definition of determinant of square matrix over ring
Topics
About: matrices space
The table of contents of this article
Starting Context
- The reader knows a definition of %ring name% matrices space.
- The reader knows a definition of \(n\)-symmetric group.
Target Context
- The reader will have a definition of determinant of square matrix over ring.
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:
\( R\): \(\in \{\text{ the rings }\}\)
\( n\): \(\in \mathbb{N} \setminus \{0\}\)
\( M\): \(\in \{\text{ the } n \times n R \text{ matrices }\}\)
\( S_n\): \(= \text{ the } n \text{ -symmetric group }\)
\(*det M\): \(= \sum_{\sigma \in S_n} sgn \sigma M^1_{\sigma_1} ... M^n_{\sigma_n}\)
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Conditions:
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2: Note
When \(R\) is commutative, \(det M = \sum_{\sigma \in S_n} sgn \sigma M^{\sigma_1}_1 ... M^{\sigma_n}_n\).
Let us see the fact.
The factors of \(M^1_{\sigma_1} ... M^n_{\sigma_n}\) can be reordered to be \(M^{\sigma^{-1}_1}_1 ... M^{\sigma^{-1}_n}_n\): when \(\sigma_j = l\), \(\sigma (j) = l\), so, \(j = \sigma^{-1} (l)\), so, \(M^j_{\sigma_j} = M^{\sigma^{-1} (l)}_l\).
\(sgn \sigma^{-1} = sgn \sigma\).
So, \(det M = \sum_{\sigma \in S_n} sgn \sigma^{-1} M^{\sigma^{-1}_1}_1 ... M^{\sigma^{-1}_n}_n\).
But when \(\sigma\) goes round \(S_n\), \(\sigma^{-1}\) goes round \(S_n\), so, \(det M = \sum_{\sigma^{-1} \in S_n} sgn \sigma^{-1} M^{\sigma^{-1}_1}_1 ... M^{\sigma^{-1}_n}_n\).
That equals \(\sum_{\sigma \in S_n} sgn \sigma M^{\sigma_1}_1 ... M^{\sigma_n}_n\), because that is just changing the symbol \(\sigma^{-1}\) to \(\sigma\).
