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description/proof of that for 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 orthogonal, 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
Target Context
-
The reader will have a description and a proof of 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 \(1\).
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:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean vectors space with the Euclidean inner product }\), regarded as the set of the rows
\(S\): \(= \{b^1, ..., b^k\} \in \{\text{ the orthonormal subsets of } \mathbb{R}^d\}\)
\(S'\): \(= \{b'^1, ..., b'^k\} \in \{\text{ the orthonormal subsets of } \mathbb{R}^d\}\)
//
Statements:
\(\exists f: \mathbb{R}^d \to \mathbb{R}^d, \in \{\text{ the orthogonal linear maps }\} (\forall j \in \{1, ..., k\} (f (b^j) = b'^j) \land M, \text{ the representative matrix of } f \text{ with respect to the canonical basis of } \mathbb{R}^d, \in \{\text{ the orthogonal matrices }\})\)
\(\land\)
(
\(k \lt d\)
\(\implies\)
\(f \text{ can be chosen such that } det M = 1\)
)
//
2: Note
Also 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 \(1\) 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, \(f: \mathbb{R}^d \to \mathbb{R}^d, v = v_j b^j \mapsto v_j b'^j\); Step 2: see that \(f\) is orthogonal and that the canonical representative matrix is orthogonal; Step 3: suppose that \(k \lt d\); Step 4: see that \(det M = - 1 \text{ or } 1\) and take \(\widetilde{B'} = (b'^1, ..., b'^k, b'^{k + 1}, ..., (det M)^{- 1} b'^d)\), 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 \(\mathbb{R}^d\), 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, \(B' = (b'^1, ..., b'^k, b'^{k + 1}, ..., b'^d)\) for \(\mathbb{R}^d\), which is possible, likewise.
Let us define the map, \(f: \mathbb{R}^d \to \mathbb{R}^d, v = v_j b^j \mapsto v_j b'^j\), 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 \(v = v_j b^j \in \mathbb{R}^d\), \(\Vert f (v) \Vert = \Vert f (v_j b^j) \Vert = \Vert v_j b'^j \Vert = \sqrt{\langle v_j b'^j, v_l b'^l \rangle} = \sqrt{v_j v_l \langle b'^j, b'^l \rangle} = \sqrt{v_j v_l \delta^{j, l}} = \sqrt{v_j v_l \langle b^j, b^l \rangle} = \sqrt{\langle v_j b^j, v_l b^l \rangle} = \Vert v_j b^j \Vert = \Vert v \Vert\).
So, its canonical representative matrix, \(M\), is orthogonal, by the proposition that for any linear map from any Euclidean-normed Euclidean vectors space into itself, the map is orthogonal if and only if the canonical representative matrix is orthogonal.
Step 3:
Let us suppose that \(k \lt d\).
Step 4:
\(det M = - 1 \text{ or } 1\), 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: any orthogonal matrix is unitary and its determinant is real.
When \(det M = 1\), there is nothing more to do; the following is for when \(det M = - 1\).
Let us take \(\widetilde{B'} = (\widetilde{b'}^1 = b'^1, ..., \widetilde{b'}^k = b'^k, \widetilde{b'}^{k + 1} = b'^{k + 1}, ..., \widetilde{b'}^d = (det M)^{- 1} b'^d)\) instead of \(B'\).
\(\widetilde{B'}\) is orthonormal, because for each \(j \lt d\), \(\langle b'^j, (det M)^{- 1} b'^d \rangle = (det M)^{- 1} \langle b'^j, b'^d \rangle = (det M)^{- 1} 0 = 0\) and \(\langle (det M)^{- 1} b'^d, (det M)^{- 1} b'^d \rangle = (det M)^{- 1} (det M)^{- 1} \langle b'^d, b'^d \rangle = 1 1 = 1\).
Let us define \(\widetilde{f}: \mathbb{R}^d \to \mathbb{R}^d, v = v_j b^j \mapsto v_j \widetilde{b'}^j\).
By Step 2, \(\widetilde{f}\) is represented by an orthogonal \(\widetilde{M}\).
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 \(\widetilde{N'}^j_d = N'^j_d det M\), because \(\widetilde{N'}^j_l \widetilde{b'}^l = \sum_{l \in \{1 ..., d - 1\}} \widetilde{N'}^j_l \widetilde{b'}^l + \widetilde{N'}^j_d \widetilde{b'}^d = \sum_{l \in \{1 ..., d - 1\}} \widetilde{N'}^j_l b'^l + \widetilde{N'}^j_d (det M)^{- 1} b'^d = e^j = N'^j_l b'^l = \sum_{l \in \{1 ..., d - 1\}} N'^j_l b'^l + N'^j_d b'^d\), which implies that \(\widetilde{N'}^j_l = N'^j_l b'^l\) for each \(l \in \{1, ..., d - 1\}\) and \(\widetilde{N'}^j_d (det M)^{- 1} = N'^j_d\), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
\(det \widetilde{N'} = det N' det M\), 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, \(det \widetilde{N'}^{- 1} = (det \widetilde{N'})^{- 1} = (det N' det M)^{- 1} = (det M)^{- 1} det N'^{- 1}\).
So, \(det \widetilde{M} = det (N I \widetilde{N'}^{- 1}) = det N det I det \widetilde{N'}^{- 1} = det N det I (det M)^{- 1} det N'^{- 1} = det N det I det N'^{- 1} (det M)^{- 1} = det (N I N'^{- 1}) (det M)^{- 1} = det M (det M)^{- 1} = 1\).
So, we can take \(\widetilde{B'}\), then, \(\widetilde{f}\) is an orthogonal linear map that maps \(S\) to \(S'\) whose representative matrix is orthogonal with determinant \(1\).
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for quaternion Euclidean module and \(2\) same-cardinality orthonormal subsets, there is orthogonal linear map that maps one of subsets to other whose canonical representative matrix is symplectic
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any quaternion Euclidean module 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 symplectic.
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:
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{H}^n\): \(= \text{ the quaternion Euclidean module }\), regarded as the set of the rows
\(S\): \(= \{b^1, ..., b^k\} \in \{\text{ the orthonormal subsets of } \mathbb{H}^n\}\): \(\forall b^j, b^l \in S (b^j {b^l}^* = \delta^j_l)\)
\(S'\): \(= \{b'^1, ..., b'^k\} \in \{\text{ the orthonormal subsets of } \mathbb{H}^n\}\) \(\forall b'^j, b'^l \in S' (b'^j {b'^l}^* = \delta^j_l)\)
//
Statements:
\(\exists f: \mathbb{H}^n \to \mathbb{H}^n, \in \{\text{ the orthogonal linear maps }\} (\forall j \in \{1, ..., k\} (f (b^j) = b'^j) \land M, \text{ the representative matrix of } f \text{ with respect to the canonical basis of } \mathbb{H}^d, \in \{\text{ the symplectic matrices }\})\)
//
"being an orthogonal linear map" here means that \(\forall m \in \mathbb{H}^n (f (m) f (m)^* = m m^*)\).
2: Note
In fact, "orthonormal subset" has been really defined only for any vectors space with any inner product, which \(\mathbb{H}^n\) is not, so, "orthonormal subset" is an abuse of the term.
Likewise, "orthogonal linear map" has been really defined only between any normed vectors spaces, which \(\mathbb{H}^n\) is not, so, "orthogonal linear map" is an abuse of the term.
We dare to abuse the terms, because descriptions without the terms would be cumbersome and the terms will be guessed what they mean.
3: Proof
Whole Strategy: Step 1: take any orthonormal bases, \(B = (b^1, ..., b^k, b^{k + 1}, ..., b^n)\) and \(B' = (b'^1, ..., b'^k, b'^{k + 1}, ..., b'^n)\), and define the linear map, \(f: \mathbb{H}^n \to \mathbb{H}^n, v = v_j b^j \mapsto v_j b'^j\); Step 2: see that \(f\) is orthogonal and that the canonical representative matrix is symplectic.
Step 1:
Let us take any orthonormal basis, \(B = (b^1, ..., b^k, b^{k + 1}, ..., b^n)\) of \(\mathbb{H}^n\), which is possible, by the proposition that for any quaternion Euclidean module, any orthogonal subset is linearly independent, the proposition that for any module over any division ring with any finite basis, any linearly independent subset can be supplemented with some elements of the basis to be a basis, and the definition of Gram-Schmidt orthonormalization of countable subset of quaternion Euclidean module.
Let us take any orthonormal basis, \(B' = (b'^1, ..., b'^k, b'^{k + 1}, ..., b'^n)\) of \(\mathbb{H}^n\), which is possible, likewise.
Let us define the map, \(f: \mathbb{H}^d \to \mathbb{H}^d, m = m_j b^j \mapsto m_j b'^j\), which is linear, because for each \(m_1 = {m_1}_j b^j, m_2 = {m_2}_j b^j \in \mathbb{H}^d\) and each \(r_1, r_2 \in \mathbb{H}\), \(f (r_1 m_1 + r_2 m_2) = f (r_1 {m_1}_j b^j + r_2 {m_2}_j b^j) = f ((r_1 {m_1}_j + r_2 {m_2}_j) b^j) = (r_1 {m_1}_j + r_2 {m_2}_j) b'^j = r_1 ({m_1}_j b'^j) + r_2 ({m_2}_j b'^j) = r_1 f (m_1) + r_2 f (m_2)\)
\(f (m) = m M\), by the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_1 \times d_2\) ring matrix: that is with respect to the canonical basis of \(\mathbb{H}^n\).
Step 2:
\(f\) is orthogonal, because for each \(m = m_j b^j \in \mathbb{H}^n\), \(f (m) f (m)^* = f (m_j b^j) f (m_l b^l)^* = (m_j b'^j) (m_l b'^l)^* = (m_j b'^j) ({b'^l}^* {m_l}^*)\), by the proposition that the quaternion-conjugate-transpose of the product of any quaternion matrices is the product of the quaternion-conjugate-transposes of the constituents in the reverse order, \(= m_j (b'^j {b'^l}^*) {m_l}^*= m_j \delta^j_l {m_l}^*\), because \(B'\) is orthonormal, \(= m_j (b^j {b^l}^*) {m_l}^*\), because \(B\) is orthonormal, \(= (m_j b^j) ({b^l}^* {m_l}^*) = (m_j b^j) (m_l b^l)^*\), by the proposition that the quaternion-conjugate-transpose of the product of any quaternion matrices is the product of the quaternion-conjugate-transposes of the constituents in the reverse order, \(= m m^*\).
So, \(m m^* = f (m) f (m)^* = (m M) (m M)^* = (m M) M^* m^*\), by the proposition that the quaternion-conjugate-transpose of the product of any quaternion matrices is the product of the quaternion-conjugate-transposes of the constituents in the reverse order, \(= m (M M^*) m^*\).
\(M M^* = I\), by the proposition that for any \(n \times n\) quaternion matrix, if and only if for each \(n\)-column-or-row element, the matrix multiplied by the element and the quaternion-conjugate-transpose of the element is the element multiplied by the quaternion-conjugate-transpose of the element, the matrix is identity.
That means that \(M\) is symplectic: refer to Note for the definition of \(n \times n\) symplectic matrices group.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for quaternion Euclidean module, orthogonal subset is linearly independent
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any quaternion Euclidean module, any orthogonal subset is linearly independent.
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:
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{H}^n\): \(= \text{ the quaternion Euclidean module }\), regarded as the set of the rows
\(S\): \(\in \{\text{ the orthogonal subsets of } \mathbb{H}^n\}\): \(\forall s \in S (s \neq 0) \land \forall s_1, s_2 \in S \text{ such that } s_1 \neq s_2 (s_1 {s_2}^* = 0)\)
//
Statements:
\(S \in \{\text{ the linearly independent subsets of } \mathbb{H}^n\}\)
//
2: Note
In fact, "orthogonal subset" has been really defined only for any vectors space with any inner product, which \(\mathbb{H}^n\) is not, so, "orthogonal subset" is an abuse of the term.
We dare to abuse the term, because descriptions without the term would be cumbersome and the term will be guessed what it means.
3: Proof
Whole Strategy: Step 1: take any finite subset, \(S^` = \{s^1, ..., s^m\} \subseteq S\), let \(h_1 s^1 + ... + h_m s^m = 0\), and see that all the \(h_j\) s are \(0\).
Step 1:
Let \(S^` = \{s^1, ..., s^m\} \subseteq S\) be any finite subset.
Let \(h_1 s^1 + ... + h_m s^m = 0\) where \(h_1, ..., h_m \in \mathbb{H}\) are any.
For each \(j \in \{1, ..., m\}\), \((h_1 s^1 + ... + h_m s^m) {s^j}^* = 0 {s^j}^* = 0\), but the left hand side is \(h_1 s^1 {s^j}^* + ... + h_m s^m {s^j}^* = h_j s^j {s^j}^*\) where \(s^j {s^j}^* \neq 0\), because \(S\) is orthogonal, so, \(h_j (s^j {s^j}^*) = 0\), and \(h_j (s^j {s^j}^*) (s^j {s^j}^*)^{- 1} = 0 (s^j {s^j}^*)^{- 1}\), which is possible, because \(\mathbb{H}\) is a division ring, so, \(h_j = h_j 1 = 0\).
So, \(S\) is linearly independent.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of canonical representative matrix of linear map between finite-product-of-copies-of-ring modules
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of canonical representative matrix of linear map between finite-product-of-copies-of-ring modules.
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:
\( R\): \(\in \{\text{ the rings }\}\)
\( n_1\): \(\in \mathbb{N} \setminus \{0\}\)
\( n_2\): \(\in \mathbb{N} \setminus \{0\}\)
\( R^{n_1}\): \(= \text{ the } R \text{ module with the canonical } n_1 \text{ -elements basis }\)
\( R^{n_2}\): \(= \text{ the } R \text{ module with the canonical } n_2 \text{ -elements basis }\)
\( B_1\): \(= \{b_1^1, ..., b_1^{n_1}\}\), \(= \text{ the canonical basis of } R^{n_1}\)
\( B_2\): \(= \{b_2^1, ..., b_2^{n_2}\}\), \(= \text{ the canonical basis of } R^{n_2}\)
\( f\): \(: R^{n_1} \to R^{n_2}\), \(\in \{\text{ the linear maps }\}\)
\(*M\): \(\in \{\text{ the } n_1 \times n_2 R \text{ matrices }\}\), such that \(M^j_l = f ({b_1}^j)_l\)
//
Conditions:
//
"the canonical basis" means that \({b_m}^j = (0, ..., 0, 1, 0, ..., 0)\) where \(1\) is the \(j\)-th component.
2: Note
\(R^{n_m}\) is indeed an \(R\) module with the canonical basis, by the proposition that any ring is canonically a module with a \(1\)-element basis and the proposition that any finite product module of modules with bases has this basis.
\(M\) indeed represents \(f\), by the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_1 \times d_2\) ring matrix.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
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
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 \(1\).
References
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description/proof of that for square complex matrix, determinant of Hermitian conjugate of matrix is complex conjugate of determinant of matrix
Topics
About:
field
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of 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.
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:
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(M\): \(\in \{\text{ the } n \times n \mathbb{C} \text{ matrices }\}\)
//
Statements:
\(det M^* = \overline{det M}\)
//
2: Note
As an immediate corollary, for any unitary \(M\), \(\vert det M \vert = 1\), because as \(M^* M = I\), \(det (M^* M) = det I = 1\), but the left hand side is \(det M^* det M = \overline{det M} det M = \vert det M \vert^2\).
3: Proof
Whole Strategy: Step 1: see that \(det M^* = \overline{\sum_{\sigma \in S_n} sgn (\sigma) M^{\sigma_1}_1 ... M^{\sigma_n}_n} = \overline{det M^t} = \overline{det M}\).
Step 1:
\(det M^* = \sum_{\sigma \in S_n} sgn (\sigma) {M^*}^1_{\sigma_1} ... {M^*}^n_{\sigma_n}\), by the definition of determinant, \(= \sum_{\sigma \in S_n} sgn (\sigma) \overline{M^{\sigma_1}_1} ... \overline{M^{\sigma_n}_n} = \overline{\sum_{\sigma \in S_n} sgn (\sigma) M^{\sigma_1}_1 ... M^{\sigma_n}_n} = \overline{\sum_{\sigma \in S_n} sgn (\sigma) {M^t}^1_{\sigma_1} ... {M^t}^n_{\sigma_n}} = \overline{det M^t}\).
\(= \overline{det M}\), by the proposition that for any square commutative ring matrix, the determinant of the transpose of the matrix is the determinant of the matrix.
References
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description/proof of that for square commutative ring matrix, determinant of transpose of matrix is determinant of matrix
Topics
About:
ring
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any square commutative ring matrix, the determinant of the transpose of the matrix is the determinant of the matrix.
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:
\(R\): \(\in \{\text{ the commutative rings }\}\)
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(M\): \(\in \{\text{ the } n \times n R \text{ matrices }\}\)
//
Statements:
\(det M^t = det M\)
//
2: Proof
Whole Strategy: Step 1: see that \(det M^t = \sum_{\sigma \in S_n} sgn (\sigma) M^{\sigma_1}_1 ... M^{\sigma_n}_n = \sum_{\sigma \in S_n} sgn (\sigma) M^1_{\sigma^{- 1}_1} ... M^n_{\sigma^{- 1}_n} = \sum_{\sigma^{- 1} \in S_n} sgn (\sigma^{- 1}) M^1_{\sigma^{- 1}_1} ... M^n_{\sigma^{- 1}_n} = det M\).
Step 1:
\(det M^t = \sum_{\sigma \in S_n} sgn (\sigma) {M^t}^1_{\sigma_1} ... {M^t}^n_{\sigma_n}\), by the definition of determinant, \(= \sum_{\sigma \in S_n} sgn (\sigma) M^{\sigma_1}_1 ... M^{\sigma_n}_n\).
As \((\sigma_1, ..., \sigma_n)\) is a permutation result of \((1, ..., n)\), \(M^{\sigma_1}_1 ... M^{\sigma_n}_n\) can be reordered to be \(M^1_{j_1} ... M^n_{j_n}\), which is possible, because \(R\) is commutative.
For each \(M^l_{j_l}\), \(M^l_{j_l} = M^{\sigma_m}_m\) for an \(m \in \{1, ..., n\}\), so, \(l = \sigma_m = \sigma (m)\) and \(j_l = m\), so, \(j_l = m = \sigma^{- 1} (l) = \sigma^{- 1}_l\), so, \(M^l_{j_l} = M^l_{\sigma^{- 1}_l}\).
So, \(det M^t = \sum_{\sigma \in S_n} sgn (\sigma) M^1_{\sigma^{- 1}_1} ... M^n_{\sigma^{- 1}_n} = \sum_{\sigma \in S_n} sgn (\sigma^{- 1}) M^1_{\sigma^{- 1}_1} ... M^n_{\sigma^{- 1}_n}\), by the proposition that for any permutation of any finite sequence, the sign of the inverse of the permutation is the sign of the permutation.
\(= \sum_{\sigma^{- 1} \in S_n} sgn (\sigma^{- 1}) M^1_{\sigma^{- 1}_1} ... M^n_{\sigma^{- 1}_n}\), by the proposition that for any group, the set of the inverses of the elements of the group is the group with no duplication: for each \(\sigma \in S_n\), indeed \(\sigma^{- 1} \in S_n\), and as \(\sigma\) goes round \(S_n\), \(\sigma^{- 1}\) goes round \(S_n\).
\(= \sum_{\sigma' \in S_n} sgn (\sigma') M^1_{\sigma'_1} ... M^n_{\sigma'_n}\), by the proposition that for any group, the set of the inverses of the elements of the group is the group with no duplication: \(\sigma^{- 1} = \sigma' \in S_n\).
\(= det M\).
References
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description/proof of that for group, set of inverses of elements is group with no duplication
Topics
About:
group
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any group, the set of the inverses of the elements of the group is the group with no duplication.
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:
\(G\): \(\in \{\text{ the groups }\}\), \(= \{g_j \vert j \in J\}\), where \(J\) is any index set, with no duplication
\(G^{- 1}\): \(= \{{g_j}^{- 1} \vert j \in J\}\)
//
Statements:
\(G^{- 1} = G\)
\(\land\)
\(\forall j, l \in J \text{ such that } j \neq l ({g_j}^{- 1} \neq {g_l}^{- 1})\)
//
2: Proof
Whole Strategy: Step 1: see that \(G^{- 1} \subseteq G\) and \(G \subseteq G^{- 1}\); Step 2: see that \(G^{- 1}\) has no duplication.
Step 1:
\(G^{- 1} \subseteq G\), because by the definition of group, the inverse of each element is contained in the group.
\(G \subseteq G^{- 1}\), because for each \(g \in G\), \(g = {g^{- 1}}^{- 1}\), and as \(g^{- 1} \in G\), \({g^{- 1}}^{- 1} \in G^{- 1}\).
So, \(G^{- 1} = G\).
Step 2:
For each \(j, l \in J\) such that \(j \neq l\), \({g_j}^{- 1} \neq {g_l}^{- 1}\), because if \({g_j}^{- 1} = {g_l}^{- 1}\), \({{g_j}^{- 1}}^{- 1} = {{g_l}^{- 1}}^{- 1}\), but the left hand side was \(g_j\) and the right hand side was \(g_l\), and so, \(g_j = g_l\), which would mean that \(j = l\), a contradiction.
That means that \(G^{- 1}\) has no duplication.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for permutation of finite sequence, sign of inverse of permutation is sign of permutation
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any permutation of any finite sequence, the sign of the inverse of the permutation is the sign of the permutation.
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:
\(S\): \(\in \{\text{ the finite subsets of } \mathbb{N}\}\)
\(f\): \(\in \{\text{ the sequences from } S\}\)
\(\sigma\): \(: S \to S\), \(\in \{\text{ the permutations of } f\}\)
//
Statements:
\(\sigma^{- 1} \in \{\text{ the permutations of } f\}\)
\(\land\)
\(sgn (\sigma) = sgn (\sigma^{- 1})\)
//
2: Proof
Whole Strategy: Step 1: see that \(\sigma^{- 1}\) is a bijection; Step 2: see that there is a bijection from \(\{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma^{- 1} (s_2) \lt \sigma^{- 1} (s_1)\}\) to \(\{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma (s_2) \lt \sigma (s_1)\}\).
Step 1:
\(\sigma\) has indeed \(\sigma^{- 1}: S \to S\), by the proposition that any map is a bijection if and only if it has an inverse.
\(\sigma^{- 1}\) is a bijection, by the proposition that the inverse of any bijection is a bijection.
So, \(\sigma^{- 1}\) is a permutation.
Step 2:
\(sgn (\sigma^{- 1})\) is determined by the cardinality of \(\{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma^{- 1} (s_2) \lt \sigma^{- 1} (s_1)\}\) and \(sgn (\sigma)\) is determined by the cardinality of \(\{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma (s_2) \lt \sigma (s_1)\}\).
Let us see that \(f: \{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma^{- 1} (s_2) \lt \sigma^{- 1} (s_1)\} \to \{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma (s_2) \lt \sigma (s_1)\}, (s_1, s_2) \mapsto (\sigma^{- 1} (s_2), \sigma^{- 1} (s_1))\) is a bijection.
\(f\) is indeed into \(\{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma (s_2) \lt \sigma (s_1)\}\), because \(\sigma^{- 1} (s_2) \lt \sigma^{- 1} (s_1)\) and \((\sigma (\sigma^{- 1} (s_1)) = s_1 \lt s_2 = \sigma (\sigma^{- 1} (s_2))\).
\(f\) is an injection, obviously.
\(f\) is a surjection, because for each \((s_1, s_2) \in \{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma (s_2) \lt \sigma (s_1)\}\), \((\sigma (s_2), \sigma (s_1)) \in \{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma^{- 1} (s_2) \lt \sigma^{- 1} (s_1)\}\), because \(\sigma (s_2) \lt \sigma (s_1)\) and \(\sigma^{- 1} (\sigma (s_1)) = s_1 \lt s_2 = \sigma^{- 1} (\sigma (s_2))\), and \(f ((\sigma (s_2), \sigma (s_1))) = (\sigma^{- 1} (\sigma (s_1)), \sigma^{- 1} (\sigma (s_2))) = (s_1, s_2)\).
So, \(f\) is a bijection.
So, \(\vert \{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma^{- 1} (s_2) \lt \sigma^{- 1} (s_1)\} \vert = \vert \{(s_1, s_2) \in \{\text{ the ordered pairs of } S\} \vert s_1 \lt s_2 \land \sigma (s_2) \lt \sigma (s_1)\} \vert\).
That implies that \(sgn (\sigma) = sgn (\sigma^{- 1})\).
References
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