Understanding Linear Maps and Exterior Algebras

[Diffy]

Homework Statement

Please, I just am trying to understand the question. I wish to prove it on my own, but the way the question is phrased makes no sense.

So here it is:

Let us define the linear map
$$\phi : V^{*} \otimes \bigwedge^{i} V \rightarrow \bigwedge^{i-1} V$$

by the formula

$$\ell \otimes v_1 \wedge ... \wedge v_i \mapsto \sum_{s=1}^{i} (-1)^{s-1} \ell (v_s) v_1 \wedge ... \wedge \hat{v_s} \wedge ... \wedge v_s$$

Prove that the map $$\phi$$ is well defined and does not depend on the choice of basis.

Homework Equations

Well all the usual definition of exterior algebras, and tensor products are needed.

The Attempt at a Solution

As I stated, I haven't started solving yet, I am simply trying to understand the question. I don't see how it goes to wedge i-1. What exactly is v hat sub s? Does that make i wedges?
I don't think this formula is going to i-1 wedges. Please help me to understand what is going on here.

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Dick]

What they are trying to notate by v hat sub s, is that the vector v_s is NOT included in the wedge product. All the other v_i's are. That's why it's an i-1 form.

 

[Diffy]

Now it makes sense!

Now I just have to solve the problem. Any hints are appreciated.

 

[lurflurf]

Well there are two main ways to prove something is invariant
1 show it does not change with form
ie
let x and x' be different forms of x
if
f(x)=f(x')
f is invariant

2 provide an invariant definition
ie
define
f(x)
so that the form of x is not a factor

Since you have been provided a noninvariant definition
let Av be a change of basis
you want to show
l(v1)^v2^...^vn=l(Av1)^Av2^...^Avn=det(a)*l(v1)^v2^...^vn
with det(A)=1