Residual Classes and Quotient Spaces in Linear Algebra 📂Linear Algebra

Residual Classes and Quotient Spaces in Linear Algebra

Definition¹

$V$ is a $F$-vector space and $W \le V$ is a subspace. For $v \in V$, the set

$$ \left\{ v \right\} + W := \left\{ v + w : w \in W \right\} $$

is called the coset of $W$ containing $v$ . The left-hand $+$ is the sum of sets.

Explanation

Often $\left\{ v \right\} + W$ is simply denoted by $v + W$.

Let $\left\{ v + W : v \in V \right\}$ denote the set of all cosets of $W$. Define addition and scalar multiplication (by $F$) as follows.

$$ (v_{1} + W) + (v_{2} + W) = (v_{1} + v_{2}) + W,\quad \forall v_{1}, v_{2} \in V $$

$$ a(v + W) = av + W\quad \forall v \in V \text{ and } a \in F $$

Then this set becomes a $F$-vector space again. Denote this vector space by $V/W$, and call it the quotient space of $V$ modulo $W$ . Even if one accepts the definitions so far, it is not immediately obvious why this vector space is called the quotient space and why it is denoted like division by $V / W$. Let us understand this with a detailed explanation. Let the whole space and subspace be as follows.

$$ V = \mathbb{R}^{5} = \left\{ (a,b,c,d,e) \right\} $$

$$ W = \mathbb{R}^{2} = \left\{ (a,b,0,0,0) \right\} $$

Then one can see that a basis generating the quotient space is as follows.

$$ \left\{ (0,0,1,0,0) + W, (0,0,0,1,0) + W, (0,0,0,0,1) + W \right\} $$

Thus $V/W = \left\{ v + W : v \in V \right\}$ is a 3-dimensional vector space.

$$ V / W \cong \mathbb{R}^{3} $$

Written as below, the notation for the quotient space becomes intuitively appropriate, as if it reduces the dimension by the laws of exponents.

$$ V/W = \dfrac{V}{W} = \dfrac{\mathbb{R}^{5}}{\mathbb{R}^{2}} = \mathbb{R}^{5-2} = \mathbb{R}^{3} $$

In fact, the following equation holds. (See here.)

$$ \dim(V/W) = \dim(V) - \dim(W) $$

Proposition

(a) $v + W$ is a subspace of $V$ if and only if $v \in W$. (Algebraic proof)

(b) For $v_{1}, v_{2} \in V$, $v_{1} + W = v_{2} + W$ is equivalent to $v_{1} - v_{2} \in W$. (Algebraic proof)

(c) $V/W$ is a vector space, and its zero vector is $0_{V} + W = W$. ($0_{V}$ is the zero vector of $V$.)

Proof

(a)

$(\Longrightarrow)$
Assume $v + W$ is a subspace of $V$. Then, letting $0_{V}$ be the zero vector of $V$, we have $0_{V} \in v + W$. Hence for some $w \in W$, $0_{V} = v + w$, and $w = -v \in W$. Since $W$ is a subspace of $V$, it is closed under scalar multiplication, so $v = -(-v) \in W$.
$(\Longleftarrow)$
Assume $v \in W$. To show that $v + W$ is a subspace of $V$, it suffices to show closure under addition and scalar multiplication. (See) Let $v + w_{1}, v + w_{2} \in v + W$. Adding these two gives
$$ (v + w_{1}) + (v_{1} + w_{2}) = v + (v + w_{1} + w_{2}) $$
Because $W$ is a subspace, it is closed under addition, and by the assumption that $v$ is an element of $W$, there exists some $w_{3} \in W$ such that the following holds.
$$ v + (v + w_{1} + w_{2}) = v + w_{3} \in W $$
Now let $a \in F$. Similarly, by the assumption there exists some $w_{4} \in W$ such that the following holds.
$$ a(v + w) = v + \left( (a-1)v + aw \right) = v + w_{4} \in W $$

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(b)

$(\Longrightarrow)$
Assume $v_{1} + W = v_{2} + W$. Then for the zero vector $0_{V}$ of $V$ and some $w \in W$, the following holds.
$$ v_{1} + 0_{V} = v_{2} + w \implies v_{1} - v_{2} = w \in W $$
$(\Longleftarrow)$
Assume $v_{1} - v_{2} \in W$. Then
$$ \begin{align*} v_{2} + W &= \left\{ v_{2} + w : w \in W \right\} \\ &= \left\{ v_{2} + \left( (v_{1} - v_{2}) + w \right) : w \in W \right\} \\ &= \left\{ v_{1} + w : w \in W \right\} \\ &= v_{1} + W \end{align*} $$

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