Quotient Spaces

Quotient spaces are not a curiosity confined to pure algebra; they are a working tool wherever linear structure needs to be simplified by discarding a chosen subspace.

• Linear algebra itself: the First Isomorphism Theorem for vector spaces states that for any linear map $T: V \to Z$, the induced map $\overline T : V/\ker T \to \operatorname{im} T$ from the Corollary above is an isomorphism — this is how ranks, images, and kernels are unified conceptually.
• Functional analysis: quotients of Banach or Hilbert spaces by closed subspaces are essential in defining spaces like $L^p/\{\text{a.e.-zero functions}\}$, and quotient norms make $V/W$ itself a normed space.
• Algebraic topology and homological algebra: homology and cohomology groups are literally defined as quotients $\ker \partial_n / \operatorname{im} \partial_{n+1}$ of vector spaces (or modules) — cycles modulo boundaries.
• Representation theory: quotient representations of a group action on $V/W$ arise whenever $W$ is an invariant subspace, letting one study representations by induction on dimension.
• Coding theory: in the theory of linear codes, quotient spaces $\mathbb F_q^n / C$ (for a linear code $C$) organize cosets used in syndrome decoding, directly exploiting that cosets partition the ambient space.
• Physics (gauge theories): physical states are often defined as equivalence classes of field configurations modulo gauge transformations, a quotient-space idea (in an appropriately generalized, often nonlinear, setting) whose linear shadow is exactly the vector-space quotient here.
• Statistics: in linear regression, the residual space is naturally described via a quotient of the response space by the column space of the design matrix, connecting directly to orthogonal projections and least squares.

Ask the tutor: How does the dimension formula $\dim(V/W) = \dim V - \dim W$ get proved in full, and how does it feed into the First Isomorphism Theorem for linear maps?

Think of $W$ as a set of directions you have decided not to care about. Two vectors $v_1$ and $v_2$ should be considered "the same" if the only way they differ is by moving along $W$ — that is, if $v_1 - v_2 \in W$. This is exactly the relation $v_1 \sim v_2 \iff v_1 - v_2 \in W$, and its equivalence classes are precisely the cosets $v + W$. Geometrically, if $V = \mathbb{R}^2$ and $W$ is a line through the origin, each coset $v + W$ is a line parallel to $W$ (or $W$ itself, when $v \in W$). These parallel lines tile the plane without overlapping — they partition $V$. The quotient space $V/W$ treats each entire line as a single point. Adding two cosets means: pick any point on each line, add the points, and take the parallel line through the sum; the well-definedness claim is that it does not matter which points you picked. A second useful picture: if you choose a complement $U$ with $V = W \oplus U$ (via Direct Sums of Subspaces), then every coset $v+W$ contains exactly one vector of $U$, and the map sending a coset to that unique representative gives a vector space isomorphism $V/W \cong U$. This shows $V/W$ "looks like" a complement of $W$ — but crucially, $V/W$ is defined without ever choosing one.

Example 1 — A concrete subspace of $\mathbb{R}^3$ with asymmetric data. Let $V = \mathbb{R}^3$ and let $W = \{(x,y,z) \in \mathbb{R}^3 : 2x - y + 3z = 0\}$, a plane through the origin (indeed a subspace since it is the kernel of the linear functional $(x,y,z)\mapsto 2x-y+3z$). Take $v_1 = (1, 5, 1)$ and $v_2 = (4, 2, 0)$. We first check whether $v_1$ and $v_2$ lie in the same coset:

Test membership in $W$: $2(-3) - 3 + 3(1) = -6-3+3 = -6 \ne 0$. So $v_1 - v_2 \notin W$, and $v_1+W \ne v_2+W$: these are genuinely different cosets. Now compute the sum of cosets $(v_1+W) + (v_2+W)$ using the definition:

Check: $2(5) - 7 + 3(1) = 10-7+3=6\ne 0$, confirming $(5,7,1)\notin W$, consistent with the sum being a nontrivial coset. To double check well-definedness concretely, replace $v_1$ by another representative of the same coset. Since $(0,2,0)\cdot$ — let's find $w\in W$ directly: take $w=(1,2,0)$; check $2(1)-2+0=0$, so $w \in W$. Then $v_1' = v_1 + w = (2,7,1)$ is another representative of $v_1+W$. Also pick $w' = (3,6,0) \in W$ (check: $2(3)-6+0=0$ ✓.) so $v_2' = v_2+w' = (7,8,0)$ represents $v_2+W$. Recompute the sum with the new representatives:

The theorem asserts $(9,15,1)+W = (5,7,1)+W$. Check by subtracting:

and test membership: $2(4)-8+0 = 0$. Indeed $(4,8,0) \in W$, confirming the two representatives give the same coset — well-definedness holds as guaranteed by the theorem. Finally, exhibit the quotient concretely. Since $W$ is a 2-dimensional plane (a hyperplane) in $\mathbb{R}^3$, and $2x-y+3z$ is a linear functional vanishing exactly on $W$, the map

is well-defined (it factors the functional $T(x,y,z)=2x-y+3z$ through $\pi$ exactly as in the Universal Property Corollary, since $\ker T = W$) and is a linear isomorphism $\mathbb{R}^3/W \xrightarrow{\sim} \mathbb{R}$. This exhibits $\dim(\mathbb{R}^3/W) = 1 = 3 - 2 = \dim V - \dim W$, foreshadowing Dimension of Quotient Spaces.

Example 2 — A quotient of a polynomial space by a non-obvious subspace. Let $V = \mathcal{P}_3(\mathbb{R})$, real polynomials of degree at most $3$, a 4-dimensional space with basis $\{1,x,x^2,x^3\}$. Let

This is a subspace: it's the kernel of the linear map $L: \mathcal P_3 \to \mathbb R^2$, $L(p) = (p(1), p'(1))$, and kernels of linear maps are always subspaces (this follows directly from linearity: if $L(p)=L(q)=0$ then $L(cp+q) = cL(p)+L(q) = 0$). We compute $\dim W$. Write $p(x) = a_0+a_1x+a_2x^2+a_3x^3$. Then

The conditions $p(1)=0, p'(1)=0$ are two independent linear equations in $(a_0,a_1,a_2,a_3)$ — independent because, e.g., setting $(a_2,a_3)=(1,0)$ and $(a_2,a_3)=(0,1)$ produces different, non-proportional solutions for $(a_0,a_1)$ (specifically $(a_0,a_1) = (0,-2)$ vs. $(a_0,a_1)=(0,-3)$ after solving), so the map $L$ has rank $2$. Hence $\dim W = 4 - 2 = 2$ by rank-nullity. A concrete basis for $W$: try $q_1(x) = (x-1)^2 = x^2-2x+1$. Check: $q_1(1) = 0$, $q_1'(x) = 2x-2$, $q_1'(1)=0$. Good, $q_1 \in W$. Try $q_2(x) = (x-1)^2 x = x^3-2x^2+x$. Check: $q_2(1) = 0$; $q_2'(x) = 3x^2-4x+1$, $q_2'(1) = 3-4+1=0$. Good, $q_2\in W$. Since $q_1, q_2$ are linearly independent (different degrees, 2 and 3) and $\dim W = 2$, $\{q_1,q_2\}$ is a basis of $W$: indeed $W = (x-1)^2 \cdot \mathcal P_1(\mathbb R)$, the multiples of $(x-1)^2$ of degree $\le 3$. Now consider two specific polynomials $p_1(x) = x^3 + x + 1$ and $p_2(x) = 2x^3 - x^2 + 3$. Are $p_1 + W$ and $p_2+W$ the same coset? Compute

Test membership in $W$ by checking $(p_1-p_2)(1)$ and $(p_1-p_2)'(1)$:

Since this already fails, $p_1-p_2 \notin W$, so $p_1+W \ne p_2+W$ — distinct cosets. By the Corollary (Universal Property), the map $L(p) = (p(1),p'(1))$ has $\ker L = W$ exactly, so it factors as an injective linear map $\overline L : \mathcal P_3/W \to \mathbb R^2$, $\overline L(p+W) = (p(1),p'(1))$. Since $\dim(\mathcal P_3/W) = 4-2=2 = \dim \mathbb R^2$ and $\overline L$ is injective between equal finite dimensions, $\overline L$ is a linear isomorphism. Concretely,

These differ (as they must, since the cosets are distinct), giving a fully computed isomorphism $\mathcal P_3/W \cong \mathbb R^2$ carrying $p_1+W \mapsto (3,4)$ and $p_2+W\mapsto(4,4)$.