description/proof of that for finite-dimensional vectors space, transition of dual bases for covectors space w.r.t. bases for vectors space is this
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of covectors (dual) space of vectors space.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space, the transition of the dual bases for the covectors space with respect to any bases for the vectors space is this.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\(V^*\): \(= \text{ the covectors space of } V\)
\(B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_1, ..., b_d\}\)
\(B'\): \(\in \{\text{ the bases for } V\}\), \(= \{b'_1, ..., b'_d\}\)
\(B^*\): \(= \text{ the dual basis of } B\), \(= \{b^1, ..., b^d\}\)
\(B'^*\): \(= \text{ the dual basis of } B'\), \(= \{b'^1, ..., b'^d\}\)
//
Statements:
\(b'_j = b_k M^k_j\)
\(\implies\)
\(b'^j = {M^{-1}}^j_k b^k\)
//
2: Proof
Whole Strategy: Step 1: let \(b'^j = N^j_k b^k\); Step 2: let the both hand sides operate on \(b_l\), and see that \(N^j_k = {M^{-1}}^j_k\).
Step 1:
Let \(b'^j = N^j_k b^k\), which is possible because \(B^*\) is a basis for \(V^*\) and \(b'^j \in V^*\).
Step 2:
\(b'^j (b_l) = N^j_k b^k (b_l) = N^j_k \delta^k_l = N^j_l\).
The matrix, \(M\), has the inverse, \(M^{-1}\), because otherwise, \(B'\) would not be linearly independent.
From \(b'_j = b_k M^k_j\), \(b'_j {M^{-1}}^j_l = b_k M^k_j {M^{-1}}^j_l = b_k \delta^k_l = b_l\).
So, \(N^j_k = b'^j (b_k) = b'^j (b'_l {M^{-1}}^l_k) = {M^{-1}}^l_k b'^j (b'_l) = {M^{-1}}^l_k \delta^j_l = {M^{-1}}^j_k\).
