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
- The reader knows a definition of quaternion Euclidean module.
- 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-ring modules.
- The reader knows a definition of \(n \times n\) symplectic matrices Topological group.
- The reader knows a definition of Gram-Schmidt orthonormalization of countable subset of quaternion Euclidean module.
- The reader admits the proposition that for any quaternion Euclidean module, any orthogonal subset is linearly independent.
- The reader admits 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.
- The reader admits 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.
- The reader admits 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.
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
\(S\): \(= \{b^1, ..., b^k\} \in \{\text{ the orthonormal subsets of } \mathbb{H}^n\}\): \(\forall b^j, b^l \in S (b^j {b^l}^* = \delta^j_l)\)
\(S'\): \(= \{b'^1, ..., b'^k\} \in \{\text{ the orthonormal subsets of } \mathbb{H}^n\}\) \(\forall b'^j, b'^l \in S' (b'^j {b'^l}^* = \delta^j_l)\)
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Statements:
\(\exists f: \mathbb{H}^n \to \mathbb{H}^n, \in \{\text{ the orthogonal linear maps }\} (\forall j \in \{1, ..., k\} (f (b^j) = b'^j) \land M, \text{ the representative matrix of } f \text{ with respect to the canonical basis of } \mathbb{H}^d, \in \{\text{ the symplectic matrices }\})\)
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"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.
