Residual Classes and Quotient Spaces in Linear Algebra 📂Linear Algebra

Residual Classes and Quotient Spaces in Linear Algebra

Definition¹

Let’s refer to $V$ as $F$-vector space, and $W \le V$ as subspace. For $v \in V$, the following set

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

is called the coset of $W$ containing $v$. $+$ is the sum of sets.

Explanation

We often abbreviate $\left\{ v \right\} + W$ as $v + W$.

Considering the set of all cosets of $W$, denoted by $\left\{ v + W : v \in V \right\}$, we 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 again forms a $F$-vector space. This vector space is denoted by $V/W$, and is called the quotient space of $V$ modulo $W$.

Theorem

(a) $v + W$ being a subspace of $V$ is equivalent to $v \in W$. (proof in algebra)

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

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

Proof

(a)

Assuming $(\Longrightarrow)$
Assume $v + W$ is a subspace of $V$. Then, considering $0_{V}$ as the zero vector of $V$, $0_{V} \in v + W$ holds. Therefore, for some $w \in W$, $0_{V} = v + w$ and $w = -v \in W$ hold. $W$ being a subspace of $V$ is closed under scalar multiplication, so $v = -(-v) \in W$ holds.
Assuming $(\Longleftarrow)$
Assume $v \in W$. To show $v + W$ is a subspace of $V$, it suffices to show closure under addition and scalar multiplication. Let’s say $v + w_{1}, v + w_{2} \in v + W$. Adding these two gives:
$$ (v + w_{1}) + (v_{1} + w_{2}) = v + (v + w_{1} + w_{2}) $$
Since $W$ is a subspace, it’s closed under addition, and by assumption $v$ is an element of $W$, so for some $w_{3} \in W$, the following holds:
$$ v + (v + w_{1} + w_{2}) = v + w_{3} \in W $$
Now, let’s say $a \in F$. Then similarly, by assumption, for some $w_{4} \in W$, the following holds:
$$ a(v + w) = v + \left( (a-1)v + aw \right) = v + w_{4} \in W $$

■

(b)

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

■