description/proof of that for bijective 'normed vectors space' isometry, inverse is 'normed vectors space' isometry
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of 'normed vectors space' isometry.
- The reader knows a definition of bijection.
Target Context
- The reader will have a description and a proof of the proposition that for any bijective 'normed vectors space' isometry, the inverse is a 'normed vectors space' isometry.
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
\(V_1\): \(\in \{\text{ the } F \text{ normed vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ normed vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the bijective 'normed vectors space' isometries }\}\)
\(f^{-1}\): \(: V_2 \to V_1\), \(= \text{ the inverse of } f\)
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Statements:
\(f^{-1} \in \{\text{ the 'normed vectors space' isometries }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(\Vert f^{-1} (v) \Vert = \Vert v \Vert\).
Step 1:
Let \(v \in V_2\) be any.
\(\Vert f^{-1} (v) \Vert = \Vert f \circ f^{-1} (v) \Vert\), because \(f\) is isometric, \(= \Vert v \Vert\).
