Linear Independence of \overline{w} and \overline{v} in R4/U

[smerhej]

Homework Statement

Well it isn't so much the problem as it is the notation used within the problem. But here is the question:

Determine whether or not \overline{w} and \overline{v} are linearly independent in R⁴/U

Homework Equations

If v \in V then \overline{v} = v + U

The Attempt at a Solution

I don't understand the notation R⁴/U (I understand R⁴ and the subspace U, but don't understand the slash between them)

 

[Deveno]

it's called a "quotient space" and its elements are called "cosets" and consist of "parallel translates of a subspace by a vector".

that's why v has the overline: it's the SET:

{v+u: u in U}.

which can be thought of as the entire subspace U, moved in the direction/distance of v.

another way to think of it, is as regarding the entire space (in this case R⁴) losing dim(U) dimensions, by regarding all points in U as "equivalent (essentially 0)".

if dim(U) = 1, each coset is a parallel line, and you need a 3-vector to tell you "which line".

if dim(U) = 2, each coset is a parallel plane, and you need a 2 vector to tell you "which plane".

higher dimensions are harder to visualize, but the same sort of logic applies.

simce v+U is a set, v is just a "representative", and the same coset v+U can have different representatives.

one common way quotient spaces arise is in analyzing linear maps: often, we don't care about the kernel of a linear map (because everything in it just maps to the 0-vector), so we "mod it out". the resulting quotient space is isomorphic to the image space (this is pretty much equivalent to the rank-nullity theorem, but in a more abstract setting).

you calculate with elements in R⁴/U pretty much like you do with elements in R⁴, but with a "+U" along for the ride:

v+U + w+U = (v+w)+U
a(v+U) = av+U

the overline notation is a bit "cleaner" but hides some of what is going on.