description/proof of that for normed vectors space, if norm is induced by inner product, inner product that induces norm is unique as this
Topics
About: vectors space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any normed vectors space, if the norm is induced by any inner product, the inner product that induces the norm is unique as this.
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
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with any norm, \(\Vert \bullet \Vert\)
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Statements:
\(\Vert \bullet \Vert \in \{\text{ the norms induced by any inner products }\}\)
\(\implies\)
\(\Vert \bullet \Vert\) is induced by \(\langle v_1, v_2 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v_1 + i^j v_2 \Vert^2\)
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2: Proof
Whole Strategy: Step 1: suppose that \(\Vert \bullet \Vert\) is induced by any inner product, \(\langle \bullet, \bullet \rangle\); Step 2: see that \(\langle v_1, v_2 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v_1 + i^j v_2 \Vert^2\).
Step 1:
Let us suppose that \(\Vert \bullet \Vert\) is induced by any inner product, \(\langle \bullet, \bullet \rangle\).
Step 2:
Let us see that for each \(v_1, v_2 \in V\), \(\langle v_1, v_2 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v_1 + i^j v_2 \Vert^2\).
\(1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v_1 + i^j v_2 \Vert^2 = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \langle v_1 + i^j v_2, v_1 + i^j v_2 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j (\langle v_1, v_1 + i^j v_2 \rangle + i^j \langle v_2, v_1 + i^j v_2 \rangle) = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j (\langle v_1, v_1 \rangle + \overline{i^j} \langle v_1, v_2 \rangle + i^j \langle v_2, v_1 \rangle + i^j \overline{i^j} \langle v_2, v_2 \rangle)\).
\(= 1 / 4 (1 (\langle v_1, v_1 \rangle + \overline{1} \langle v_1, v_2 \rangle + 1 \langle v_2, v_1 \rangle + 1 \overline{1} \langle v_2, v_2 \rangle) + i (\langle v_1, v_1 \rangle + \overline{i} \langle v_1, v_2 \rangle + i \langle v_2, v_1 \rangle + i \overline{i} \langle v_2, v_2 \rangle) + (-1) (\langle v_1, v_1 \rangle + \overline{-1} \langle v_1, v_2 \rangle + (-1) \langle v_2, v_1 \rangle + (-1) \overline{-1} \langle v_2, v_2 \rangle) + (- i) (\langle v_1, v_1 \rangle + \overline{- i} \langle v_1, v_2 \rangle + (- i) \langle v_2, v_1 \rangle + (- i) \overline{- i} \langle v_2, v_2 \rangle)) = 1 / 4 ((\langle v_1, v_1 \rangle + \langle v_1, v_2 \rangle + 1 \langle v_2, v_1 \rangle + \langle v_2, v_2 \rangle) + i (\langle v_1, v_1 \rangle + (- i) \langle v_1, v_2 \rangle + i \langle v_2, v_1 \rangle + i (- i) \langle v_2, v_2 \rangle) + (-1) (\langle v_1, v_1 \rangle + (-1) \langle v_1, v_2 \rangle + (-1) \langle v_2, v_1 \rangle + (-1) (-1) \langle v_2, v_2 \rangle) + (- i) (\langle v_1, v_1 \rangle + i \langle v_1, v_2 \rangle + (- i) \langle v_2, v_1 \rangle + (- i) i \langle v_2, v_2 \rangle)) = 1 / 4 ((\langle v_1, v_1 \rangle + \langle v_1, v_2 \rangle + \langle v_2, v_1 \rangle + \langle v_2, v_2 \rangle) + i (\langle v_1, v_1 \rangle - i \langle v_1, v_2 \rangle + i \langle v_2, v_1 \rangle + \langle v_2, v_2 \rangle) + (-1) (\langle v_1, v_1 \rangle - \langle v_1, v_2 \rangle -1 \langle v_2, v_1 \rangle + \langle v_2, v_2 \rangle) - i (\langle v_1, v_1 \rangle + i \langle v_1, v_2 \rangle - i \langle v_2, v_1 \rangle + \langle v_2, v_2 \rangle)) = 1 / 4 (\langle v_1, v_2 \rangle + \langle v_1, v_2 \rangle + \langle v_1, v_2 \rangle + \langle v_1, v_2 \rangle) = \langle v_1, v_2 \rangle\).
So, \(\langle v_1, v_2 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v_1 + i^j v_2 \Vert^2\).
So, the inner product is uniquely determined, because the norm is determined.
