description/proof of for commutative ring, some expansions of determinant of product of rectangular matrices
Topics
About: matrices space
The table of contents of this article
Starting Context
- The reader knows a definition of %ring name% matrices space.
Target Context
- The reader will have a description and a proof of the proposition that for any commutative ring, some expansions of the determinant of the product of any rectangular matrices whose product is square hold.
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:
\(R\): \(\in \{\text{ the commutative rings }\}\)
\(M\): \(\in \{\text{ the } m \times n R \text{ matrices }\}\)
\(N\): \(\in \{\text{ the } n \times m R \text{ matrices }\}\)
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Statements:
\(det (M N) = \sum_{l_1 \in \{1, ..., n\}} ... \sum_{l_m \in \{1, ..., n\}} M^1_{l_1} ... M^m_{l_m} det \begin{pmatrix} N^{l_1}_1 & ... & N^{l_1}_m \\ ... \\ N^{l_m}_1 & ... & N^{l_m}_m \end{pmatrix}\)
\(\land\)
\(det (M N) = \sum_{l_1 \in \{1, ..., n\}} ... \sum_{l_m \in \{1, ..., n\}} N^{l_1}_1 ... N^{l_m}_m det \begin{pmatrix} M^1_{l_1} & ... & M^1_{l_m} \\ ... \\ M^m_{l_1} & ... & M^m_{l_m} \end{pmatrix}\)
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2: Note
The ring needs to be commutative, because "well-known properties of determinant" used in Proof requires it.
The expansions claim that when \(n \lt m\), the determinant is inevitably \(0\), because each \(det \begin{pmatrix} N^{l_1}_1 & ... & N^{l_1}_m \\ ... \\ N^{l_m}_1 & ... & N^{l_m}_m \end{pmatrix}\) cannot be nonzero because some rows are inevitably duplicated.
For example, when \(m = 2\) and \(n = 1\), \(M = \begin{pmatrix} M^1_1 \\ M^2_1 \end{pmatrix}\) and \(N = \begin{pmatrix} N^1_1 & N^1_2 \end{pmatrix}\), and \(det (M N) = det \begin{pmatrix} M^1_1 N^1_1 & M^1_1 N^1_2 \\ M^2_1 N^1_1 & M^2_1 N^1_2 \end{pmatrix} = M^1_1 N^1_1 M^2_1 N^1_2 - M^2_1 N^1_1 M^1_1 N^1_2 = 0\): the commutativity required.
As any square matrix is a rectangular matrix, \(M\) and \(N\) can be \(m \times m\) matrices.
In that case, also \(det (M N) = det M det N\) holds.
For the \(m \neq n\) case, "\(det (M N) = det M det N\)" does not make sense, because each of \(det M\) and \(det N\) is not defined.
3: Proof
Whole Strategy: Step 1: see the components of \(M N\); Step 2: expand \(det (M N)\) with respect to the 1st row, then with respect to the 2nd row, ...; Step 3: use \(det (M N) = det ((M N)^t) = det (N^t M^t)\) and apply the result of the Step 2.
Step 1:
\(M N = \begin{pmatrix} \sum_{l \in \{1, ..., n\}} M^j_l N^l_o \end{pmatrix} = \begin{pmatrix} \sum_{l \in \{1, ..., n\}} M^1_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^1_l N^l_m \\ ... \\ \sum_{l \in \{1, ..., n\}} M^m_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^m_l N^l_m \end{pmatrix}\).
Step 2:
The 1st row of \(M N\) can be written as \((\sum_{l_1 \in \{1, ..., n\}} M^1_{l_1} N^{l_1}_1, ... , \sum_{l_1 \in \{1, ..., n\}} M^1_{l_1} N^{l_1}_m) = \sum_{l_1 \in \{1, ..., n\}} (M^1_{l_1} N^{l_1}_1, ... , M^1_{l_1} N^{l_1}_m)\).
By a well-known property of determinant, \(det (M N) = \sum_{l_1 \in \{1, ..., n\}} det \begin{pmatrix} M^1_{l_1} N^{l_1}_1 & ... & M^1_{l_1} N^{l_1}_m \\ \sum_{l \in \{1, ..., n\}} M^2_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^2_l N^l_m \\ ... \\ \sum_{l \in \{1, ..., n\}} M^m_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^m_l N^l_m \end{pmatrix}\).
By a well-known property of determinant, \(det (M N) = \sum_{l_1 \in \{1, ..., n\}} M^1_{l_1} det \begin{pmatrix} N^{l_1}_1 & ... & N^{l_1}_m \\ \sum_{l \in \{1, ..., n\}} M^2_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^2_l N^l_m \\ ... \\ \sum_{l \in \{1, ..., n\}} M^m_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^m_l N^l_m \end{pmatrix}\).
For each of the new determinants, the 2nd row can be written as \((\sum_{l_2 \in \{1, ..., n\}} M^2_{l_2} N^{l_2}_1 ... \sum_{{l_2} \in \{1, ..., n\}} M^2_{l_2} N^{l_2}_m) = \sum_{l_2 \in \{1, ..., n\}} (M^2_{l_2} N^{l_2}_1, ... , M^2_{l_2} N^{l_2}_m)\).
By a well-known property of determinant, \(det \begin{pmatrix} N^{l_1}_1 & ... & N^{l_1}_m \\ \sum_{l \in \{1, ..., n\}} M^2_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^2_l N^l_m \\ ... \\ \sum_{l \in \{1, ..., n\}} M^m_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^m_l N^l_m \end{pmatrix} = \sum_{l_2 \in \{1, ..., n\}} det \begin{pmatrix} N^{l_1}_1 & ... & N^{l_1}_m \\ M^2_{l_2} N^{l_2}_1 & ... & M^2_{l_2} N^{l_2}_m \\ \sum_{l \in \{1, ..., n\}} M^3_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^3_l N^l_m \\ ... \\ \sum_{l \in \{1, ..., n\}} M^m_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^m_l N^l_m \end{pmatrix}\).
By a well-known property of determinant, \(= \sum_{l_2 \in \{1, ..., n\}} M^2_{l_2} det \begin{pmatrix} N^{l_1}_1 & ... & N^{l_1}_m \\ N^{l_2}_1 & ... & N^{l_2}_m \\ \sum_{l \in \{1, ..., n\}} M^3_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^3_l N^l_m \\ ... \\ \sum_{l \in \{1, ..., n\}} M^m_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^m_l N^l_m \end{pmatrix}\).
So, \(det (M N) = \sum_{l_1 \in \{1, ..., n\}} M^1_{l_1} \sum_{l_2 \in \{1, ..., n\}} M^2_{l_2} det \begin{pmatrix} N^{l_1}_1 & ... & N^{l_1}_m \\ N^{l_2}_1 & ... & N^{l_2}_m \\ \sum_{l \in \{1, ..., n\}} M^3_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^3_l N^l_m \\ ... \\ \sum_{l \in \{1, ..., n\}} M^m_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^m_l N^l_m \end{pmatrix} = \sum_{l_1 \in \{1, ..., n\}} \sum_{l_2 \in \{1, ..., n\}} M^1_{l_1} M^2_{l_2} det \begin{pmatrix} N^{l_1}_1 & ... & N^{l_1}_m \\ N^{l_2}_1 & ... & N^{l_2}_m \\ \sum_{l \in \{1, ..., n\}} M^3_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^3_l N^l_m \\ ... \\ \sum_{l \in \{1, ..., n\}} M^m_l N^l_1 & ... & \sum_{l \in \{1, ..., n\}} M^m_l N^l_m \end{pmatrix}\).
And so on, after all, \(det (M N) = \sum_{l_1 \in \{1, ..., n\}} ... \sum_{l_m \in \{1, ..., n\}} M^1_{l_1} ... M^m_{l_m} det \begin{pmatrix} N^{l_1}_1 & ... & N^{l_1}_m \\ ... \\ N^{l_m}_1 & ... & N^{l_m}_m \end{pmatrix}\).
Step 3:
\(det (M N) = det ((M N)^t) = det (N^t M^t)\).
By the result of Step 2, \(det (N^t M^t) = \sum_{l_1 \in \{1, ..., n\}} ... \sum_{l_m \in \{1, ..., n\}} N^{l_1}_1 ... N^{l_m}_m det \begin{pmatrix} M^1_{l_1} & ... & M^m_{l_1} \\ ... \\ M^1_{l_m} & ... & M^m_{l_m} \end{pmatrix} = \sum_{l_1 \in \{1, ..., n\}} ... \sum_{l_m \in \{1, ..., n\}} N^{l_1}_1 ... N^{l_m}_m det \begin{pmatrix} M^1_{l_1} & ... & M^1_{l_m} \\ ... \\ M^m_{l_1} & ... & M^m_{l_m} \end{pmatrix}\).
