description/proof of that for linear map from complex-Euclidean-normed complex Euclidean vectors space into itself, map is orthogonal iff canonical representative matrix is unitary
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of complex-Euclidean-normed complex Euclidean vectors space.
- The reader knows a definition of orthogonal linear map.
- The reader knows a definition of canonical representative matrix of linear map between finite-product-of-copies-of-field vectors spaces.
- The reader knows a definition of unitary matrix.
- The reader admits the proposition that for any complex polynomial with any finite number of variables and their complex conjugates, if the polynomial is constantly \(0\), the coefficients are \(0\).
Target Context
- The reader will have a description and a proof of the proposition that for any linear map from any complex-Euclidean-normed complex Euclidean vectors space into itself, the map is orthogonal if and only if the canonical representative matrix 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:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(f\): \(: \mathbb{C}^d \to \mathbb{C}^d\), \(\in \{\text{ the linear maps }\}\)
\(M\): \(= \text{ the canonical representative matrix of } f\)
//
Statements:
\(f \in \{\text{ the orthogonal maps }\}\)
\(\iff\)
\(M \in \{\text{ the unitary matrices }\}\)
//
2: Proof
Whole Strategy: apply the proposition that for any complex polynomial with any finite number of variables and their complex conjugates, if the polynomial is constantly \(0\), the coefficients are \(0\); Step 1: suppose that \(f\) is an orthogonal map; Step 2: see that \(M\) is a unitary matrix; Step 3: suppose that \(M\) is a unitary matrix; Step 4: see that \(f\) is an orthogonal map.
Step 1:
Let us suppose that \(f\) is an orthogonal map.
Step 2:
Let us see that \(M\) is a unitary matrix.
Let \(v \in \mathbb{C}^d\) be any.
\(\Vert f (v) \Vert^2 = (M v^t)^t \overline{M v^t}\), by the definition of complex Euclidean norm, \(= v M^t \overline{M} \overline{v^t} = \Vert v \Vert^2\), because \(f\) is orthogonal, but \(= v \overline{v^t} = \sum_{j \in \{1, ..., d\}} v^j \overline{v^j} = \sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} v^l \delta^l_j \overline{v^j}\).
Let \(N := M^t \overline{M}\).
\(v N \overline{v^t} = \sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} v^l N^l_j \overline{v^j}\).
So, \(\sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} v^l N^l_j \overline{v^j} = \sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} v^l \delta^l_j \overline{v^j}\), and \(\sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} (v^l N^l_j \overline{v^j} - v^l \delta^l_j \overline{v^j}) = 0\), so, \(\sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} (N^l_j - \delta^l_j) v^l \overline{v^j} = 0\).
That holds constantly with respect to \((v^1, ..., v^d)\), so, by the proposition that for any complex polynomial with any finite number of variables and their complex conjugates, if the polynomial is constantly \(0\), the coefficients are \(0\), \(N^l_j - \delta^l_j = 0\), so, \(N^l_j = \delta^l_j\).
So, \(M^t \overline{M} = I\).
\(\overline{M^t \overline{M}} = \overline{I} = I\), but the left hand side is \(\overline{M^t} M\), so, \(M^* M = \overline{M^t} M = I\).
So, \(M\) is a unitary matrix.
Step 3:
Let us suppose that \(M\) is a unitary matrix.
Step 4:
Let \(v \in \mathbb{C}^d\) be any.
\(\Vert f (v) \Vert^2 = \Vert M v^t \Vert^2 = (M v^t)^t \overline{M v^t} = v M^t \overline{M} \overline{v^t} = v \overline{\overline{M^t} M} \overline{v^t} = v \overline{M^* M} \overline{v^t} = v \overline{I} \overline{v^t} = v I \overline{v^t} = v \overline{v^t} = \Vert v \Vert^2\).
So, \(f\) is an orthogonal map.
