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: Note
Do not wantonly generalize this proposition as though "for any linear map from any complex Euclidean vectors space with any norm into itself, the map is orthogonal if and only if the representative matrix with respect to any basis is unitary", which is not true.
This proposition holds because the complex-Euclidean norm and the canonical basis are used.
For example, let \(d = 2\), \(\mathbb{C}^d\) have any basis, \(B = \{b^1, b^2\}\), and \(f\) have the representative matrix with respect to \(B\), \(M = \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix}\); also \(B' = \{1 / 2 (b^1 + b^2), b^1 - b^2\}\) is a basis for \(\mathbb{C}^d\) and let \(\mathbb{C}^d\) have the norm with respect to \(B'\) defined in the proposition that for any real or complex vectors space, taking any basis and the sum of the absolute components for each vector is a norm.
Then, \(M\) is unitary, but \(f\) is not orthogonal, because \(f (b^1 + b^2) = b^1 - b^2\), but \(\Vert b^1 + b^2 \Vert = \Vert 2 1 / 2 (b^1 + b^2) + 0 (b^1 - b^2) \Vert = \vert 2 \vert + \vert 0 \vert = 2\) and \(\Vert b^1 - b^2 \Vert = \Vert 0 (1 / 2 (b^1 + b^2)) + 1 (b^1 - b^2) \Vert = \vert 0 \vert + \vert 1 \vert = 1\), so, \(\Vert f (b^1 + b^2) \Vert \neq \Vert b^1 + b^2 \Vert\).
On the other hand, if the representative matrix with respect to a basis of an orthogonal \(f\) is unitary, the representative matrix with respect to another basis of \(f\) may not be unitary, by the proposition that for any module with any \(2\) bases of any same finite cardinality and any module endomorphism, the transition of the endomorphism matrices with respect to the change of the bases is this and the proposition that the conjugate of a unitary matrix by an invertible matrix is not necessarily unitary.
3: 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.
