# Quotient vector space/New page

## Definition

Let [ilmath]\mathbb{F} [/ilmath] be a field and let [ilmath](V,\mathbb{F})[/ilmath] be a vector space over that field. Let [ilmath]U\subseteq V[/ilmath] be a vector subspace of [ilmath]V[/ilmath]. Then:

• [ilmath]V/U[/ilmath] considered as a quotient group of the additive group [ilmath](V,+)[/ilmath] by the subgroup [ilmath]U[/ilmath] is actually a vector space in its own right.
• The addition operation is automatic from the quotient group structure: [ilmath][u]+[v]\eq [u+v][/ilmath]
• The scalar multiplication is: [ilmath]\lambda[v]\eq[\lambda v][/ilmath] (see Claim 1 for proof of this)
TODO: Flesh this out

## Proof of claims

### Claim 1: Scalar multiplication can be defined on [ilmath]V/U[/ilmath]

 We wish to factor [ilmath]\pi\circ(\cdot)[/ilmath] through [ilmath](\text{Id},\pi)[/ilmath] [ilmath]\xymatrix{ \mathbb{F}\times V \ar[rr]^{(\cdot)} \ar[d]_{(\text{Id},\pi)} \ar@{-->}@/^1.7ex/[drr]^(.7){\pi\circ(\cdot)} & & V \ar[d]^{\pi} \\ {\mathbb{F}\times\frac{V}{U}} \ar@{.>}[rr]_{\overline{(\cdot)}} & & \frac{V}{U} }[/ilmath]
Let [ilmath]\pi:V\rightarrow\frac{U}{V} [/ilmath] denote the canonical projection of the quotient group.

We wish to factor scalar multiplication (and the projection of the result) through to [ilmath]\mathbb{F}\times\frac{V}{U} [/ilmath]. This is hidden slightly by the functions, remember [ilmath]\pi(\lambda v)\eq[\lambda v][/ilmath] and we wish to define:

• [ilmath]\overline{(\cdot)}:\mathbb{F}\times\frac{V}{U}\rightarrow\frac{V}{U} [/ilmath] taking [ilmath](\lambda,[v])[/ilmath] to some element of [ilmath]\frac{V}{U} [/ilmath]
Or, more obviously written, takes [ilmath]\lambda[v] [/ilmath] to some element of [ilmath]\frac{V}{U} [/ilmath]
• Such that [ilmath]\lambda[v]\eq [\lambda v][/ilmath] - the diagram commutes.

Recall in order to factor we require:

• [ilmath]\forall (\lambda,u),(\mu,v)\in \mathbb{F}\times V[(\text{Id},\pi)(\lambda,u)\eq(\text{Id},\pi)(\mu,v)\implies \pi(\lambda u)\eq\pi(\mu v)][/ilmath], that is to say:
• [ilmath]\forall (\lambda,u),(\mu,v)\in \mathbb{F}\times V\big[(\lambda,[u])\eq(\mu,[v])\implies [\lambda u]\eq [\mu v]\big][/ilmath]

Proof

• Recall by definition of an ordered pair that [ilmath](\lambda,[u])\eq(\mu,[v])\iff\big[\lambda\eq\mu\wedge [u]\eq[v]\big][/ilmath], so already we see that we can drop [ilmath]\mu[/ilmath] and deal only with [ilmath]\lambda[/ilmath]. Thus:
[ilmath]\forall (\lambda,u),(\lambda,v)\in \mathbb{F}\times V\big[ [u]\eq[v]\implies [\lambda u]\eq [\lambda v]\big][/ilmath]

With this in mind:

• Let [ilmath]\lambda\in\mathbb{F} [/ilmath] be given. Let [ilmath]u,v\in V[/ilmath] be given also.
• Suppose [ilmath]\pi(u)\neq\pi(v)[/ilmath] (that is [ilmath][u]\neq[v][/ilmath]), then by the nature of logical implication we're done. We do not care about the truth or falsity of the RHS.
• Suppose [ilmath]\pi(u)\eq\pi(v)[/ilmath] (that is [ilmath][u]\eq[v][/ilmath]) - we must show in this case that we have [ilmath][\lambda u]\eq[\lambda v][/ilmath]
• Well [ilmath][u]\eq [v][/ilmath] means [ilmath]u\in [v][/ilmath], means [ilmath]u\in v+U[/ilmath] (where [ilmath]v+U[/ilmath] denotes a coset and:
• [ilmath]u\in v+U\iff\exists w_1\in U[u\eq v+w_1][/ilmath]
• Taking [ilmath]u\eq v+w_1[/ilmath] we see that [ilmath]\lambda u\eq \lambda(v+w_1)\eq \lambda v+\lambda w_1[/ilmath]
• This is okay to do as we are working in a vector space and these are all elements of [ilmath]V[/ilmath] remember
• As [ilmath]U[/ilmath] is a vector subspace of [ilmath]V[/ilmath] and [ilmath]w_1\in U[/ilmath] we see that [ilmath]\lambda w_1\in U[/ilmath] also
• Define [ilmath]w_2:\eq \lambda w_1[/ilmath], note that [ilmath]w_2\in U[/ilmath]
• We see now that [ilmath]\lambda u\eq \lambda v+ w_2[/ilmath]
• So [ilmath]\lambda u\in \lambda v+ U[/ilmath]
• So [ilmath]\lambda u\in [\lambda v][/ilmath]
• So [ilmath][\lambda u]\eq[\lambda v][/ilmath]
• As required.

We apply factoring to yield a scalar multiplication on the elements of [ilmath]V/U[/ilmath], and this can be written unambiguously as:

• [ilmath]\lambda[u]\eq[\lambda u][/ilmath]