description/proof of that for self-adjoint linear map from dense subspace of vectors space with inner product with induced topology into same vectors space, if map is surjective, map is bijective
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of adjoint map of map from dense subset of vectors space with inner product with induced topology into same vectors space.
- The reader knows a definition of linear map.
- The reader knows a definition of surjection.
- The reader knows a definition of bijection.
- The reader admits the proposition that for any map from any dense subset of any vectors space with any inner product with the induced topology into the same vectors space, the kernel of the adjoint of the map is the orthogonal complement of the range of the map.
- The reader admits the proposition that any linear map is injective if and only if its kernel is the \(0\)-subspace.
Target Context
- The reader will have a description and a proof of the proposition that for any self-adjoint linear map from any dense subspace of any vectors space with any inner product with the induced topology into the same vectors space, if the map is surjective, the map is bijective.
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:
\(F\): \(\in \{\mathbb{R}, \mathbb{H}\}\), with the canonical field structure
\(V'\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with any inner product, with the topology induced by the metric induced by the inner product
\(V\): \(\in \{\text{ the dense subspaces of } V'\}\)
\(f\): \(: V \to V'\), such that \(f^* = f\)
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Statements:
\(f \in \{\text{ the surjections }\}\)
\(\implies\)
\(f \in \{\text{ the bijections }\}\)
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2: Note
\(V\) needs to be a vectors subspace, not just a subset, because otherwise, being linear would not make sense.
3: Proof
Whole Strategy: Step 1: see that \(Ker (f^*) = (Ran (f))^\perp\); Step 2: see that \(Ker (f^*) = Ker (f)\) and \((Ran (f))^\perp = \{0\}\); Step 3: conclude the proposition.
Step 1:
\(Ker (f^*) = (Ran (f))^\perp\), by the proposition that for any map from any dense subset of any vectors space with any inner product with the induced topology into the same vectors space, the kernel of the adjoint of the map is the orthogonal complement of the range of the map.
Step 2:
\(Ker (f^*) = Ker (f)\), by the supposition.
\((Ran (f))^\perp = \{0\}\), because \((Ran (f))^\perp = V^\perp\), and for each \(v \in V^\perp\), for each \(v' \in V\), \(\langle v, v' \rangle = 0\), especially, \(\langle v, v \rangle = 0\), which implies that \(v = 0\).
Step 3:
So, \(Ker (f) = \{0\}\).
So, \(f\) is injective, by the proposition that any linear map is injective if and only if its kernel is the \(0\)-subspace.
So, \(f\) is bijective.
