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
- The reader knows a definition of determinant of square matrix over ring.
- The reader admits the proposition that for any permutation of any finite sequence, the sign of the inverse of the permutation is the sign of the permutation.
- The reader admits the proposition that for any group, the set of the inverses of the elements of the group is the group with no duplication.
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:
\(R\): \(\in \{\text{ the commutative rings }\}\)
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(M\): \(\in \{\text{ the } n \times n R \text{ matrices }\}\)
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Statements:
\(det M^t = det M\)
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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: \(\sigma^{- 1} = \sigma' \in S_n\).
\(= det M\).
