Quotient Vector Space is Vector Space

Theorem

Let $K$ be a field.

Let $X$ be a vector space over $K$.

Let $N$ be a linear subspace of $X$.

Define:

$X/N = \set {x + N : x \in X}$

where $x + N$ is the Minkowski sum of $x$ and $N$.

Define:

$\paren {x + N} +_{X/N} \paren {y + N} = \paren {x + y} + N$

for $x, y \in X$, and:

$\alpha \circ_{X/N} {x + N} = \paren {\alpha x} + N$

for $\alpha \in K$ and $x \in X$.

Then $\paren {X/N, +_{X/N}, \circ_{X/N} }_K$ is a vector space.

Proof

Lemma

Let $K$ be a field.

Let $X$ be a vector space over $K$.

Let $N$ be a linear subspace of $X$.

Let $z \in N$.

Then:

$z + N = N$

$\Box$

Vector Addition is Well-Defined

To show that $+_{X/N}$ is well-defined, we show that if $x, x', y, y' \in N$ are such that:

$x + N = x' + N$

and:

$y + N = y' + N$

we have:

$\paren {x + y} + N = \paren {x' + y'} + N$

Since $x + N = x' + N$, we have:

$x' \in x + N$

and so there exists $z \in N$ such that $x + z = x'$.

Then $z = x' - x$, and so $x' - x \in N$.

Similarly, we have that $y' - y \in N$.

So we have $\paren {x' + y'} - \paren {x + y} \in N$, since $N$ is a linear subspace of $X$, so that:

$N = \paren {x' + y'} - \paren {x + y} + N$

by the lemma.

So that:

$\paren {x + y} + N = \paren {x' + y'} + N$

$\Box$

Scalar Multiplication is Well-Defined

Let $\alpha \in K$.

Suppose that $x, x' \in N$ are such that:

$x + N = x' + N$

We need to show that:

$\alpha x + N = \alpha x' + N$

As argued in proving vector addition is well-defined, we have that $x' - x \in N$.

Since $N$ is a linear subspace of $X$, we have:

$\alpha \paren {x' - x} = \alpha x' - \alpha x \in N$

So from the lemma, we have:

$N = \alpha x' - \alpha x + N$

So that:

$\alpha x + N = \alpha x' + N$

$\Box$

Proof of Vector Space Axiom $\text V 1$: Commutativity

Let $\mathbf x, \mathbf y \in X/N$.

Then there exists $x, y \in N$ such that $\mathbf x = x + N$ and $\mathbf y = y + N$.

Then we have:

\(\ds \mathbf x + \mathbf y\)

\(=\)

\(\ds \paren {x + N} + \paren {y + N}\)

\(\ds \)

\(=\)

\(\ds \paren {x + y} + N\)

\(\ds \)

\(=\)

\(\ds \paren {y + x} + N\)

using Vector Space Axiom $\text V 1$: Commutativity for $X$

\(\ds \)

\(=\)

\(\ds \paren {y + N} + \paren {x + N}\)

\(\ds \)

\(=\)

\(\ds \mathbf y + \mathbf x\)

$\Box$

Proof of Vector Space Axiom $\text V 2$: Associativity

Let $\mathbf x, \mathbf y, \mathbf z \in X/N$.

Then there exists $x, y, z \in N$ such that $\mathbf x = x + N$, $\mathbf y = y + N$ and $\mathbf z = z + N$.

Then we have:

\(\ds \paren {\mathbf x + \mathbf y} + \mathbf z\)

\(=\)

\(\ds \paren {\paren {x + N} + \paren {y + N} } + \paren {z + N}\)

\(\ds \)

\(=\)

\(\ds \paren {\paren {x + y} + N} + \paren {z + N}\)

\(\ds \)

\(=\)

\(\ds \paren {\paren {x + y} + z} + N\)

\(\ds \)

\(=\)

\(\ds \paren {x + \paren {y + z} } + N\)

using Vector Space Axiom $\text V 2$: Associativity for $X$

\(\ds \)

\(=\)

\(\ds \paren {x + N} + \paren {\paren {y + z} + N}\)

\(\ds \)

\(=\)

\(\ds \paren {x + N} + \paren {\paren {y + N} + \paren {z + N} }\)

\(\ds \)

\(=\)

\(\ds \mathbf x + \paren {\mathbf y + \mathbf z}\)

$\Box$

Proof of Vector Space Axiom $\text V 3$: Identity

Let $0_{X/N} = 0_X + N = N$.

We prove that $0_{X/N}$ satisfies the demand of Vector Space Axiom $\text V 3$: Identity.

Let $\mathbf x \in X/N$.

Then there exists $x \in X$ such that $\mathbf x = x + N$.

We have:

\(\ds 0_{X/N} + \mathbf x\)

\(=\)

\(\ds \paren {0_X + N} + \paren {x + N}\)

\(\ds \)

\(=\)

\(\ds \paren {0_X + x} + N\)

\(\ds \)

\(=\)

\(\ds x + N\)

using Vector Space Axiom $\text V 3$: Identity for $X$

\(\ds \)

\(=\)

\(\ds \paren {x + 0_X} + N\)

using Vector Space Axiom $\text V 3$: Identity for $X$

\(\ds \)

\(=\)

\(\ds \paren {x + N} + \paren {0_X + N}\)

\(\ds \)

\(=\)

\(\ds \mathbf x + \mathbf 0_{X/N}\)

$\Box$

Proof of Vector Space Axiom $\text V 4$: Inverses

Let $\mathbf x \in X/N$.

Then there exists $x \in X$ such that $\mathbf x = x + N$.

Set $-\mathbf x = -x + N$.

We then have:

\(\ds \mathbf x + \paren {-\mathbf x}\)

\(=\)

\(\ds \paren {x + N} + \paren {-x + N}\)

\(\ds \)

\(=\)

\(\ds \paren {x - x} + N\)

\(\ds \)

\(=\)

\(\ds 0_X + N\)

\(\ds \)

\(=\)

\(\ds 0_{X/N}\)

$\Box$

Proof of Vector Space Axiom $\text V 5$: Distributivity over Scalar Addition

Let $\lambda, \mu \in K$.

Let $\mathbf x \in X/N$.

Then there exists $x \in X$ such that $\mathbf x = x + N$.

Then we have:

\(\ds \paren {\lambda + \mu} \mathbf x\)

\(=\)

\(\ds \)

\(\ds \)

\(=\)

\(\ds \paren {\lambda + \mu} \paren {x + N}\)

\(\ds \)

\(=\)

\(\ds \paren {\paren {\lambda + \mu} x} + N\)

\(\ds \)

\(=\)

\(\ds \paren {\lambda x + \mu x} + N\)

Vector Space Axiom $\text V 5$: Distributivity over Scalar Addition for $X$

\(\ds \)

\(=\)

\(\ds \paren {\lambda x + N} + \paren {\mu x + N}\)

\(\ds \)

\(=\)

\(\ds \lambda \paren {x + N} + \mu \paren {x + N}\)

\(\ds \)

\(=\)

\(\ds \lambda \mathbf x + \mu \mathbf x\)

$\Box$

Proof of Vector Space Axiom $\text V 6$: Distributivity over Vector Addition

Let $\lambda \in K$.

Let $\mathbf x, \mathbf y \in X/N$.

Then there exists $x, y \in N$ such that $\mathbf x = x + N$ and $\mathbf y = y + N$.

We then have:

\(\ds \lambda \paren {\mathbf x + \mathbf y}\)

\(=\)

\(\ds \lambda \paren {\paren {x + N} + \paren {y + N} }\)

\(\ds \)

\(=\)

\(\ds \lambda \paren {\paren {x + y} + N}\)

\(\ds \)

\(=\)

\(\ds \paren {\lambda \paren {x + y} } + N\)

\(\ds \)

\(=\)

\(\ds \paren {\lambda x + \lambda y} + N\)

Vector Space Axiom $\text V 6$: Distributivity over Vector Addition for $X$

\(\ds \)

\(=\)

\(\ds \paren {\lambda x + N} + \paren {\lambda y + N}\)

\(\ds \)

\(=\)

\(\ds \lambda \paren {x + N} + \lambda \paren {y + N}\)

\(\ds \)

\(=\)

\(\ds \lambda \mathbf x + \lambda \mathbf y\)

Proof of Vector Space Axiom $\text V 7$: Associativity with Scalar Multiplication

Let $\lambda, \mu \in K$.

Let $\mathbf x \in X/N$.

Then there exists $x \in X$ such that $\mathbf x = x + N$.

We have:

\(\ds \lambda \paren {\mu \mathbf x}\)

\(=\)

\(\ds \lambda \paren {\mu \paren {x + N} }\)

\(\ds \)

\(=\)

\(\ds \lambda \paren {\mu x + N}\)

\(\ds \)

\(=\)

\(\ds \paren {\lambda \paren {\mu x} } + N\)

\(\ds \)

\(=\)

\(\ds \lambda \mu x + N\)

Vector Space Axiom $\text V 7$: Associativity with Scalar Multiplication for $X$

\(\ds \)

\(=\)

\(\ds \lambda \mu \paren {x + N}\)

\(\ds \)

\(=\)

\(\ds \lambda \mu \mathbf x\)

$\Box$

Proof of Vector Space Axiom $\text V 8$: Identity for Scalar Multiplication

Let $\mathbf x \in X/N$.

Then there exists $x \in X$ such that $\mathbf x = x + N$.

Then we have:

\(\ds 1_K \mathbf x\)

\(=\)

\(\ds 1_K \paren {x + N}\)

\(\ds \)

\(=\)

\(\ds \paren {1_K x} + N\)

\(\ds \)

\(=\)

\(\ds x + N\)

Vector Space Axiom $\text V 8$: Identity for Scalar Multiplication for $X$

$\blacksquare$