description/proof of that for complex Euclidean vectors space with canonical inner product and \(2\) same-cardinality orthonormal subsets, there is orthogonal linear map that maps one of subsets to other whose canonical representative matrix is unitary, and when cardinality is smaller than dimension, determinant can be made \(1\)
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of complex Euclidean inner product on 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 knows a definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
- The reader admits the proposition that for any vectors space with any inner product, any orthonormal subset is linearly independent.
- The reader admits the proposition that for any finite-dimensional vectors space, any linearly independent subset can be expanded to be a basis by adding a finite number of elements.
- The reader admits the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism.
- The reader admits 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.
- The reader admits the proposition that for any square complex matrix, the determinant of the Hermitian conjugate of the matrix is the complex conjugate of the determinant of the matrix.
- The reader admits the proposition that for any linear map between any modules with any finite bases, the transition of the map representative matrices with respect to changes of bases with the bases-cardinalities conserved is this.
Target Context
- The reader will have a description and a proof of the proposition that for any complex Euclidean vectors space with the complex Euclidean inner product and any \(2\) same-cardinality orthonormal subsets, there is an orthogonal linear map that maps any one of the subsets to the other whose canonical representative matrix is unitary, and when the cardinality is smaller than the dimension, the determinant can be made \(1\).
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:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{C}^d\): \(= \text{ the complex Euclidean vectors space with the complex Euclidean inner product }\), regarded as the set of the rows
\(S\): \(= \{b^1, ..., b^k\} \in \{\text{ the orthonormal subsets of } \mathbb{C}^d\}\)
\(S'\): \(= \{b'^1, ..., b'^k\} \in \{\text{ the orthonormal subsets of } \mathbb{C}^d\}\)
//
Statements:
\(\exists f: \mathbb{C}^d \to \mathbb{C}^d, \in \{\text{ the orthogonal linear maps }\} (\forall j \in \{1, ..., k\} (f (b^j) = b'^j) \land M, \text{ the representative matrix of } f \text{ with respect to the canonical basis of } \mathbb{C}^d, \in \{\text{ the unitary matrices }\})\)
\(\land\)
(
\(k \lt d\)
\(\implies\)
\(f \text{ can be chosen such that } det M = 1\)
)
//
2: Note
Also the proposition that for any Euclidean vectors space with the Euclidean inner product and any \(2\) same-cardinality orthonormal subsets, there is an orthogonal linear map that maps any one of the subsets to the other whose canonical representative matrix is orthogonal, and when the cardinality is smaller than the dimension, the determinant can be made \(1\) holds.
The reason why "\(k \lt d\)" is required is that it gives the freedom to adjust \(f\): when \(k = d\), \(f\) is uniquely determined.
3: Proof
Whole Strategy: Step 1: take any orthonormal bases, \(B = (b^1, ..., b^k, b^{k + 1}, ..., b^d)\) and \(B' = (b'^1, ..., b'^k, b'^{k + 1}, ..., b'^d)\), and define the linear map, \(f: \mathbb{C}^d \to \mathbb{C}^d, v = v_j b^j \mapsto v_j b'^j\); Step 2: see that \(f\) is orthogonal and that the canonical representative matrix is unitary; Step 3: suppose that \(k \lt d\); Step 4: see that \(det M = e^{i \theta}\) and take \(\widetilde{B'} = (b'^1, ..., b'^k, b'^{k + 1}, ..., e^{- i \theta} b'^d)\), and see that \(det M = 1\) with respect to \(B\) and \(\widetilde{B'}\).
Step 1:
Let us take any orthonormal basis, \(B = (b^1, ..., b^k, b^{k + 1}, ..., b^d)\) of \(\mathbb{C}^d\), which is possible, by the proposition that for any vectors space with any inner product, any orthonormal subset is linearly independent, the proposition that for any finite-dimensional vectors space, any linearly independent subset can be expanded to be a basis by adding a finite number of elements, and the definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
Let us take any orthonormal basis, \(B' = (b'^1, ..., b'^k, b'^{k + 1}, ..., b'^d)\) of \(\mathbb{C}^d\), which is possible, likewise.
Let us define the map, \(f: \mathbb{C}^d \to \mathbb{C}^d, v = v_j b^j \mapsto v_j b'^j\), which is linear (in fact, a 'vectors spaces - linear morphisms' isomorphism), by the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism: \(f\) maps \(b_j\) to \(b'_j\) and is expanding the mapping between the bases linearly.
Step 2:
\(f\) is orthogonal, because for each \(v = v_j b^j \in \mathbb{C}^d\), \(\Vert f (v) \Vert = \Vert f (v_j b^j) \Vert = \Vert v_j b'^j \Vert = \sqrt{\langle v_j b'^j, v_l b'^l \rangle} = \sqrt{v_j \overline{v_l} \langle b'^j, b'^l \rangle} = \sqrt{v_j \overline{v_l} \delta^{j, l}} = \sqrt{v_j \overline{v_l} \langle b^j, b^l \rangle} = \sqrt{\langle v_j b^j, v_l b^l \rangle} = \Vert v_j b^j \Vert = \Vert v \Vert\).
So, its canonical representative matrix, \(M\), is unitary, by 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.
Step 3:
Let us suppose that \(k \lt d\).
Step 4:
\(det M = e^{i \theta}\) where \(\theta \in \mathbb{R}\), by Note for the proposition that for any square complex matrix, the determinant of the Hermitian conjugate of the matrix is the complex conjugate of the determinant of the matrix.
Let us take \(\widetilde{B'} = (\widetilde{b'}^1 = b'^1, ..., \widetilde{b'}^k = b'^k, \widetilde{b'}^{k + 1} = b'^{k + 1}, ..., \widetilde{b'}^d = e^{- i \theta} b'^d)\) instead of \(B'\).
\(\widetilde{B'}\) is orthonormal, because for each \(j \lt d\), \(\langle b'^j, e^{- i \theta} b'^d \rangle = \overline{e^{- i \theta}} \langle b'^j, b'^d \rangle = \overline{e^{- i \theta}} 0 = 0\) and \(\langle e^{- i \theta} b'^d, e^{- i \theta} b'^d \rangle = e^{- i \theta} \overline{e^{- i \theta}} \langle b'^d, b'^d \rangle = 1 1 = 1\).
Let us define \(\widetilde{f}: \mathbb{C}^d \to \mathbb{C}^d, v = v_j b^j \mapsto v_j \widetilde{b'}^j\).
By Step 2, \(\widetilde{f}\) is represented by a unitary \(\widetilde{M}\).
By the proposition that for any linear map between any modules with any finite bases, the transition of the map representative matrices with respect to changes of bases with the bases-cardinalities conserved is this, \(M = N I N'^{- 1}\) and \(\widetilde{M} = N I \widetilde{N'}^{- 1}\), where \(e^j = N'^j_l b'^l\) and \(e^j = \widetilde{N'}^j_l \widetilde{b'}^l\), where \(\{e^1, ..., e^d\}\) is the canonical basis.
In fact, for each \(l \in \{1, ..., d - 1\}\), \(\widetilde{N'}^j_l = N'^j_l\) and \(\widetilde{N'}^j_d = N'^j_d e^{i \theta}\), because \(\widetilde{N'}^j_l \widetilde{b'}^l = \sum_{l \in \{1 ..., d - 1\}} \widetilde{N'}^j_l \widetilde{b'}^l + \widetilde{N'}^j_d \widetilde{b'}^d = \sum_{l \in \{1 ..., d - 1\}} \widetilde{N'}^j_l b'^l + \widetilde{N'}^j_d e^{- i \theta} b'^d = e^j = N'^j_l b'^l = \sum_{l \in \{1 ..., d - 1\}} N'^j_l b'^l + N'^j_d b'^d\), which implies that \(\widetilde{N'}^j_l = N'^j_l b'^l\) for each \(l \in \{1, ..., d - 1\}\) and \(\widetilde{N'}^j_d e^{- i \theta} = N'^j_d\), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
\(det \widetilde{N'} = det N' e^{i \theta}\), as is well known.
As \(1 = det I = det (N'^{- 1} N') = det N'^{- 1} det N'\), \(det N'^{- 1} = (det N')^{- 1}\); \(det \widetilde{N'}^{- 1} = (det \widetilde{N'})^{- 1}\), likewise.
So, \(det \widetilde{N'}^{- 1} = (det \widetilde{N'})^{- 1} = (det N' e^{i \theta})^{- 1} = e^{- i \theta} det N'^{- 1}\).
So, \(det \widetilde{M} = det (N I \widetilde{N'}^{- 1}) = det N det I det \widetilde{N'}^{- 1} = det N det I e^{- i \theta} det N'^{- 1} = det N det I det N'^{- 1} e^{- i \theta} = det (N I N'^{- 1}) e^{- i \theta} = det M e^{- i \theta} = e^{i \theta} e^{- i \theta} = 1\).
So, we can take \(\widetilde{B'}\), then, \(\widetilde{f}\) is an orthogonal linear map that maps \(S\) to \(S'\) whose representative matrix is unitary with determinant \(1\).
