Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $W \subset V$ be vector spaces. I don't understand the quotient space $V/W$. I read the Wikipedia and searched this site.

I would have thought: say the vector space operation is $+$. let $Q = V/W$. Then $V = W+Q$ by "multiplying across". So $Q$ contains elements of the form $V + (-1)W$. Why isn't this how the quotient space is defined?

share|cite|improve this question
You have $Q = \{ \{v\}+W | v \in V \}$. I suspect the term quotient came from group notation? – copper.hat Oct 21 '12 at 16:18
A Vector space is an abelian group! So the quotient space would contain the cosets of $W$! – ILoveMath Oct 21 '12 at 17:21
You can't "solve" equations for sets. If $A=B+C$ as sets, that doesn't imply that $C=A-B$. (Take $A=B$ to be any set and $C=\{0\}$, for example.) – Greg Martin Oct 21 '12 at 17:25

Any subspace $\,W\leq V\,$ (over some field $\,\Bbb F\,$) defines an equivalence relation $\,\sim_W\,$ on $\,V\,$ as follows:

$$v_1\sim_Wv_2\Longleftrightarrow v_1-v_2\in W$$

1) Show the above is an equivalence relation

2) If we denote the equivalence clases of the above relation by $\,v+W\,$ (in set theory this would usually be defined as $\,[v]\,\,,\,\,[v]_W\,$ or something similar), then we can define two operations on the set of equivalence classes, denoted by $\,V/W\,$ , as follows:

(i) Sum of classes: $\,(v_1+W)+(v_2+W):=(v_1+v_2)+W\,$

(ii) Product by scalar: for any $\,k\in\Bbb F\;\;,\;\;k(v+W):=(kv)+W\,$

3) Prove both operations above are well defined and they determine a structure of $\,\Bbb F_\,$vector space on $\,V/W\,$

If you know some group theory, the above applies mutatis mutandis to normal subgroups of a group, though the plain equivalence relation (i.e., without the operations) applies to any subgroup of a group.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.