# Thread: dual space bijection

1. ## dual space bijection

let V be finite -dimentional and T:V->V*(V* is the dual space of V with same dimension as V) ,let ei be the bases of V,e^i be the bases of V*,consider the linear bijection :K:V->V* defined K(ei)=e^i,show that this bijection depends on the original choice of basis.

My proof is :
choose ei to be the bases first,then prove K under this bases is bijetion.then choose ei' to be the bases, and then prove K under the new bases is bijection , is my proof right or wrong??? can someone give the ideas?

2. ## Re: dual space bijection

As it's phrased, it's not even clear that they want you to prove that it's a bijection. It sounds like you can assume that's already known.
What they do want you to prove is that the linear map (bijection) K between V and V* as they've defined it would have been a different map if it had been defined using a different basis.

This is actually an important observation. It means that although V and V* are isomorphic vector spaces, they aren't "naturally" isomorphic. There is no "canonical" isomorphism between them. Every isomorphism between them will be described "in terms of coordinates" (i.e. referring to some basis). "Natural" means that the construction of the isomophism doesn't happen in a nice generalizable & universal way, but rather depends on these vector space specific picking activities of choosing a basis; thus many advanced math constructions that rely on "natural" isomorphisms become more complicated, or fail outright, for V and V*.

What's interesting is that V and (V*)* ARE naturally/canonically/coordinate-free isomorphic for finite dimensional vector spaces.

Anyhow, for this problem, the point is to have two different bases for V, and then follow the description to get two K's, say K and K', and then show that they actually are different. Two functions are different if they disagree on some element ion the domain, so what you'll show is that K(v) is not equal to K'(v) in V*. But how are elements of V* different? When they disagree, as linear functionals, when applied to some vector w in V.

Thus you'll need to show that there exists v, w in V such that two base-field values K(v)(w) and K'(v)(w) are diffferent.

I imagine your curious how to choose two different bases at this level of abstraction. Here's one way:
Let Basis1 = {e1, e2, e3, ..., en}
Let Basis2 = {b1, b2, b3, ..., bn}, where b1 = e1+e2, and bi = ei for i >=2 (that is: b2 = e2, b3 = e3, ..., bn = en).
Of course, you'd need to prove that Basis2 actually is a basis, assuming that Basis1 is a basis - but that's easy enough.
Now define K as your problem described using Basis1. Define K' as your problem described using Basis2.
Then show that there exists v, w in V such that K(v)(w) is not equal to K'(v)(w) (both are images of linear functionals, so are values in the V's base field).