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 $X$ |
$\blacksquare$