description/proof of that for motion between real vectors spaces with norms induced by inner products that fixes 0, orthonormal subset of domain is mapped to orthonormal subset
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
- 4: Note
Starting Context
- The reader knows a definition of motion.
- The reader knows a definition of norm induced by inner product on real or complex vectors space.
- The reader admits the proposition that any motion is injective.
Target Context
- The reader will have a description and a proof of the proposition that for any motion between any real vectors spaces with the norms induced by any inner products that (the motion) fixes 0, any orthonormal subset of the domain is mapped to an orthonormal subset.
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:
\( V_1\): \(\in \{\text{ the normed real vectors spaces }\}\) with the norm, \(\Vert \bullet \Vert_1\), induced by any inner product, \(\langle \bullet, \bullet \rangle_1\)
\( V_2\): \(\in \{\text{ the normed real vectors spaces }\}\) with the norm, \(\Vert \bullet \Vert_2\), induced by any inner product, \(\langle \bullet, \bullet \rangle_2\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the motions }\}\)
\(S\): \(\subseteq V_1\)
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Statements:
\(f (0) = 0 \land \forall v, v' \in S ((v = v' \implies \langle v, v' \rangle_1 = 1) \land (v \neq v' \implies \langle v, v' \rangle_1 = 0))\)
\(\implies\)
\(\forall v, v' \in f (S) ((v = v' \implies \langle v, v' \rangle_2 = 1) \land (v \neq v' \implies \langle v, v' \rangle_2 = 0))\)
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2: Natural Language Description
For any normed real vectors space, \(V_1\), with the norm, \(\Vert \bullet \Vert_1\), induced by any inner product, \(\langle \bullet, \bullet \rangle_1\), any normed real vectors space, \(V_2\), with the norm, \(\Vert \bullet \Vert_2\), induced by any inner product, \(\langle \bullet, \bullet \rangle_2\), any motion, \(f: V_1 \to V_2\), such that \(f (0) = 0\), and any subset, \(S \subseteq V_1\), such that \(\forall v, v' \in S ((v = v' \implies \langle v, v' \rangle_1 = 1) \land (v \neq v' \implies \langle v, v' \rangle_1 = 0))\), \(f (S)\) satisfies that \(\forall v, v' \in f (S) ((v = v' \implies \langle v, v' \rangle_2 = 1) \land (v \neq v' \implies \langle v, v' \rangle_2 = 0))\).
3: Proof
Whole Strategy: Step 1: for each \(v, v' \in V_j\), see that \(\Vert v - v' \Vert_j^2 = \Vert v \Vert_j^2 + \Vert v' \Vert_j^2 - 2 \langle v, v' \rangle\); Step 2: using that equation, see that \(\langle v, v' \rangle_1 = \langle f (v), f (v') \rangle_2\); Step 3: for the equation of Step 2, let \(v, v' \in S\).
Step 1:
Let \(v, v' \in V_j\) be any.
\(\Vert v - v' \Vert_j^2 = \langle v - v', v - v' \rangle_j = \langle v, v - v' \rangle_j - \langle v', v - v' \rangle_j = \langle v, v \rangle_j - \langle v, v' \rangle_j - \langle v', v \rangle_j + \langle v', v' \rangle_j = \Vert v \Vert_j^2 + \Vert v' \Vert_j^2 - 2 \langle v, v' \rangle_j\). Note that we requires \(V_j\) to be a real vectors space in order to have that equation.
Step 2:
Putting \(v, v' \in V_1\) in that equation, we get \(\Vert v - v' \Vert_1^2 = \Vert v \Vert_1^2 + \Vert v' \Vert_1^2 - 2 \langle v, v' \rangle_1\).
Putting \(f (v), f (v') \in V_2\) in that equation, we get \(\Vert f (v) - f (v') \Vert_2^2 = \Vert f (v) \Vert_2^2 + \Vert f (v') \Vert_2^2 - 2 \langle f (v), f (v') \rangle_2\).
But as \(\Vert v - v' \Vert_1 = \Vert f (v) - f (v') \Vert_2\), \(\Vert v \Vert_1 = \Vert v - 0 \Vert_1 = \Vert f (v) - f (0) \Vert_2 = \Vert f (v) - 0 \Vert_2 = \Vert f (v) \Vert_2\), and likewise, \(\Vert v' \Vert_1 = \Vert f (v') \Vert_2\), \(\Vert v \Vert_1^2 + \Vert v' \Vert_1^2 - 2 \langle v, v' \rangle_1 = \Vert f (v) \Vert_2^2 + \Vert f (v') \Vert_2^2 - 2 \langle f (v), f (v') \rangle_2\) and \(- 2 \langle v, v' \rangle_1 = - 2 \langle f (v), f (v') \rangle_2\), so, \(\langle v, v' \rangle_1 = \langle f (v), f (v') \rangle_2\).
Step 3:
For each \(f (v), f (v') \in f (S) \subseteq V_2\), \(\langle f (v), f (v') \rangle_2 = \langle v, v' \rangle_1\). As \(f\) is injective by the proposition that any motion is injective, when \(f (v) = f (v')\), \(v = v'\), and when \(f (v) \neq f (v')\), \(v \neq v'\). So, when \(f (v) = f (v')\), \(\langle f (v), f (v') \rangle_2 = \langle v, v' \rangle_1 = 1\), and when \(f (v) \neq f (v')\), \(\langle f (v), f (v') \rangle_2 = \langle v, v' \rangle_1 = 0\).
4: Note
\(f\) is not supposed to be linear.
