description/proof of that for linear map between finite-dimensional vectors spaces, corresponding map between components vectors spaces is linear
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% vectors space.
- The reader knows a definition of linear map.
- The reader admits the proposition that for any finite-dimensional vectors space and any basis, the vectors space is 'vectors spaces - linear morphisms' isomorphic to the components vectors space with respect to the basis.
- The reader admits the proposition that for any linear map between any modules and any linear map from any supermodule of the codomain of the 1st map into any module, the composition of the 2nd map after the 1st map is linear.
Target Context
- The reader will have a description and a proof of the proposition that for any linear map between any finite-dimensional vectors spaces and any bases, the corresponding map between the components vectors spaces is linear.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(d_1\): \(\in \mathbb{N}\)
\(d_2\): \(\in \mathbb{N}\)
\(V_1\): \(\in \{\text{ the } d_1 \text{ -dimensional } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } d_2 \text{ -dimensional } F \text{ vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
\(B_1\): \(= \{b_{1, 1}, ..., b_{1, d_1}\}\), \(\in \{\text{ the bases for } V_1\}\)
\(B_2\): \(= \{b_{2, 1}, ..., b_{2, d_2}\}\), \(\in \{\text{ the bases for } V_2\}\)
\(F^{d_1}\): \(= \text{ the } d_1 \text{ -dimensional } F \text{ vectors space }\)
\(F^{d_2}\): \(= \text{ the } d_2 \text{ -dimensional } F \text{ vectors space }\)
\(f_1\): \(: V_1 \to F^{d_1}, v^j b_{1, j} \mapsto (v^1, ..., v^{d_1})\)
\(f_2\): \(: V_2 \to F^{d_2}, v^j b_{2, j} \mapsto (v^1, ..., v^{d_2})\)
\(f'\): \(: F^{d_1} \to F^{d_2}\), \(= f_2 \circ f \circ {f_1}^{-1}\)
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Statements:
\(f' \in \{\text{ the linear maps }\}\)
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2: Note
This proposition may seem obvious, but let us prove it once for all.
3: Proof
Whole Strategy: Step 1: see that \(f_j\) is a 'vectors spaces - linear morphisms' isomorphism; Step 2: see that \(f'\) is well-defined; Step 3: conclude the proposition.
Step 1:
\(f_j\) is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that for any finite-dimensional vectors space and any basis, the vectors space is 'vectors spaces - linear morphisms' isomorphic to the components vectors space with respect to the basis.
Step 2:
So, \(f_j\) is a bijection.
So, \(f' := f_2 \circ f \circ {f_1}^{-1}\) is well-defined.
Step 3:
\(f'\) is linear, by the proposition that for any linear map between any modules and any linear map from any supermodule of the codomain of the 1st map into any module, the composition of the 2nd map after the 1st map is linear.
