description/proof of that for commutative ring, transpose of product of matrices is product of transposes of constituents in reverse order
Topics
About: matrices space
The table of contents of this article
Starting Context
- The reader knows a definition of transpose of matrix.
Target Context
- The reader will have a description and a proof of the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order.
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:
\(R\): \(\in \{\text{ the commutative rings }\}\)
\(M_1\): \(\in \{\text{ the } m \times n R \text{ matrices } \}\)
\(M_2\): \(\in \{\text{ the } n \times o R \text{ matrices } \}\)
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Statements:
\((M_1 M_2)^t = {M_2}^t {M_1}^t\)
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2: Proof
Whole Strategy: Step 1: let \(M_1 = \begin{pmatrix} {M_1}^j_l \end{pmatrix}\) and \(M_2 = \begin{pmatrix} {M_2}^l_m \end{pmatrix}\); Step 2: see the components of \((M_1 M_2)^t\); Step 3: see the components of \({M_2}^t {M_1}^t\), and conclude the proposition.
Step 1:
Let \(M_1 = \begin{pmatrix} {M_1}^j_l \end{pmatrix}\) and \(M_2 = \begin{pmatrix} {M_2}^l_m \end{pmatrix}\).
Step 2:
\(M_1 M_2 = \begin{pmatrix} {M_1}^j_l {M_2}^l_m \end{pmatrix}\).
\({(M_1 M_2)^t}^j_m = {M_1}^m_l {M_2}^l_j\).
Step 3:
\(({M_2}^t {M_1}^t)^j_m = {{M_2}^t}^j_l {{M_1}^t}^l_m = {M_2}^l_j {M_1}^m_l = {M_1}^m_l {M_2}^l_j\), because \(R\) is a commutative ring, \(= {(M_1 M_2)^t}^j_m\), by the result of Step 2.
