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
- The reader knows a definition of top-covectors space of finite-dimensional vectors space.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\(\Lambda_d (V: F)\): \(= \text{ the top-covectors space }\)
\(t\): \(\in \Lambda_d (V: F)\)
\(t'\): \(\in \Lambda_d (V: F)\)
\((b_1, ..., b_d)\): \(\in \{\text{ the bases for } V\}\)
//
Statements:
\(t (b_1, ..., b_d) = t' (b_1, ..., b_d)\)
\(\implies\)
\(t = t'\)
//
2: Proof
Whole Strategy: Step 1: let \((v_1, ..., v_d)\) be any set of vectors of \(V\) and see that \(t (v_1, ..., v_d) = t' (v_1, ..., v_d)\).
Step 1:
Let \((v_1, ..., v_d)\) be any set of vectors of \(V\).
If \(t (v_1, ..., v_d) = t' (v_1, ..., v_d)\), \(t = t'\).
So, let us see that \(t (v_1, ..., v_d) = t' (v_1, ..., v_d)\).
\(v_j = v^{l_j}_j b_{l_j}\).
\(t (v_1, ..., v_d) = t (v^{l_1}_1 b_{l_1}, ..., v^{l_d}_d b_{l_d}) = v^{l_1}_1 ... v^{l_d}_d t (b_{l_1}, ..., b_{l_d})\).
When \((b_{l_1}, ..., b_{l_d})\) is from \((b_1, ..., b_d)\) by a permutation, \(\sigma\), \(t (b_{l_1}, ..., b_{l_d}) = sgn \sigma t (b_1, ..., b_d) = sgn \sigma t' (b_1, ..., b_d) = t' (b_{l_1}, ..., b_{l_d})\).
Otherwise, \(\{b_{l_1}, ..., b_{l_d}\}\) is not distinct, so, \(t (b_{l_1}, ..., b_{l_d}) = 0 = t' (b_{l_1}, ..., b_{l_d})\).
So, \(t (v_1, ..., v_d) = v^{l_1}_1 ... v^{l_d}_d t (b_{l_1}, ..., b_{l_d}) = v^{l_1}_1 ... v^{l_d}_d t' (b_{l_1}, ..., b_{l_d}) = t' (v^{l_1}_1 b_{l_1}, ..., v^{l_d}_d b_{l_d}) = t' (v_1, ..., v_d)\).
So, \(t = t'\).
