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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?

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    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.)2012-10-21

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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.