Quotient Spaces and Codimension

When we study a linear operator $T : X \to Y$ , its kernel $\ker(T)$ captures what $T$ “forgets.” The quotient space $X / \ker(T)$ is the space that remains after collapsing everything $T$ cannot distinguish.

Quotient Spaces¶

Definition 1 (Quotient space)

Let $X$ be a vector space and $M \subset X$ a subspace. The quotient space $X / M$ is the set of equivalence classes

[x] = x + M = \{x + m : m \in M\}

(1)

under the equivalence relation $x \sim y \iff x - y \in M$ . This is a vector space with operations

[x] + [y] = [x + y], \qquad \lambda [x] = [\lambda x].

(2)

The canonical projection $\pi : X \to X/M$ defined by $\pi(x) = [x]$ is a surjective linear map with $\ker(\pi) = M$ .

The elements of $X/M$ are the “cosets” or translates $x + M$ . Two vectors $x$ and $y$ represent the same coset precisely when $x - y \in M$ — that is, when they differ by something in $M$ .

Example 1 (Quotient in $\mathbb{R}^3$ )

Let $X = \mathbb{R}^3$ and $M = \{(x,y,0) : x, y \in \mathbb{R}\}$ (the $xy$ -plane). Then $X/M \cong \mathbb{R}$ : each coset is a horizontal plane at height $z$ , and the quotient map sends $(x,y,z) \mapsto z$ . All the $x$ - and $y$ -information is “collapsed” — only the height survives.

Example 2 (Quotient by a kernel)

Let $f : \mathbb{R}^3 \to \mathbb{R}$ be defined by $f(x,y,z) = 2x + 3y - z$ . Then $\ker(f) = \{2x + 3y - z = 0\}$ is a plane through the origin. The cosets of $\ker(f)$ are the level sets $f^{-1}(c) = \{2x + 3y - z = c\}$ — parallel planes, each labeled by the value $c$ . The quotient $\mathbb{R}^3 / \ker(f)$ is one-dimensional, reflecting the fact that $f$ maps onto $\mathbb{R}$ .

The quotient norm¶

When $X$ is a normed space and $M$ is a closed subspace, the quotient inherits a natural norm.

Definition 2 (Quotient norm)

Let $X$ be a normed space and $M \subset X$ a closed subspace. The quotient norm on $X/M$ is

\|[x]\|_{X/M} = \inf_{m \in M} \|x - m\| = \operatorname{dist}(x, M).

(3)

The closedness of $M$ ensures this is a genuine norm (not just a seminorm): if $\|[x]\|_{X/M} = 0$ , then $\operatorname{dist}(x, M) = 0$ , so $x \in \overline{M} = M$ , hence $[x] = [0]$ .

Proposition 1

If $X$ is a Banach space and $M \subset X$ is a closed subspace, then $X/M$ is a Banach space.

Proof 1

Codimension¶

Definition 3 (Codimension)

Let $M$ be a subspace of a vector space $X$ . The codimension of $M$ in $X$ is

\operatorname{codim}(M) = \dim(X / M).

(4)

Equivalently, $\operatorname{codim}(M) = k$ means there exist vectors $e_1, \ldots, e_k \in X$ such that

X = M \oplus \operatorname{span}(e_1, \ldots, e_k),

(5)

and $k$ is the smallest number for which such a decomposition exists.

Codimension measures how many dimensions are “missing” from $M$ . A codimension-1 subspace is as large as possible while still being proper — it is a hyperplane through the origin.

Example 3

In $\mathbb{R}^n$ :

A line through the origin has codimension $n - 1$ .
A plane through the origin in $\mathbb{R}^3$ has codimension 1.
The subspace $\{0\}$ has codimension $n$ .

The First Isomorphism Theorem¶

Theorem 1 (First Isomorphism Theorem)

Let $T : X \to Y$ be a linear map. Then $T$ induces an isomorphism

\widetilde{T} : X / \ker(T) \xrightarrow{\;\cong\;} \operatorname{Range}(T)

(6)

defined by $\widetilde{T}([x]) = Tx$ . This map is well-defined, linear, injective, and surjective onto $\operatorname{Range}(T)$ .

If $X$ and $Y$ are Banach spaces, $T$ is bounded, and $\operatorname{Range}(T)$ is closed, then $\widetilde{T}$ is a topological isomorphism (i.e., $\widetilde{T}$ and $\widetilde{T}^{-1}$ are both bounded). This follows from the Theorem 1 applied to the bijective bounded operator $\widetilde{T}$ .

Proof 2

The first isomorphism theorem immediately answers the question: why does a nonzero linear functional have a codimension-1 kernel?

Corollary 1

Let $f : X \to \mathbb{R}$ be a nonzero linear functional. Then $\operatorname{codim}(\ker(f)) = 1$ .

Proof 3

The key insight: the codimension equals 1 because the target is $\mathbb{R}$ , which is one-dimensional. The first isomorphism theorem tells us that the quotient $X / \ker(f)$ is isomorphic to the range, and since $f$ maps onto all of $\mathbb{R}$ , the quotient is one-dimensional — regardless of whether $X$ itself is finite- or infinite-dimensional.