Recent content by xaos

I'm just using what was given in the hypothesis. T and T' are not arbitrary, they're related by the identification T'([ b])=T(b). This is where I'm starting.

Let T'(p(U)) = T(U) s.t. p(U)=[0]. Then T(U)=T'([0])=T(0)=0 so U is in the KerT. I'm not getting the coset idea to work unless a+U =b + U -> a+U - U = b+U-U -> a=b. which might not work.

Refering to #13. For Well-defined. Maybe this works: [a]= [ b] then a=b as coset representatives then T(a)=T(b) since linear maps are well defined. Then use the identification T'([c])=T(c) to say T'([a])=T'([ b])

Yes, I concede. I've been missing a step. [edit] Try: [a-b] is in the KerT' then [a]+KerT' = [ b]+KerT' -> [a]= [ b] as coset representatives ?

Let T:V->W be a linear map. Let p:V->V/U be the canonical projection. Then the linear map T' induced by the given identication T'(p(x))=T(x) is an isomorphism onto the image T(V). I think we need T' to be linear for this to work.

Proof#1. Let T':V/U -> T(V) s.t. T'([x]) = T(x). Full stop. No further properties are assumed. This is a very strong constraint. Everything has to be amicable to this assumption for it to work. This is a hypothetical definition, not a '≡' identity definition. Proof#2. Let T': V/KerT -> T(V) s.t...

If you allow for the moment that T'([a])=T(a)... That there maps that do not have this property then T isn't one of them. True in general, maybe not. But this is where we are starting with by saying "let us define..." By doing this we are automatically making strong constraints on what T can be.

I think the proof#1 in the book is assuming a priori this defining. Once we start with a defining property, we're showing it must be an isomorphism. Proof#2 is using a different defining property. So Proof#1 != Proof#2.

T((1,1)) = T'([(1,1)] = T'([(0,1)]) = T((0,1)) So they have the same image under T. What does [T(x,y)] mean? The equivalence classes are not in W but in V/U.

[edit] v. T' is linear. T'(a[x]+b[y]) = T'([ax + by]) = T(ax+by) = aT(x) + bT(y) = aT'([x])+bT'([y]) then I can correct ii.One-to-one. 0=T(a)-T(b) = T(a-b) = T'([a-b])=T'([a]-[ b])=T'([a])-T'([ b]) Wait. Are you rejecting the proofs or the theorems in the book? I wasn't looking at the proofs...

The cosets {a + U | a in V} partition V into distinct equivalence classes under the equivalence relation [a] = [ b] iff a + U= b + U iff a-b is in U. If 'a' is in the coset 0 + U = U is another representation of [0], then [a]=[0]. If 'a' is not in U then there is a unique coset a+U with 'a'...

We're sending U to the equivalence class [0] in V/U. So by equivalence if a is in [0] then [a]=[0]. This means a + U = [a] = [0] = 0 + U = U, so a and 0 are both in U. T'([a]) = T'([0]) = T(0) = 0 so [a] is in the KerT'=[0] by 1-1 ( T'([a]) != T'([ b]) => T(a) != T(b) )

If c is in [a] then we have [a-c] = (a-c) + U = (a + U) - (c +U) = [a] - [c] = [a] - [a] = [0]. So it [a]=[c] then T'([a-c])=T'([0]) = T(0) = 0

I think its safe to assume the book is implicitly stating distinct cosets correspond to distinct coset representatives. So if b is anything in [a] then necessarily "[ b] = [a] " and defines an equivalence relation. T' receives the set U identified to [0] and sends it to the value T'([0])=T|0> =...

##(1,2)=c_{1}v_{1}+c_{2}v_{2}=(c_{1},c_{2})[v_{1},v_{2}]=[(1,2)]_β[v_{1},v_{2}]## ##[(1,2)]_β = (c_{1},c_{2}) = (1,2)[v_{1},v_{2}]^{-1}## ##(1,2)=d_{1}w_{1}+d_{2}w_{2}=(d_{1},d_{2})[w_{1},w_{2}]=[(1,2)]_{β'}[w_{1},w_{2}]## ##[(1,2)]_{β'} = (d_{1},d_{2}) =(1,2)[w_{1},w_{2}]^{-1} ## ##[(1,2)]_{β'}...