2025-05-25

1131: Top-Covector for Vectors Space Is Determined by Result for Ordered Basis for Vectors Space

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description/proof of that top-covector for vectors space is determined by result for ordered basis for vectors space

Topics


About: vectors space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that any top-covector for any vectors space is determined by the result for any ordered basis for the vectors space.

Orientation


There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

Main Body

1: Structured Description


Here is the rules of Structured Description.

Entities:
F: ∈{ the fields }
V: ∈{ the d -dimensional F vectors spaces }
Λd(V:F): = the top-covectors space 
t: ∈Λd(V:F)
t′: ∈Λd(V:F)
(b1,...,bd): ∈{ the bases for V}
//

Statements:
t(b1,...,bd)=t′(b1,...,bd)
⟹
t=t′
//

2: Proof


Whole Strategy: Step 1: let (v1,...,vd) be any set of vectors of V and see that t(v1,...,vd)=t′(v1,...,vd).

Step 1:

Let (v1,...,vd) be any set of vectors of V.

If t(v1,...,vd)=t′(v1,...,vd), t=t′.

So, let us see that t(v1,...,vd)=t′(v1,...,vd).

vj=vjljblj.

t(v1,...,vd)=t(v1l1bl1,...,vdldbld)=v1l1...vdldt(bl1,...,bld).

When (bl1,...,bld) is from (b1,...,bd) by a permutation, σ, t(bl1,...,bld)=sgnσt(b1,...,bd)=sgnσt′(b1,...,bd)=t′(bl1,...,bld).

Otherwise, {bl1,...,bld} is not distinct, so, t(bl1,...,bld)=0=t′(bl1,...,bld).

So, t(v1,...,vd)=v1l1...vdldt(bl1,...,bld)=v1l1...vdldt′(bl1,...,bld)=t′(v1l1bl1,...,vdldbld)=t′(v1,...,vd).

So, t=t′.

References
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