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
- The reader knows a definition of quaternion Euclidean module.
- The reader knows a definition of orthogonal subset of vectors space with inner product.
- The reader knows a definition of linearly independent subset of module.
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
\(S\): \(\in \{\text{ the orthogonal subsets of } \mathbb{H}^n\}\): \(\forall s \in S (s \neq 0) \land \forall s_1, s_2 \in S \text{ such that } s_1 \neq s_2 (s_1 {s_2}^* = 0)\)
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Statements:
\(S \in \{\text{ the linearly independent subsets of } \mathbb{H}^n\}\)
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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.
