description/proof of that for map from vectors space with inner product with induced topology into same vectors space, if map is unitary, it preserves norm and if map is linear surjection that preserves norm, it is unitary
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 knows a definition of surjection.
Target Context
- The reader will have a description and a proof of 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.
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
\(f\): \(: V \to V\)
//
Statements:
(
\(f \in \{\text{ the unitary maps }\}\)
\(\implies\)
\(\forall v \in V (\langle f (v), f (v) \rangle = \langle v, v \rangle)\)
\(\land\)
(
\(f \in \{\text{ the linear surjections }\} \land \forall v \in V (\langle f (v), f (v) \rangle = \langle v, v \rangle)\)
\(\implies\)
\(f \in \{\text{ the unitary maps }\}\)
)
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2: Proof
Whole Strategy: Step 1: suppose that \(f\) is unitary; Step 2: see that \(\langle f (v), f (v) \rangle = \langle v, v \rangle\); Step 3: suppose that \(f\) is a linear surjection and \(\langle f (v), f (v) \rangle = \langle v, v \rangle\); Step 4: see that for each \(v, v' \in V\), \(\langle f (v), f (v') \rangle = \langle v, v' \rangle\); Step 5: see that \(f\) is unitary.
Step 1:
Let us suppose that \(f\) is unitary.
Step 2:
Let \(v \in V\) be any.
\(\langle f (v), f (v) \rangle = \langle f^* \circ f (v), v \rangle = \langle id (v), v \rangle = \langle v, v \rangle\).
Step 3:
Let us suppose that \(f\) is a linear surjection and \(\forall v \in V (\langle f (v), f (v) \rangle = \langle v, v \rangle)\).
Step 4:
Let us see that for each \(v, v' \in V\), \(\langle f (v), f (v') \rangle = \langle v, v' \rangle\).
\(\langle f (v + v'), f (v + v') \rangle = \langle v + v', v + v' \rangle\).
The left hand side is \(\langle f (v) + f (v'), f (v) + f (v') \rangle = \langle f (v), f (v) \rangle + \langle f (v), f (v') \rangle + \langle f (v'), f (v) \rangle + \langle f (v'), f (v') \rangle\).
The right hand side is \(\langle v, v \rangle + \langle v, v' \rangle + \langle v', v \rangle + \langle v', v' \rangle\).
So, \(\langle f (v), f (v') \rangle + \langle f (v'), f (v) \rangle = \langle v, v' \rangle + \langle v', v \rangle\), but the left hand side is \(\langle f (v), f (v') \rangle + \overline{\langle f (v), f (v') \rangle} = 2 Re (\langle f (v), f (v') \rangle)\) and the right hand side is \(\langle v, v' \rangle + \overline{\langle v, v' \rangle} = 2 Re (\langle v, v' \rangle)\), so, \(Re (\langle f (v), f (v') \rangle) = Re (\langle v, v' \rangle)\).
When \(F = \mathbb{R}\), that already means that \(\langle f (v), f (v') \rangle = \langle v, v' \rangle\).
Let us suppose that \(F = \mathbb{C}\).
\(\langle f (i v + v'), f (i v + v') \rangle = \langle i v + v', i v + v' \rangle\).
The left hand side is \(\langle i f (v) + f (v'), i f (v) + f (v') \rangle = - \langle f (v), f (v) \rangle + i \langle f (v), f (v') \rangle - i \langle f (v'), f (v) \rangle + \langle f (v'), f (v') \rangle\).
The right hand side is \(- \langle v, v \rangle + i \langle v, v' \rangle - i \langle v', v \rangle + \langle v', v' \rangle\).
So, \(i \langle f (v), f (v') \rangle - i \langle f (v'), f (v) \rangle = i \langle v, v' \rangle - i \langle v', v \rangle\), but the left hand side is \(i \langle f (v), f (v') \rangle - i \overline{\langle f (v), f (v') \rangle} = i (\langle f (v), f (v') \rangle - \overline{\langle f (v), f (v') \rangle}) = 2 i Im (\langle f (v), f (v') \rangle) i\) and the right hand side is \(i \langle v, v' \rangle - i \overline{\langle v, v' \rangle} = i (\langle v, v' \rangle - \overline{\langle v, v' \rangle}) = 2 i Im (\langle v, v' \rangle) i\), so, \(Im (\langle f (v), f (v') \rangle) = Im (\langle v, v' \rangle)\).
So, \(\langle f (v), f (v') \rangle = \langle v, v' \rangle\).
Step 5:
As \(f\) is a surjection, for each \(v'' \in V\), there is a \(v \in V\) such that \(v'' = f (v)\).
By Step 4, for each \(v' \in V\), \(\langle v'', f (v') \rangle = \langle f (v), f (v') \rangle = \langle v, v' \rangle\), which means that the domain of \(f^*\) is \(V\) and \(f^* (f (v)) = v\), according to the definition of adjoint map of map from dense subset of vectors space with inner product with induced topology into same vectors space.
As that holds for each \(v \in V\), \(f^* \circ f = id\).
For each \(v' \in V\), \(f \circ f^* (v') = f \circ f^* (f (v)) = f (v) = v'\), which means that \(f \circ f^* = id\).
\(\forall v \in V (\langle f (v), f (v) \rangle = \langle v, v \rangle)\) means that \(f\) is bounded.
So, \(f\) is a unitary map.
