Quotient vector space/New page

Definition

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

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

TODO: Flesh this out

Proof of claims

Claim 1: Scalar multiplication can be defined on $V/U$

$\xymatrix{ \mathbb{F}\times V \ar[rr]^{(\cdot)} \ar[d]_{(\text{Id},\pi)} \ar@{-->}@/^1.7ex/[drr]^(.7){\pi\circ(\cdot)} & & V \ar[d]^{\pi} \\ {\mathbb{F}\times\frac{V}{U}} \ar@{.>}[rr]_{\overline{(\cdot)}} & & \frac{V}{U} }$
We wish to factor $\pi\circ(\cdot)$ through $(\text{Id},\pi)$

Let $\pi:V\rightarrow\frac{U}{V}$ denote the canonical projection of the quotient group.

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

• $\overline{(\cdot)}:\mathbb{F}\times\frac{V}{U}\rightarrow\frac{V}{U}$ taking $(\lambda,[v])$ to some element of $\frac{V}{U}$

  Or, more obviously written, takes $\lambda[v]$ to some element of $\frac{V}{U}$

• Such that $\lambda[v]\eq [\lambda v]$ - the diagram commutes.

Recall in order to factor we require:

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

Proof

• Recall by definition of an ordered pair that $(\lambda,[u])\eq(\mu,[v])\iff\big[\lambda\eq\mu\wedge [u]\eq[v]\big]$, so already we see that we can drop $\mu$ and deal only with $\lambda$. Thus:

  $\forall (\lambda,u),(\lambda,v)\in \mathbb{F}\times V\big[ [u]\eq[v]\implies [\lambda u]\eq [\lambda v]\big]$

With this in mind:

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

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

• $\lambda[u]\eq[\lambda u]$