Bilinear mapping between quotient spaces

[lineintegral1]

Problem: Let L and M be finite dimensional linear spaces over the field K and let g: L\times M \rightarrow K be a bilinear mapping. Let L_0 be the left kernel of g and let M_0 be the right kernel of g.

a) Prove that dim L/L_0 = dim M/M_0.

b) Prove that g induces the bilinear mapping g': L/L_0 \times M/M_0 \rightarrow K, g'(l+L_0, m+M_0) = g(l,m), for which the left and right kernels are zero.

I am trying to prove b) before a) (I think this would make a) relatively trivial). However, I am still getting used to the notion of quotient spaces as sets of equivalence classes. I am unsure as to how to deal with the bilinear mapping of equivalence classes. All that I have seen so far deal with vector spaces. Perhaps someone can give me some insight as to how to show that this mapping is well defined? How can I visualize this better? My professor is of little help, so I appreciate any insights anyone can give me.

Thanks all!

 

[micromass]

You must prove that if l+L_0=l^\prime +L_0 and m+M_0=m^\prime+M_0, then g^\prime(l+L_0,m+M_0) = g^\prime(l^\prime+L_0,m^\prime+M_0). So you must actually prove g(l,m)=g(l^\prime,m^\prime)...

 

[Deveno]

but...this is obvious because g is linear in l and m, and l-l' is in L₀ and m-m' is in M₀

(g(l,m) = g(l',m') means g(l,m) - g(l',m') = 0.

but g(l,m) - g(l',m') = g(l,m) - g(l',m) + g(l',m) - g(l',m')

= g(l-l',m) + g(l',m-m') =...?)

 

[micromass]

And what do you know about l-l' and m-m' ?

 

[Deveno]

we are assuming l + L₀ = l' + L₀

so l-l' is in 0 + L₀ = L₀.

the same for m-m', with regard to M₀.

we have to have these conditions, for g' to be well-defined as a function

(we are defining g' over cosets, by using g defined on elements.

if we do not have the same image for g(l',m') and g(l,m) when l,l' and m,m'

are in the same cosets of L₀ and M₀,

g' isn't even a function, much less bilinear).

 

[lineintegral1]

Wow, okay, so that makes it obvious. Thanks for the help guys, I appreciate it. Neat stuff!