description/proof of that for complex-conjugate-linear bijection, inverse is complex-conjugate-linear
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of complex-conjugate-linear map.
- The reader knows a definition of bijection.
Target Context
- The reader will have a description and a proof of the proposition that for any complex-conjugate-linear bijection, the inverse is complex-conjugate-linear.
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
\(V_1\): \(\in \{\text{ the } \mathbb{C} \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } \mathbb{C} \text{ vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the complex-conjugate-linear bijections }\}\)
\(f^{-1}\): \(: V_2 \to V_1\), \(= \text{ the inverse of } f\)
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Statements:
\(f^{-1} \in \{\text{ the complex-conjugate-linear maps }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(f^{-1} (r v + r' v') = \overline{r} f^{-1} (v) + \overline{r'} f^{-1} (v')\).
Step 1:
Let \(v, v' \in V_2\) and \(r, r' \in \mathbb{C}\) be any.
Let us see that \(f^{-1} (r v + r' v') = \overline{r} f^{-1} (v) + \overline{r'} f^{-1} (v')\).
\(f (\overline{r} f^{-1} (v) + \overline{r'} f^{-1} (v')) = \overline{\overline{r}} f (f^{-1} (v)) + \overline{\overline{r'}} f (f^{-1} (v'))\), because \(f\) is complex-conjugate-linear, \(= r v + r' v'\).
That means that \(f^{-1} (r v + r' v') = \overline{r} f^{-1} (v) + \overline{r'} f^{-1} (v')\).
