description/proof of that for real or complex vectors space, inner product can be chosen
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 real or complex vectors space, an inner product can be chosen.
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
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
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Statements:
\(\exists \langle \bullet, \bullet \rangle: V \times V \to F \in \{\text{ the inner products for } V\}\)
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2: Note
Of course, the inner product produced in this proposition is not the unique option.
This is not about choosing an inner product that induces a predetermined norm, which is not always possible.
3: Proof
Whole Strategy: present an inner product; Step 1: take any basis for \(V\), \(B = \{b_j\}\); Step 2: define \(\langle v' = 0 + \sum_{j \in J} v'^j b_j, v = 0 + \sum_{l \in L} v^l b_l \rangle\) as \(0 + \sum_{(j, l) \in J \times L} v'^j \overline{v^l} \delta_{j, l}\); Step 3: see that it is an inner product.
Step 1:
Let us take any basis for \(V\), \(B = \{b_j\}\), which is possible, by the proposition that any vectors space has a basis.
Step 2:
For each \(v, v' \in V\), \(v = 0 + \sum_{l \in L} v^l b_l\) and \(v' = 0 + \sum_{j \in J} v'^j b_j\), which is the unique decompositions with the coefficients restricted to be nonzero, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique: "\(0 +\)" is there for when \(J\) or \(L\) is empty.
Let us define \(\langle v', v \rangle := 0 + \sum_{(j, l) \in J \times L} v'^j \overline{v^l} \delta_{j, l}\): "\(0 +\)" is there for when \(J\) or \(L\) is empty.
That is well-defined, because the decompositions are unique.
Note that that formula is valid even when some coefficients are allowed to be zero, because the zero coefficients produce only zero terms, which is used later.
Step 3:
Let us see that it is indeed an inner product.
Let \(v_1, v_2, v_3 \in V\) and \(r_1, r_2 \in F\) be any.
1) \((0 \le \langle v_1, v_1 \rangle)\) \(\land\) \((0 = \langle v_1, v_1 \rangle \iff v_1 = 0)\): \(\langle v_1, v_1 \rangle = 0 + \sum_{(j, l) \in J \times J} {v_1}^j \overline{{v_1}^l} \delta_{j, l} = 0 + \sum_{j \in J} \vert {v_1}^j \vert^2\), so, \(0 \le \langle v_1, v_1 \rangle\); suppose that \(v_1 = 0\), then, \(\langle v_1, v_1 \rangle = 0 + \sum_{j \in J} \vert {v_1}^j \vert^2 = 0\) (in fact, \(J = \emptyset\)); suppose that \(0 + \sum_{j \in J} \vert {v_1}^j \vert^2 = 0\), then, \(J = \emptyset\), which means that \(v_1 = 0\).
2) \(\langle v_1, v_2 \rangle = \overline{\langle v_2, v_1 \rangle}\): \(\langle v_1, v_2 \rangle = 0 + \sum_{(j, l) \in J \times L} {v_1}^j \overline{{v_2}^l} \delta_{j, l} = 0 + \overline{\sum_{(j, l) \in J \times L} \overline{{v_1}^j} {v_2}^l \delta_{j, l}} = \overline{0 + \sum_{(l, j) \in L \times J} {v_2}^l \overline{{v_1}^j} \delta_{l, j}} = \overline{\langle v_2, v_1 \rangle}\).
3) \(\langle r_1 v_1 + r_2 v_2, v_3 \rangle = r_1 \langle v_1, v_3 \rangle + r_2 \langle v_2, v_3 \rangle\): \(\langle r_1 v_1 + r_2 v_2, v_3 \rangle = \langle r_1 (0 + \sum_{j_1 \in J_1} {v_1}^{j_1} b_{j_1}) + r_2 (0 + \sum_{j_2 \in J_2} {v_2}^{j_2} b_{j_2}), 0 + \sum_{l \in L} {v_3}^l b_l \rangle\), which can be expressed as \(\langle r_1 (0 + \sum_{j \in J} {v_1}^j b_j) + r_2 (0 + \sum_{j \in J} {v_2}^j b_j), 0 + \sum_{l \in L} {v_3}^l b_l \rangle\) where \(J := J_1 \cup J_2\) with some coefficients possibly zero, \(= \langle 0 + \sum_{j \in J} r_1 {v_1}^j b_j + 0 + \sum_{j \in J} r_2 {v_2}^j b_j, 0 + \sum_{l \in L} {v_3}^l b_l \rangle = \langle 0 + \sum_{j \in J} (r_1 {v_1}^j + r_2 {v_2}^j) b_j, 0 + \sum_{l \in L} {v_3}^l b_l \rangle = 0 + \sum_{(j, l) \in J \times L} (r_1 {v_1}^j + r_2 {v_2}^j) \overline{{v_3}^l} \delta_{j, l} = 0 + \sum_{(j, l) \in J \times L} (r_1 {v_1}^j \overline{{v_3}^l} \delta_{j, l} + r_2 {v_2}^j \overline{{v_3}^l} \delta_{j, l}) = 0 + \sum_{(j, l) \in J \times L} (r_1 {v_1}^j \overline{{v_3}^l} \delta_{j, l}) + 0 + \sum_{(j, l) \in J \times L} (r_2 {v_2}^j \overline{v^l} \delta_{j, l}) = 0 + r_1 \sum_{(j, l) \in J \times L} ({v_1}^j \overline{{v_3}^l} \delta_{j, l}) + 0 + r_2 \sum_{(j, l) \in J \times L} ({v_2}^j \overline{v^l} \delta_{j, l}) = r_1 \langle v_1, v_3 \rangle + r_2 \langle v_2, v_3 \rangle\).
So, it is an inner product.
