- #1
Somali_Physicist
- 117
- 13
Moved from a technical forum, so homework template missing
The theorem is as follows:
All finite dimensional vector spaces of the same dimension are isomorphic
Attempt:
If T is a linear map defined as :
T : V →W
: dim(V) = dim(W) = x < ∞
& V,W are vector spaces
It would be sufficient to prove T is a bijective linear map:
let W := {wi}ni
like wise let : let V:= {vi}ni
let ω ∈ W & ζ ∈ V
It can be shown:
ω = ∑iwiki
ζ = ∑ivioi
The above is a result of the definition of a vector, note ki and oi are of an arbitary vector field.
now:
T(ω) = ζ
T(∑iwiki) = ∑ivioi
∑iT(wiki) = ∑ivioi
thus:
∑i(T(wiki) - vioi) = 0
therefore:(a bit iffy)
T(wiki) - vioi=0
=>
T(wiki) = vioi
T(ωi)=ζi
Hence Surjective
a linear transform is by nature injective, therefore bijective
i have a hunch i am wrong.Any advice :)
All finite dimensional vector spaces of the same dimension are isomorphic
Attempt:
If T is a linear map defined as :
T : V →W
: dim(V) = dim(W) = x < ∞
& V,W are vector spaces
It would be sufficient to prove T is a bijective linear map:
let W := {wi}ni
like wise let : let V:= {vi}ni
let ω ∈ W & ζ ∈ V
It can be shown:
ω = ∑iwiki
ζ = ∑ivioi
The above is a result of the definition of a vector, note ki and oi are of an arbitary vector field.
now:
T(ω) = ζ
T(∑iwiki) = ∑ivioi
∑iT(wiki) = ∑ivioi
thus:
∑i(T(wiki) - vioi) = 0
therefore:(a bit iffy)
T(wiki) - vioi=0
=>
T(wiki) = vioi
T(ωi)=ζi
Hence Surjective
a linear transform is by nature injective, therefore bijective
i have a hunch i am wrong.Any advice :)