v + U$

for each $v \in V$ .

We can write that

 $\pi(v_1) = \pi(v_2) \iff v_1 - v_2 \in U$

The quotient map has two essential properties:

The null space of $\pi$ is exactly the subspace $U$ , because $v+U=0+U \iff v-0 \in U \iff v \in U$
The range of $\pi$ is the entire quotient space $V/U$

Quotient Space Is a Vector Space

First define the addition and scalar multiplication operations:

3.102 definition: addition and scalar multiplication on $V/U$

Suppose $U$ is a subspace of $V$ . Then addition and scalar multiplication are defined on $V/U$ by

 $\begin{align*} (v + U) + (w + U) &= (v + w) + U \\ \lambda(v + U) &= (\lambda v) + U \end{align*}$

for all $v, w \in V$ and $\lambda \in \mathbf{F}$ .

$v+U$ is not the unique way to represent a member in $V/U$ , because there may exist $v'\ne v$ that $u + U = v' + U$ . The operations make sense only when the choice of $v$ to represent a translate makes no differences.

Specifically, suppose $v_1, v_2, w_1, w_2 \in V$ such that

 $v_1 + U = v_2 + U \quad\textrm{and}\quad w_1 + U = w_2 + U$

From the addition definition:

 $\begin{align*} (v_1+U) + (w_1+U) &= (v_1 + w_1) + U \\ (v_2+U) + (w_2+U) &= (v_2 + w_2) + U \end{align*}$

The left side of the two equations indeed are the different representation of the same equation, so we must show that the right side equal: $(v_1 + w_1)+U=(v2+w2)+U$ .

This applies to scalar multiplication as well:

 $\begin{align*} \lambda(v_1 + U) &= (\lambda v_1) + U \\ \lambda(v_2 + U) &= (\lambda v_2) + U \end{align*}$

We must show that $(\lambda v_1) + U = (\lambda v_2) + U$ .

Dimension

The dimension of the quotient space is given by a simple subtraction, relating the dimension of $V/U$ to the “lost” dimension of $U$ :

3.105 dimension of quotient space

Suppose $V$ is finite-dimensional and $U$ is a subspace of V. Then

 $\text{dim } V/U = \text{dim }V - \text{dim }U.$

Linear Map from V/(null T) to W

3.106 notation: $\widetilde{T}$

Suppose $T \in \mathcal{L}(V, W)$ . Define $\widetilde{T}: V/(\text{null } T) \to W$ by

 $\widetilde{T}(v + \text{null } T) = Tv.$

Think of merging inputs having the same output. These inputs will be the same input in the quotient space $V/(\text{null } T)$ .

For any $v_1, v_2 \in V$ that $Tv_1 = Tv_2$ , $v_1 + \mathrm{null}\, T$ and $v_2 + \mathrm{null}\, T$ are the same value in $V/(\mathrm{null}\, T)$ . This makes $\widetilde{T}$ injective. Because $\mathrm{range}\,\widetilde{T}=\mathrm{range}\, T$ , $\widetilde{T}$ is also surjective on to $\mathrm{range}\, T$ .

3.63 invertibility $\iff$ injectivity and surjectivity

A linear map is invertible if and only if it is injective and surjective.

3.63 shows us that $\widetilde{T}$ is invertible, and according to the definition of isomorphic, $V/(\mathrm{null}\, T)$ and $\mathrm{range}\,T$ are isomorphic vector spaces and $\widetilde{T}$ is their isomorphism.

3.69 definition: isomorphism, isomorphic

An isomorphism is an invertible linear map.
Two vector spaces are called isomorphic if there is an isomorphism from one vector space onto the other one.

One of the key uses of $\widetilde{T}$ is demonstrating a canonical isomorphism. For any linear map $T \in \mathcal{L}(V, W)$ , the quotient space $V/(\text{null } T)$ is isomorphic to the image space $\text{range } T$ . This shows that the quotient space $V/(\text{null } T)$ serves as a way to “mod out” the non-injective part of $T$ .