# Quotient space example

1. Nov 6, 2008

### jamjar

Can anyone post me a clear example of how to compute the quotient space U/V from a vector space U and subspace V?
I've seen many formal definitions but I'm a little stuck on practical use.
I'm particularly interested in an example that shows how U and V being over the reals (for example) can result in the quotient being over a finite field.

2. Nov 6, 2008

### Office_Shredder

Staff Emeritus
The quotient space should always be over the same field as your original vector space. To 'counterprove' your desired example, if U/V is over a finite field, the field has characteristic p, which means that for some u not in V, p*u is in V. But V is a vector space.

An example of a quotient space.... I think in R2 is the best example. U is R2 V is the linear span of (1,0). Then U/V is going to be the set of all (a,b) + V. But this is ((a,0) + (0,b)) + V (a,0) is in V for all a, so (a,0)+V = 0+V which implies that (a,b) + V = (0,b) + V

In general for a finite dimensional U, if V has basis v1,....,vm[/sub] and this is extended to a basis of U by adding u1... un Then if u is in U, $$u = \sum_{1}^{m}a_iv_i + \sum_{1}^{n}b_iu_i$$ where the a's and b's are in the field. Then any vi+V = 0+V so $$u+V = \sum_{1}^{n}b_iu_i + V$$ and it's easy to see that U/V is isomorphic to the linear span of u1,...,un

3. Nov 6, 2008

### HallsofIvy

A "quotient space" is always a collection of equivalence classes. Given a subspace V, we can define "u~ v" if and only if u- v is in V.

For a simple example, let U be R2, the set of all pairs (x, y) with the usual addition and multiplication and V the subspace {(x, 0)}. Two vectors are (x1, y1) and (x2, y2) are equivalent if and only if (x1,y1)- (x2,y2) is in (x1-x2, y1-y2) is in V which means y1=y2. So an equivalence class consists of all {(x,y0} where x is a variable and y0 is a specific number. As points, those are all the points on the line parallel to the y axis, y= y0. Of course, in order to make that a vector space you have to define an addition and scalar multiplication. The standard way of "adding" equivalence classes is to choose one member from each equivalence class. For example, there is an equivalence class consisting of all pairs of the form (x, 1) and another equivalence class consisting of all pairs of the form (x, 5). One "representative" of the first class is (3, 1) and a "representative" of the second is (4, 5). Their sum is (3, 1)+ (4, 5)= (7, 6). That is in the equivalence class of all pairs (x, 6) so {(x, 1)}+ {(x, 5)}= {(x, 6)}. To multiply by a scalar, you can do the same thing: a*{(u,y0)}= {( u, ay0)}. Of course, the pair (0, y0) is always in the equivalence class {(x, y0} so you can always use that pair and you can think of this quotient space being the space of points (0, y) which is a subspace of R2[/sub].

If you like, you can think of that as "collapsing" the line y= y0 to the single point (0,y0) so it's easy to see that quotient space is really the same as the subspace {(0,y)}.

Having said all that, I come to your last question! The definition of "quotient space" I have given- which I think is standard- clearly gives the quotient space as a vector spaced over the samefield as U and V. I can't think of any way of starting with U a vector space over R, say, and U/V being a vector space over a finite field.

Last edited by a moderator: Nov 6, 2008