# Chapter 27. Quotient Spaces

# Chapter 27. Quotient Spaces

A quotient space is a vector space obtained by identifying all vectors that differ by a vector in a fixed subspace. If \(U\) is a subspace of \(V\), then the quotient space is written

$$
V/U.
$$

It is read as “\(V\) modulo \(U\)” or “\(V\) by \(U\).” Informally, the construction collapses the whole subspace \(U\) to zero and keeps track of what remains outside \(U\). More formally, its elements are equivalence classes, also called cosets, of the form \(v+U\).

## 27.1 The Basic Idea

Let \(V\) be a vector space over a field \(F\), and let \(U\subseteq V\) be a subspace.

Two vectors \(v,w\in V\) are regarded as equivalent modulo \(U\) when their difference lies in \(U\):

$$
v \sim w
\quad\Longleftrightarrow\quad
v-w\in U.
$$

This means that \(v\) and \(w\) are considered the same after we ignore all movement inside \(U\).

The quotient space \(V/U\) is the set of all equivalence classes under this relation.

## 27.2 Cosets

The equivalence class of a vector \(v\in V\) is

$$
[v]=\{w\in V : w-v\in U\}.
$$

This class is more commonly written as

$$
v+U.
$$

Thus

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

The set \(v+U\) is called a coset of \(U\) in \(V\).

If \(v\in U\), then

$$
v+U=U.
$$

So every vector in \(U\) belongs to the same coset as \(0\). In the quotient space, all vectors of \(U\) become the zero element.

## 27.3 The Quotient Space

The quotient space is

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

Its elements are not individual vectors of \(V\). Its elements are cosets.

This distinction is important. A vector \(v\) belongs to \(V\). A coset \(v+U\) belongs to \(V/U\).

The zero vector of \(V/U\) is

$$
0+U=U.
$$

Thus the subspace \(U\) itself becomes the zero element in the quotient.

## 27.4 Equality of Cosets

Two cosets are equal exactly when their representatives differ by an element of \(U\):

$$
v+U=w+U
\quad\Longleftrightarrow\quad
v-w\in U.
$$

This gives the practical test for equality in a quotient space.

For example, if

$$
V=\mathbb{R}^2
$$

and

$$
U=\operatorname{span}
\left(
\begin{bmatrix}
1\\
0
\end{bmatrix}
\right),
$$

then two vectors

$$
\begin{bmatrix}
a\\
b
\end{bmatrix},
\qquad
\begin{bmatrix}
c\\
d
\end{bmatrix}
$$

belong to the same coset if their difference lies in \(U\). That means

$$
\begin{bmatrix}
a-c\\
b-d
\end{bmatrix}
\in U.
$$

This happens exactly when

$$
b=d.
$$

So in \(\mathbb{R}^2/U\), only the second coordinate remains relevant.

## 27.5 Operations on Cosets

Addition in the quotient space is defined by

$$
(v+U)+(w+U)=(v+w)+U.
$$

Scalar multiplication is defined by

$$
c(v+U)=cv+U.
$$

These definitions are natural: choose representatives, perform the operation in \(V\), then take the resulting coset.

The operations are well-defined because changing representatives by elements of \(U\) does not change the final coset. This relies on \(U\) being a subspace.

## 27.6 Well-Defined Addition

We must check that the definition of addition does not depend on the chosen representatives.

Suppose

$$
v+U=v'+U
$$

and

$$
w+U=w'+U.
$$

Then

$$
v-v'\in U,
\qquad
w-w'\in U.
$$

Since \(U\) is closed under addition,

$$
(v+w)-(v'+w')=(v-v')+(w-w')\in U.
$$

Therefore

$$
(v+w)+U=(v'+w')+U.
$$

So addition is well-defined.

## 27.7 Well-Defined Scalar Multiplication

Suppose

$$
v+U=v'+U.
$$

Then

$$
v-v'\in U.
$$

For any scalar \(c\),

$$
cv-cv'=c(v-v')\in U,
$$

because \(U\) is closed under scalar multiplication.

Therefore

$$
cv+U=cv'+U.
$$

So scalar multiplication is well-defined.

## 27.8 Vector Space Structure

With the operations

$$
(v+U)+(w+U)=(v+w)+U
$$

and

$$
c(v+U)=cv+U,
$$

the quotient \(V/U\) is a vector space over \(F\).

Its zero element is

$$
U.
$$

The additive inverse of \(v+U\) is

$$
(-v)+U.
$$

Indeed,

$$
(v+U)+((-v)+U)=0+U=U.
$$

The vector space axioms follow from the corresponding axioms in \(V\).

## 27.9 Geometric Picture in \(\mathbb{R}^2\)

Let

$$
V=\mathbb{R}^2
$$

and let

$$
U=\operatorname{span}
\left(
\begin{bmatrix}
1\\
0
\end{bmatrix}
\right).
$$

Then \(U\) is the \(x\)-axis.

For a vector

$$
v=
\begin{bmatrix}
a\\
b
\end{bmatrix},
$$

the coset \(v+U\) is

$$
\left\{
\begin{bmatrix}
a+t\\
b
\end{bmatrix}
:t\in\mathbb{R}
\right\}.
$$

This is the horizontal line through height \(b\).

Thus \(\mathbb{R}^2/U\) can be viewed as the set of all horizontal lines. Each horizontal line is one element of the quotient space. Since only the height \(b\) matters, this quotient is naturally isomorphic to \(\mathbb{R}\).

## 27.10 Geometric Picture in \(\mathbb{R}^3\)

Let

$$
V=\mathbb{R}^3
$$

and let

$$
U=\operatorname{span}
\left(
\begin{bmatrix}
1\\
0\\
0
\end{bmatrix},
\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
\right).
$$

Then \(U\) is the \(xy\)-plane.

For

$$
v=
\begin{bmatrix}
a\\
b\\
c
\end{bmatrix},
$$

the coset \(v+U\) is

$$
\left\{
\begin{bmatrix}
a+s\\
b+t\\
c
\end{bmatrix}
:s,t\in\mathbb{R}
\right\}.
$$

This is the horizontal plane at height \(c\).

Therefore \(\mathbb{R}^3/U\) can be identified with the remaining coordinate \(c\). Hence

$$
\mathbb{R}^3/U \cong \mathbb{R}.
$$

The quotient collapses the entire \(xy\)-plane direction and keeps only the vertical direction.

## 27.11 Quotient by a Coordinate Subspace

Let

$$
V=\mathbb{R}^n
$$

and let

$$
U=\operatorname{span}(e_1,\ldots,e_m).
$$

Then \(U\) consists of all vectors whose last \(n-m\) coordinates are zero.

Two vectors in \(\mathbb{R}^n\) are equivalent modulo \(U\) exactly when their last \(n-m\) coordinates agree.

Thus

$$
\mathbb{R}^n/U \cong \mathbb{R}^{n-m}.
$$

The quotient removes the first \(m\) coordinate directions and preserves the remaining \(n-m\) directions.

## 27.12 Dimension of a Quotient Space

If \(V\) is finite-dimensional and \(U\) is a subspace of \(V\), then

$$
\dim(V/U)=\dim V-\dim U.
$$

This number is called the codimension of \(U\) in \(V\):

$$
\operatorname{codim}(U)=\dim(V/U).
$$

For example, if

$$
\dim V=5
$$

and

$$
\dim U=2,
$$

then

$$
\dim(V/U)=3.
$$

The quotient keeps the directions not already accounted for by \(U\).

## 27.13 Proof of the Dimension Formula

Let

$$
\dim U=k.
$$

Choose a basis of \(U\):

$$
u_1,\ldots,u_k.
$$

Extend it to a basis of \(V\):

$$
u_1,\ldots,u_k,v_1,\ldots,v_r.
$$

Then

$$
\dim V=k+r.
$$

We claim that

$$
v_1+U,\ldots,v_r+U
$$

form a basis of \(V/U\).

First, they span \(V/U\). Any vector \(v\in V\) can be written as

$$
v=a_1u_1+\cdots+a_ku_k+b_1v_1+\cdots+b_rv_r.
$$

In the quotient, the terms from \(U\) vanish:

$$
v+U=b_1(v_1+U)+\cdots+b_r(v_r+U).
$$

Second, they are linearly independent. If

$$
b_1(v_1+U)+\cdots+b_r(v_r+U)=U,
$$

then

$$
b_1v_1+\cdots+b_rv_r\in U.
$$

So

$$
b_1v_1+\cdots+b_rv_r =
a_1u_1+\cdots+a_ku_k
$$

for some scalars \(a_i\). This gives a linear relation among the basis vectors of \(V\). Hence all coefficients are zero, in particular

$$
b_1=\cdots=b_r=0.
$$

Therefore

$$
\dim(V/U)=r=\dim V-\dim U.
$$

## 27.14 The Quotient Map

There is a natural linear map

$$
\pi:V\to V/U
$$

defined by

$$
\pi(v)=v+U.
$$

This is called the quotient map or canonical projection.

It is linear because

$$
\pi(v+w)=(v+w)+U=(v+U)+(w+U),
$$

and

$$
\pi(cv)=cv+U=c(v+U).
$$

It is surjective because every element of \(V/U\) has the form \(v+U\).

## 27.15 Kernel of the Quotient Map

The kernel of the quotient map

$$
\pi:V\to V/U
$$

is exactly \(U\).

Indeed,

$$
\ker \pi=\{v\in V:\pi(v)=U\}.
$$

But

$$
\pi(v)=U
$$

means

$$
v+U=0+U.
$$

This holds exactly when

$$
v\in U.
$$

Therefore

$$
\ker \pi=U.
$$

So quotienting by \(U\) is precisely the operation of making \(U\) into the zero subspace.

## 27.16 Quotients and Linear Maps

Let

$$
T:V\to W
$$

be a linear map.

If \(U\subseteq \ker T\), then \(T\) is constant on each coset of \(U\). That is, if

$$
v+U=w+U,
$$

then

$$
v-w\in U\subseteq \ker T,
$$

so

$$
T(v-w)=0.
$$

Therefore

$$
T(v)=T(w).
$$

This means \(T\) descends to a well-defined map on the quotient:

$$
\overline{T}:V/U\to W
$$

given by

$$
\overline{T}(v+U)=T(v).
$$

The quotient space is designed exactly to support this kind of construction.

## 27.17 First Isomorphism Theorem

Let

$$
T:V\to W
$$

be a linear map. Then

$$
V/\ker T \cong \operatorname{Im}(T).
$$

The isomorphism is

$$
\overline{T}:V/\ker T\to \operatorname{Im}(T)
$$

defined by

$$
\overline{T}(v+\ker T)=T(v).
$$

This is well-defined, linear, one-to-one, and onto.

This theorem says that after collapsing exactly the vectors killed by \(T\), what remains is the image of \(T\).

## 27.18 Example: Projection onto One Coordinate

Define

$$
T:\mathbb{R}^2\to\mathbb{R}
$$

by

$$
T(x,y)=y.
$$

The kernel is

$$
\ker T=\{(x,0):x\in\mathbb{R}\},
$$

the \(x\)-axis.

The image is all of \(\mathbb{R}\).

By the first isomorphism theorem,

$$
\mathbb{R}^2/\ker T \cong \mathbb{R}.
$$

This matches the geometric picture: quotienting \(\mathbb{R}^2\) by the \(x\)-axis leaves only height.

## 27.19 Quotients and Direct Sums

Suppose

$$
V=U\oplus W.
$$

Then every vector \(v\in V\) can be written uniquely as

$$
v=u+w,
$$

where

$$
u\in U,
\qquad
w\in W.
$$

In the quotient \(V/U\),

$$
v+U=(u+w)+U=w+U.
$$

Thus every coset has a unique representative in \(W\). Therefore

$$
V/U\cong W.
$$

So when \(V\) splits as a direct sum, quotienting by \(U\) leaves the complementary part \(W\).

## 27.20 Summary

A quotient space \(V/U\) is formed by identifying vectors of \(V\) that differ by an element of the subspace \(U\). Its elements are cosets \(v+U\). The whole subspace \(U\) becomes the zero vector of the quotient.

The key ideas are:

| Concept | Meaning |
|---|---|
| Equivalence modulo \(U\) | \(v\sim w\) when \(v-w\in U\) |
| Coset | \(v+U=\{v+u:u\in U\}\) |
| Quotient space | Set of all cosets \(v+U\) |
| Zero coset | \(U=0+U\) |
| Quotient map | \(\pi(v)=v+U\) |
| Kernel of quotient map | \(\ker \pi=U\) |
| Dimension formula | \(\dim(V/U)=\dim V-\dim U\) |
| First isomorphism theorem | \(V/\ker T\cong\operatorname{Im}(T)\) |

Quotient spaces express the idea of ignoring a subspace. They appear whenever a linear construction identifies several vectors as equivalent and keeps only the information that remains after that identification.