description/proof of a way to construct complex-conjugate-linear map from linear map
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of linear map.
- The reader knows a definition of complex-conjugate-linear map.
- The reader knows a definition of basis of module.
- The reader admits the proposition that any vectors space has a basis.
- The reader admits the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
Target Context
- The reader will have a description and a proof of the proposition that a way to construct a complex-conjugate-linear map from any linear map, which depends on the basis for the domain, holds.
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:
\(\mathbb{C}\): with the canonical field structure
\(V_1\): \(\in \{\text{ the } \mathbb{C} \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } \mathbb{C} \text{ vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
\(B\): \(\in \{\text{ the bases for } V_1\}\), \(= \{b_j \vert j \in J\}\), where \(J\) is any possibly uncountable index set
\(f'\): \(: V_1 \to V_2, v = \sum_{j \in J_v} v^j b_j \mapsto \sum_{j \in J_v} \overline{v^j} f (b_j)\)
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Statements:
\(f' \in \{\text{ the complex-conjugate-linear maps }\}\)
\(\land\)
\(f \neq 0\) \(\implies\) \(f'\) depends on \(B\)
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2: Note
You cannot define \(f'\) like for each \(v \in V_1\) and each \(r \in \mathbb{C}\), \(f' (r v) = \overline{r} f (v)\), because that is not well-defined: \(f' (v) = f' (1 v) = \overline{1} f (v) = 1 f (v) = f (v)\) but \(f' (v) = f' (1 / i i v) = \overline{1 / i} f (i v) = \overline{- i} f (i v) = i f (i v) = i^2 f (v) = - f (v)\).
3: Proof
Whole Strategy: Step 1: see that \(f'\) is well-defined; Step 2: see that \(f'\) is complex-conjugate-linear; Step 3: see that \(f'\) depends on \(B\).
Step 1:
Let us see that \(f'\) is well-defined.
Let \(v \in V_1\) be any.
A basis exists, by the proposition that any vectors space has a basis.
\(v\) can be expressed as \(\sum_{j \in J_v} v^j b_j\) where \(J_v \subseteq J\) is a finite subset, by the definition of basis of module.
When \(v = 0\), \(v = \sum_{j \in J_v} v^j b_j\) implies that all the components are \(0\), because \(B\) is linearly independent.
Then, \(\sum_{j \in J_v} \overline{v^j} f (b_j)\) is uniquely determined to be \(0\).
When \(v \neq 0\), \(v = \sum_{j \in J_v} v^j b_j\) is unique with nonzero components, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
With respect to the unique decomposition, \(\sum_{j \in J_v} \overline{v^j} f (b_j)\) is uniquely determined.
Even when \(v = \sum_{j \in J_v} v^j b_j\) has some zero components, the zero terms do not influence \(\sum_{j \in J_v} \overline{v^j} f (b_j)\), so, \(\sum_{j \in J_v} \overline{v^j} f (b_j)\) is uniquely determined anyway.
So, \(f'\) is well-defined.
Step 2:
Let us see that \(f'\) is complex-conjugate-linear.
Let \(v, v' \in V_1\) and \(r, r' \in \mathbb{C}\) be any.
\(v = \sum_{j \in J_v} v^j b_j\) and \(v' = \sum_{j \in J_{v'}} v'^j b_j\).
\(f' (r v + r' v') = f' (r \sum_{j \in J_v} v^j b_j + r' \sum_{j \in J_{v'}} v'^j b_j) = f' (r (\sum_{j \in J_v \setminus J_{v'}} v^j b_j + \sum_{j \in J_v \cap J_{v'}} v^j b_j) + r' (\sum_{j \in J_{v'} \setminus J_v} v'^j b_j + \sum_{j \in J_v \cap J_{v'}} v'^j b_j)) = f' (\sum_{j \in J_v \setminus J_{v'}} r v^j b_j + \sum_{j \in J_{v'} \setminus J_v} r' v'^j b_j + \sum_{j \in J_v \cap J_{v'}} (r v^j + r' v'^j) b_j) = \sum_{j \in J_v \setminus J_{v'}} \overline{r v^j} f (b_j) + \sum_{j \in J_{v'} \setminus J_v} \overline{r' v'^j} f (b_j) + \sum_{j \in J_v \cap J_{v'}} \overline{r v^j + r' v'^j} f (b_j) = \sum_{j \in J_v \setminus J_{v'}} \overline{r v^j} f (b_j) + \sum_{j \in J_{v'} \setminus J_v} \overline{r' v'^j} f (b_j) + \sum_{j \in J_v \cap J_{v'}} (\overline{r v^j} + \overline{r' v'^j}) f (b_j) = \sum_{j \in J_v \setminus J_{v'}} \overline{r v^j} f (b_j) + \sum_{j \in J_{v'} \setminus J_v} \overline{r' v'^j} f (b_j) + \sum_{j \in J_v \cap J_{v'}} \overline{r v^j} f (b_j) + \sum_{j \in J_v \cap J_{v'}} \overline{r' v'^j} f (b_j) = \sum_{j \in J_v \setminus J_{v'}} \overline{r v^j} f (b_j) + \sum_{j \in J_v \cap J_{v'}} \overline{r v^j} f (b_j) + \sum_{j \in J_{v'} \setminus J_v} \overline{r' v'^j} f (b_j) + \sum_{j \in J_v \cap J_{v'}} \overline{r' v'^j} f (b_j) = \sum_{j \in J_v} \overline{r v^j} f (b_j) + \sum_{j \in J_{v'}} \overline{r' v'^j} f (b_j) = \overline{r} \sum_{j \in J_v} \overline{v^j} f (b_j) + \overline{r'} \sum_{j \in J_{v'}} \overline{v'^j} f (b_j) = \overline{r} f' (v) + \overline{r'} f' (v')\).
So, \(f'\) is complex-conjugate-linear.
Step 3:
Let us see that when \(f \neq 0\), \(f'\) depends on \(B\).
As \(f \neq 0\), there is a \(b_l \in B\) such that \(f (b_l) \neq 0\).
Let \(B' := (B \setminus \{b_l\}) \cup \{i b_l\}\).
\(B'\) is a basis for \(V_1\), because \(B'\) keeps being linearly independent and spanning \(V_1\).
Let us define the complex-conjugate-linear \(f'': V_1 \to V_2\) with respect to \(B'\).
\(f' (b_l) = f (b_l)\).
\(f'' (b_l) = f'' (1 / i i b_l) = \overline{1 / i} f (i b_l) = \overline{- i} f (i b_l) = i f (i b_l) = i^2 f (b_l)\), because \(f\) is linear, \(= - f (b_l) \neq f (b_l) = f' (b_l)\), because \(f (b_l) \neq 0\).
So, \(f' \neq f''\).
If \(f = 0\), \(f' = f'' = 0\).
