Dual Space Bijection: Proving Dependence on Basis

[cummings12332]

Homework Statement

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.

2. The attempt at a solution
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?

 

[mighty2000]

Thank you for your post. I would like to respond to your inquiry.

Your proof is on the right track, but there are a few things that could be clarified.

First, it is important to note that the bijection K: V -> V* is defined as K(ei) = e^i, where ei is the basis of V and e^i is the basis of V*. This means that the bijection is defined in terms of the original choice of basis, and any change in the basis will result in a different bijection.

To prove that K is a bijection, you can show that it is both injective and surjective. To show injectivity, you can assume that K(v) = K(w) for some vectors v and w in V, and then show that this implies that v = w. To show surjectivity, you can take an arbitrary element in V* and show that it can be written as K(v) for some v in V.

Next, you can choose a different basis ei' for V and show that the bijection K is still well-defined under this new basis. This means that the value of K(ei') is uniquely determined by the basis ei', and the same holds true for any other vector in V. This will show that K is independent of the choice of basis.

In conclusion, your proof is correct, but it would be helpful to provide more details and explanations. I hope this helps clarify your understanding of the problem.