Quotient Spaces

New vector spaces can be constructed from other vector spaces. One of the important contructions is the quotient spaces which is base on equivalence relation of sets.

Definition. Let $\sim$ be an equivalence relation on a set $X$. The set of equivalence classes $X/{\sim} = \Set{[a] : a \in X}$ is called the quotient set of $X$ by $\sim$.

Proposition. Let $V$ be a vector space over $\mathbb{F}$ and $Y$ be a subspace of $V$. Then the relation $\sim$ on the set $V$ defined by

\[v \sim v' \qquad \text{if} \qquad v - v' \in Y\]

is an equivalence relation. The quotient set $V/{\sim}$ is a vector space over $\mathbb{F}$, with addition $[v] + [v’] = [v + v’]$ and scalar multiplication $\lambda [v] = [\lambda v]$.

Proof.

$v - v = 0 \in Y$ so $\sim$ is reflexive. $v - v’ = -(v’ - v) \in Y$ so $\sim$ is symmetric. $v - v’, v’ - v’’ \in Y$ implies $v - v’ + v’ - v’’ = v - v’’ \in Y$ so $\sim$ is transitive.

For $[v] = [w]$ and $[v’] = [w’]$, $v - w, v’ - w’ \in Y$. $v - w + v’ - w’ = (v + v’) - (w + w’) \in Y$ so $[v + v’] = [w + w’]$ and addition is well-defined.

For $[v] = [v’]$, $v - v’ \in Y$. $\lambda (v - v’) = \lambda v - \lambda v’ \in Y$ so $[\lambda v] = [\lambda v’$ and scalar multiplication is well-defined.

The vector space axioms can be easily verified using the definitions.

Definition. The vector space $V/{\sim}$ defined above is called the quotient space of $V$ by $Y$, denoted by $V/Y$. The equivalence classes $[v]$ are denoted by $v + Y$ and therefore $(v + Y) + (v’ + Y) = v + v’ + Y$ and $\lambda (v + Y) = \lambda v + Y$.

Proposition. Suppose that $Y$ is finite dimensional, then

\[\dim V = \dim Y + \dim V/Y\]

Proof.

The mapping $\alpha: V \to V/Y$ defined by $\alpha(v) = [v]$ is linear and surjective. Obviously, $n(\alpha) = \dim Y$ and $r(\alpha) = \dim V/Y$. Hence, by rank-nullity theorem (base on direct proof), $\dim V = \dim Y + \dim V/Y$.

We can now investigate what this construction means to linear transformations and matrices.

Proposition. Let $\alpha \in \mathcal{L}(V, V)$ and $Y$ be a subspace of $V$ that is invariant relative to $\alpha$, i.e. $\alpha(Y) \subset Y$. Then the mappings

\[\alpha_Y : Y \to Y, \alpha_Y(y) = \alpha(y) \qquad \text{and} \qquad \alpha_{V/Y} : V/Y \to V/Y, \alpha_{V/Y}(v + Y) = \alpha(v) + Y\]

are linear, i.e. $\alpha_Y \in \mathcal{L}(Y, Y)$ and $\alpha_{V/Y} \in \mathcal{L}(V/Y, V/Y)$.

Proof.

$\alpha_Y \in \mathcal{L}(Y, Y)$ since $Y$ is invariant relative to $\alpha$.

If $v + Y = v’ + Y$, then $v - v’ \in Y$ implies $\alpha(v) - \alpha(v’) = \alpha(v - v’) \in Y$, so $\alpha(v) + Y = \alpha(v’) + Y$ and $\alpha_{V/Y}$ is well-defined. Also,

\[\begin{align*} \alpha_{V/Y}(\lambda (v + Y) + \mu (v' + Y)) &= \alpha_{V/Y}(\lambda v + \mu v' + Y) \\ &= \alpha(\lambda v + \mu v') + Y \\ &= \lambda \alpha(v) + \mu \alpha(v') + Y \\ &= \lambda \alpha_{V/Y}(v + Y) + \mu \alpha_{V/Y}(v' + Y) \end{align*}\]

so $\alpha_{V/Y} \in \mathcal{L}(V/Y, V/Y)$.

Definition. The linear map $\alpha_Y : Y \to Y, \alpha_Y(y) = \alpha(y)$ is called the restriction of $I$ to $Y$.

Definition. The linear map $\alpha_{V/Y} : V/Y \to V/Y, \alpha_{V/Y}(v + Y) = \alpha(v) + Y$ is called the transformation on $V/Y$ induced by $\alpha$.

Proposition. Let $Y$ be a nonzero subspace of a finite dimensional vector space $V$ which is invariant relative to a linear map $\alpha \in \mathcal{L}(V, V)$. Let $\Set{e_1, …, e_k}$ be a basis of $Y$ and extend it to $\Set{e_1, …, e_n}$ as a basis for $V$. If $Y \not= V$, then $k < n$ and $\Set{e_{k+1} + Y, …, e_n + Y}$ is a basis of $V/Y$. Let $A$ be the matrix of $\alpha$ with respect to the basis $\Set{e_1, …, e_n}$. Then $A$ has the form

\[A = \begin{pmatrix} A_1 & A_3 \\ 0 & A_2 \\ \end{pmatrix}\]

where $A_1$ is the $k$-by-$k$ matrix of $\alpha_Y$ with respect to $\Set{e_1, …, e_k}$ and $A_2$ is the $(n-k)$-by-$(n-k)$ matrix of $\alpha_{V/Y}$ with respect to $\Set{e_{k+1} + Y, …, e_n + Y}$.

Proof.

For any $v \in V$, we have $v = \sum \lambda_i e_i$, therefore

\[[v] = \sum_{i=1}^n \lambda_i [e_i] = \sum_{i=k+1}^n \lambda_i [e_i]\]

and $\langle [e_{k+1}], …, [e_n] \rangle = V/Y$.

Suppose that $\sum_{i=k+1}^n \lambda_i [e_i] = [0]$. Then $\sum_{i=k+1}^n \lambda_i e_i \in Y$ and we have

\[\sum_{i=k+1}^n \lambda_i e_i = \sum_{j=1}^k \mu_j e_j\]

which implies $\lambda_i = \mu_j = 0$ and $\Set{e_{k+1} + Y, …, e_n + Y}$ are linearly independent.

The matrix of $\alpha$ is the mapping of basis vectors between domain and codomain. For $Y$ to be invariant, we must have $\alpha(e_i) = \sum_{j=1}^k A_{ji} e_j$ for $1 \le i \le k$ which results with the form of first column. For $k+1 \le i \le n$, $\alpha(e_i) = \sum_{j=1}^k A_{ji} e_j + \sum_{j=k+1}^n A_{ji} e_j$ so $\alpha_{V/Y}([e_i]) = [\alpha(e_i)] = \sum_{j=k+1}^n A_{ji} [e_j]$ which is $A_2$.

Reference