description/proof of that for principal integral domain, rectangle matrix over domain, and square matrix over domain, sum of principal ideals by specified-dimensional subdeterminants of product is contained in sum of principal ideals by same-dimensional subdeterminants of rectangle matrix
Topics
About: ring
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of principal integral domain.
Target Context
- The reader will have a description and a proof of the proposition that for any principal integral domain, any rectangle matrix over the domain, and any rectangular-matrix-columns-dimensional or rectangular-matrix-rows-dimensional square matrix over the domain, the sum of the principal ideals by the specified-dimensional subdeterminants of the product is contained in the sum of the principal ideals by the same-dimensional subdeterminants of the rectangle 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:
\(R\): \(\in \{\text{ the principal integral domains }\}\)
\(M\): \(\in \{\text{ the m x n matrices over } R\}\)
\(N\): \(\in \{\text{ the n x n matrices over } R\}\)
\(O\): \(\in \{\text{ the m x m matrices over } R\}\)
\(k\): \(\in \mathbb{N}\), \(1 \le k \le min (m, n)\)
\(I_k (M)\): \(= \text{ the sum of the principal ideals by the } k \text{ -dimensional subdeterminants of } M\)
\(I_k (M N)\): \(= \text{ the sum of the principal ideals by the } k \text{ -dimensional subdeterminants of } M N\)
\(I_k (O M)\): \(= \text{ the sum of the principal ideals by the } k \text{ -dimensional subdeterminants of } O M\)
//
Statements:
\(I_k (M N) \subseteq I_k (M)\)
\(\land\)
\(I_k (O M) \subseteq I_k (M)\)
//
As an immediate corollary, \(I_k (O M N) \subseteq I_k (M)\): \(I_k (O M N) \subseteq I_k (M N) \subseteq I_k (M)\).
2: Natural Language Description
For any principal integral domain, \(R\), any \(m x n\) matrix over \(R\), \(M\), any \(n x n\) matrix over \(R\), \(N\), any \(m x m\) matrix over \(R\), \(O\), any natural number, \(k\), such that \(1 \le k \le min (m, n)\), the sum of the principal ideals by the \(k\)-dimensional subdeterminants of \(M\), \(I_k (M)\), the sum of the principal ideals by the \(k\)-dimensional subdeterminants of \(M N\), \(I_k (M N)\), and the sum of the principal ideals by the \(k\)-dimensional subdeterminants of \(O M\), \(I_k (O M)\), \(I_k (M N) \subseteq I_k (M)\) and \(I_k (O M) \subseteq I_k (M)\).
3: Proof
Generally, for any matrix, \(A\), \(I_k (A)\) denotes the sum of the principal ideals by the \(k\) -dimensional subdeterminants of \(A\).
Hereafter, let us denote the \((j, l)\) component of any matrix, \(A\), as \(A^j_l\); the \(l\)-th column of \(A\) as \(A_l\); the submatrix of \(A\) with the \(B = \{r_1, ..., r_k\}\) rows as \(A (B)\).
\(M N = \begin{pmatrix} \sum_{j_1 \in \{1, ..., n\}} M_{j_1} N^{j_1}_1, ..., \sum_{j_m \in \{1, ..., n\}} M_{j_m} N^{j_m}_m \end{pmatrix}\).
Let us take the \(k x k\) submatrix of \(M N\) by the \(B := \{r_1, ..., r_k\}\)-th rows and the \(\{c_1, ..., c_k\}\)-columns: it is \(\begin{pmatrix} \sum_{j_1 \in \{1, ..., n\}} M (B)_{j_1} N^{j_1}_{c_1}, ..., \sum_{j_k \in \{1, ..., n\}} M (B)_{j_k} N^{j_k}_{c_k} \end{pmatrix}\).
Its determinant is \(\vert \begin{pmatrix} \sum_{j_1 \in \{1, ..., n\}} M (B)_{j_1} N^{j_1}_{c_1}, ..., \sum_{j_k \in \{1, ..., n\}} M (B)_{j_k} N^{j_k}_{c_k} \end{pmatrix} \vert = \sum_{j_1 \in \{1, ..., n\}} N^{j_1}_{c_1} \vert \begin{pmatrix} M (B)_{j_1}, ..., \sum_{j_k \in \{1, ..., n\}} M (B)_{j_k} N^{j_k}_{c_k} \end{pmatrix} \vert\) (because determinant is linear with respect to each column) \(= ... = \sum_{j_1 \in \{1, ..., n\}} ... \sum_{j_k \in \{1, ..., n\}} N^{j_1}_{c_1} ... N^{j_k}_{c_k} \vert \begin{pmatrix} M (B)_{j_1}, ..., M (B)_{j_k} \end{pmatrix} \vert\).
When \(\{j_1, ..., j_k\}\) has a duplication, the term is \(0\); otherwise, the term is a multiple of the \(k x k\) subdeterminant of \(M\).
So, each \(k x k\) subdeterminant of \(M N\) is a linear combination of some \(k x k\) subdeterminants of \(M\).
So, while each element of \(I_k (M N)\) is a linear combination of the \(k x k\) subdeterminants of \(M N\), it is a linear combination of the \(k x k\) subdeterminants of \(M\), which means that it is an element of \(I_k (M)\).
The set of the \(k x k\) subdeterminants of \(O M\) equals the set of the \(k x k\) subdeterminants of \((O M)^t\), where \((O M)^t\) is the transposition of \(O M\), because the transposition of any square matrix has the same determinant with the original matrix. \((O M)^t = M^t O^t\). \(I_k (O M) = I_k (M^t O^t) \subseteq I_k (M^t) = I_k (M)\).
