description/proof of that orthogonal linear map between same-finite-dimensional normed vectors spaces is 'vectors spaces - linear morphisms' isomorphism and inverse is orthogonal linear map
Topics
About: vectors space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of orthogonal linear map.
- The reader knows a definition of dimension of vectors space.
- The reader admits the proposition that any orthogonal linear map is a motion.
- The reader admits the proposition that any motion is injective.
- The reader admits the proposition that any linear injection between any same-finite-dimensional vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any orthogonal linear map between any same-finite-dimensional normed vectors spaces is a 'vectors spaces - linear morphisms' isomorphism and the inverse is an orthogonal linear map.
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:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V_1\): \(\in \{\text{ the } d \text{ -dimensional normed } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } d \text{ -dimensional normed } F \text{ vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the orthogonal linear maps }\}\)
//
Statements:
\(f \in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
\(\land\)
\(f^{-1} \in \{\text{ the orthogonal linear maps }\}\)
//
2: Natural Language Description
For any \(F \in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure, any \(d\)-dimensional normed \(F\) vectors spaces, \(V_1, V_2\), and any orthogonal linear map, \(f: V_1 \to V_2\), \(f\) is a 'vectors spaces - linear morphisms' isomorphism, and \(f^{-1}\) is an orthogonal linear map.
3: Proof
Whole Strategy: Step 1: see that \(f\) is a motion, so, \(f\) is injective, and so, \(f\) is a 'vectors spaces - linear morphisms' isomorphism; Step 2: see that for each \(v \in V_2\), \(\Vert v \Vert = \Vert f^{-1} (v) \Vert\).
Step 1:
\(f\) is a motion, by the proposition that any orthogonal linear map is a motion.
\(f\) is injective, by the proposition that any motion is injective.
\(f\) is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any linear injection between any same-finite-dimensional vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
So, \(f^{-1}\) exists.
Step 2:
Let \(v \in V_2\) be any. \(\Vert v \Vert = \Vert f^{-1} (v) \Vert\)?
\(\Vert v \Vert = \Vert f \circ f^{-1} (v) \Vert = \Vert f^{-1} (v) \Vert\). So, yes.
