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 $T$.