description/proof of that for unitary map from vectors space with inner product with induced topology into same vectors space, adjoint of map is unitary and double adjoint of map is map
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of unitary map from vectors space with inner product with induced topology into same vectors space.
- The reader admits the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
- The reader admits the proposition that for any map from any vectors space with any inner product with the induced topology into the same vectors space, if the map is unitary, it preserves norm and if the map is any linear surjection that preserves norm, it is unitary.
Target Context
- The reader will have a description and a proof of the proposition that for any unitary map from any vectors space with any inner product with the induced topology into the same vectors space, the adjoint of the map is unitary and the double adjoint of the map is 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:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), 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 norm induced by the inner product
\(f\): \(: V \to V\), \(\in \{\text{ the unitary maps }\}\)
\(f^*\): \(\text{ = the adjoint of } f\)
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Statements:
\(f^* \in \{\text{ the unitary maps }\}\)
\(\land\)
\({f^*}^* = f\)
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2: Proof
Whole Strategy: Step 1: see that \(\langle f^* (v), f^* (v) \rangle = \langle v, v \rangle\) and conclude that \(f^*\) is unitary; Step 2: see that \({f^*}^*\) is unitary and \({f^*}^* = {f^*}^{-1} = {f^{-1}}^{-1} = f\).
Step 1:
\(f^* \circ f = f \circ f^* = id\) means that \(f^* = f^{-1}\).
\(f^* = f^{-1}\) is a linear surjection, by the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
\(\langle f^* (v), f^* (v) \rangle = \langle f^{-1} (v), f^{-1} (v) \rangle\).
By the proposition that for any map from any vectors space with any inner product with the induced topology into the same vectors space, if the map is unitary, it preserves norm and if the map is any linear surjection that preserves norm, it is unitary, \(\langle f (f^{-1} (v)), f (f^{-1} (v)) \rangle = \langle f^{-1} (v), f^{-1} (v) \rangle\).
But the left hand side is \(\langle v, v \rangle\), so, \(\langle f^* (v), f^* (v) \rangle = \langle v, v \rangle\).
By the proposition that for any map from any vectors space with any inner product with the induced topology into the same vectors space, if the map is unitary, it preserves norm and if the map is any linear surjection that preserves norm, it is unitary, \(f^*\) is unitary.
Step 2:
By Step 1, \({f^*}^*\) is unitary.
So, \({f^*}^* \circ f^* = f^* \circ {f^*}^* = id\), which means that \({f^*}^* = {f^*}^{-1}\). But as \(f^* = f^{-1}\), \({f^*}^{-1} = {f^{-1}}^{-1} = f\).
