# Quotient spaces in linear algebra

There's a statement in some notes I'm reading that goes like this: "...$V/U$ is a 'simplified version' of $V$ where the elements of $U$ are ignored" ($V$ and $U$ are vector spaces).

I'm still trying to understand this idea: can someone shed some light on why we ignore $U$ in $V/U$? I mean I understand that everything belonging to $U$ falls in the equivalence class $[0]$, but is that the sense in which we are "ignoring" $U$? Is a quotient space a set of equivalence classes? That is to say that, if we have equivalent affine subsets $v_1 + U = v_2 + U$, are $[v_1]$ and $[v_2]$ the same elements in $U/V$? If this is the case, then aren't we ignoring certain elements of $V$ (like we would ignore $[v_2]$ in this case) because all equivalent affine subsets are redundancies and are therefore collapsed into one respective equivalence class?

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You'll understand it more as you compute more examples. Roughly speaking, for each vector, the component that belongs to $U$ is ignored. – TBrendle Oct 11 '13 at 2:26

## 3 Answers

Assume that everything is a finite dimensional vector space. Your vector subspace $U$ has a basis $\{u_1,\ldots,u_n\}$. You can extend this to a basis of $V$, $\{u_1,\ldots,u_n,v_1,\ldots,v_{k}\}$. Then a basis for $V/U$ is given by the equivalence classes $\{[v_1],\ldots,[v_k]\}$. In this way you might be able to see how we are "ignoring" $U$.

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Yes, you are correct that $V/U$ is a set of equivalence classes of the form $v+U$. When we say that we are ignoring $U$, what we mean is that if two vectors are equivalent up to an element in $U$, then they are the same in $V/U$. So if $U$ is all that is distinguishing two vectors, i.e. $v - w \in U$, then we ignore that and consider them the same.

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In the quotient all elements of $U$ are zero. So in essence we are looking at things outside $U$. Two elements of $V$ are the same in $V/U$ when they differ by an element in $U$. Not only are we looking at things outside $U$, but if they differ by an element of $U$ we consider them the same. So adding elements of $U$ doesn't change anything. It is ignoring $U$ completely, Elements of $U$ are zero and adding by an element of $U$ does nothing.

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