description/proof of that for map from dense subset of vectors space with inner product with induced topology into same vectors space, kernel of adjoint of map is orthogonal complement of range of map
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 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.
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
\(S\): \(\in \{\text{ the dense subsets of } V\}\)
\(f\): \(: S \to V\)
\(f^*\): \(= \text{ the adjoint of } f\), \(: S^* \to V\)
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Statements:
\(Ker (f^*) = (Ran (f))^\perp\)
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2: Proof
Whole Strategy: Step 1: see that \(Ker (f^*) \subseteq (Ran (f))^\perp\); Step 2: see that \((Ran (f))^\perp \subseteq Ker (f^*)\).
Step 1:
Let \(v \in Ker (f^*)\) be any.
For each \(s \in S\), \(\langle v, f (s) \rangle = \langle f^* (v), s \rangle = \langle 0, s \rangle = 0\).
That means that \(v \in (Ran (f))^\perp\).
Step 2:
Let \(v \in (Ran (f))^\perp\) be any.
For each \(s \in S\), \(\langle v, f (s) \rangle = 0 = \langle 0, s \rangle\), which means that \(v \in S^*\) and \(f^* (v) = 0\).
So, \(v \in Ker (f^*)\).
