description/proof of that for finite-dimensional vectors space, transition of components of vector w.r.t. change of bases is this
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% 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 components of any vector with respect to any change of bases 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(B\): \(\in \{\text{ the bases for } V\} = \{b_s \vert 1 \le s \le dim V\}\)
\(B'\): \(\in \{\text{ the bases for } V\} = \{b'_s = b_j M_s^j \vert 1 \le s \le dim V\}\)
//
Statements:
\(\forall v = v^j b_j = v'^j b'_j \in V\)
(
\(v'^j = {M^{-1}}^j_l v^l\)
)
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2: Proof
Whole Strategy: Step 1: for \(b'_j v'^j = b_j v^j\), expand \(b'_j\) with \(b_l\) s, and compare the coefficients of \(b_l\) s on the both hand sides.
Step 1:
From \(b'_j v'^j = b_j v^j\), \(b'_j v'^j = b'_l v'^l = b_j M_l^j v'^l = b_j v^j\).
So, \(M_l^j v'^l = v^j\).
So, \(v'^j = {M^{-1}}^j_l v^l\).
