description/proof of that for motion between same-finite-dimensional real vectors spaces with norms induced by inner products, motion is bijective
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 motion.
- The reader knows a definition of dimension of vectors space.
- The reader knows a definition of norm induced by inner product on real or complex vectors space.
- The reader knows a definition of bijection.
- The reader admits the proposition that any finite composition of motions is a motion.
- The reader admits the proposition that for any motion between any same-finite-dimensional real vectors spaces with the norms induced by any inner products that (the motion) fixes 0, the motion is an orthogonal linear map.
- 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.
- The reader admits the proposition that any finite composition of bijections is a bijection, if the codomains of the constituent bijections equal the domains of the succeeding bijections.
Target Context
- The reader will have a description and a proof of the proposition that for any motion between any same-finite-dimensional real vectors spaces with the norms induced by any inner products, the motion is bijective.
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:
\( V_1\): \(\in \{\text{ the } d \text{ -dimensional normed real vectors spaces }\}\) with the norm, \(\Vert \bullet \Vert_1\), induced by any inner product, \(\langle \bullet, \bullet \rangle_1\)
\( V_2\): \(\in \{\text{ the } d \text{ -dimensional normed real vectors spaces }\}\) with the norm, \(\Vert \bullet \Vert_2\), induced by any inner product, \(\langle \bullet, \bullet \rangle_2\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the motions }\}\)
//
Statements:
\(f \in \{\text{ the bijections }\}\)
//
2: Natural Language Description
For any \(d\)-dimensional normed real vectors space, \(V_1\), with the norm, \(\Vert \bullet \Vert_1\), induced by any inner product, \(\langle \bullet, \bullet \rangle_1\), any \(d\)-dimensional normed real vectors space, \(V_2\), with the norm, \(\Vert \bullet \Vert_2\), induced by any inner product, \(\langle \bullet, \bullet \rangle_2\), and any motion, \(f: V_1 \to V_2\), \(f\) is a bijection.
3: Proof
Whole Strategy: Step 1: for \(v_0 = f (0)\), take the map, \(f': V_2 \to V_2, v \mapsto v - v_0\), and see that \(f'\) is a bijective motion; Step 2: see that \(f' \circ f: V_1 \to V_2\) is a motion such that \(f' \circ f (0) = 0\) and see that \(f' \circ f\) is a bijection; Step 3: see that \(f = f'^{-1} \circ f' \circ f\) is a bijection.
Step 1:
Let \(v_0 = f (0)\).
Let us take the map, \(f': V_2 \to V_2, v \mapsto v - v_0\).
Let us see that \(f'\) is bijective.
For each \(v, v' \in V_2\) such that \(v \neq v'\), \(f' (v) = v - v_0 \neq v' - v_0 = f' (v')\). For each \(v \in V_2\), \(f' (v + v_0) = v + v_0 - v_0 = v\).
Let us see that \(f'\) is a motion.
For each \(v, v' \in V_2\), \(\Vert v - v' \Vert_2 = \Vert v - v_0 - (v' - v_0) \Vert_2 = \Vert f' (v) - f' (v') \Vert_2\).
Step 2:
\(f' \circ f: V_1 \to V_2\) is a motion, by the proposition that any finite composition of motions is a motion.
\(f' \circ f\) is injective, by the proposition that any motion is injective.
\(f' \circ f (0) = f' (v_0) = v_0 - v_0 = 0\).
\(f' \circ f\) is an orthogonal linear map, by the proposition that for any motion between any same-finite-dimensional real vectors spaces with the norms induced by any inner products that (the motion) fixes 0, the motion is an orthogonal linear map.
\(f' \circ f\) 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, especially a bijection.
Step 3:
\(f = f'^{-1} \circ f' \circ f\) is a bijection, by the proposition that any finite composition of bijections is a bijection, if the codomains of the constituent bijections equal the domains of the succeeding bijections.
