---
title: "quotient space"
type: concept
related: [Operation, Additive Identity, Closed, Inverses, Quotient Space]
source: https://www.jemoka.com/posts/kbhquotient_space/
confidence: high
status: active
---

A quotient space is the set of all affine subsets of \(V\) parallel to some subspace \(U\). This should be reminiscent of quotient groups.
constituents vector space \(V\) a subspace \(U \subset V\) requirements \begin{equation} V / U = \{v+U : v \in V \} \end{equation}
additional information operations on quotient space Addition and scalar multiplication on the quotient space is defined in the expected way:
given \((v+U), (w+U) \in V / U\), and \(\lambda \in \mathbb{F}\):
\begin{equation} \begin{cases} (v+U) + (w+U) = ((v+w)+U) \\ \lambda (v+U) = ((\lambda v)+U) \end{cases} \end{equation}
quotient space operations behave uniformly on equivalent affine subsets The tricky thing about quotient space operations is that there are multiple ways of representing a single affine subset parallel to \(U\); the one-liner about this is that if you think about shifting a parallel line with a vector: shifting the line along any perpendicular vector to the line with the same magnitude will get you the same shifted line.
For the operations above to work, we have to make sure that they behave in the same way on distinct representations of the same affine subset, which we endeavor to proof here:
Suppose we have \(v,w \in V\), \(v&rsquo;,w&rsquo; \in V\), and that \(v+U = v&rsquo;+U\); \(w+U = w&rsquo;+U\). We desire that the operations above behave the same way to any addition groupings: that WLOG \((v+U)+(w+U) = (v&rsquo;+U)+(w&rsquo;+U)\) &mdash; that is, we have to show that \((v+w)+U = (v&rsquo;+w&rsquo;)+U\).
By the fact that two affine subsets parallel to \(U\) are either equal or disjoint, we have that \(v-v&rsquo;, w-w&rsquo; \in U\). And so, \((v-v&rsquo;)+(w-w&rsquo;) \in U\). Commuting these things under \(V\), we now have that \((v+w)-(v&rsquo;+w&rsquo;) \in U\). Therefore, invoking the same result again, \((v+w)+U = (v&rsquo;+w&rsquo;)+U\), as desired.
The same logic can be used for scalar multiplication. Suppose we have \(v, v&rsquo; \in V\), \(\lambda \in \mathbb{F}\), and that \(v+U = v&rsquo;+U\). We desire that WLOG \(\lambda (v+U) = \lambda (v&rsquo;+U)\) &mdash; that is, we have to show that \((\lambda v)+U = (\lambda v&rsquo;)+U\).
Again invoking the two affine subsets parallel to \(U\) are either equal or disjoint result, we have that \(v-v&rsquo; \in U\). Now, this means that \(\lambda (v-v&rsquo;) = \lambda v-\lambda v&rsquo; \in U\) because closure of scalar multiplication in \(U\). Invoking the result again, we now have that \(\lambda v + U = \lambda v&rsquo; +U\), as desired.
Having shown both operations make sense, we can declare that they make sense indeed. \(\blacksquare\)
quotient space is a vector space Given the name! (jk)
Bleh I give up just prove it yourself given the above operations and the fact that the additive identity is \(0+U = U\), the additive inverse is \(-v+U\).
&ldquo;instead of the elements single vectors, we fuse the whole affine subset together. instead of counting the contents, we count the bucket.&rdquo;
dimension of a quotient space is the difference between dimensions of its constituents that is,
\begin{equation} \dim V / U = \dim V - \dim U \end{equation}
for finite dimensional \(V\).
Proof:
Let \(\pi: V \to V /U\). By definition, \(null\ \pi =U\); and, given the input is any \(v \in V\), \(range\ \pi = V / U\). rank-nullity theorem then tells us that:
\begin{equation} \dim V = \dim U + \dim V / U \end{equation}
now subtract and get \(\dim V /U\) by itself. \(\blacksquare\)
