description/proof of that for ring, multiplications of matrices over ring are distributive w.r.t. addition
Topics
About: ring
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 ring, the multiplications of any matrices over the ring are distributive with respect to addition.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } n_1 \times n_2 \text{ matrices over } R\}\)
\(M_2\): \(\in \{\text{ the } n_2 \times n_3 \text{ matrices over } R\}\)
\(M_3\): \(\in \{\text{ the } n_2 \times n_3 \text{ matrices over } R\}\)
\(M_4\): \(\in \{\text{ the } n_3 \times n_4 \text{ matrices over } R\}\)
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Statements:
\(M_1 (M_2 + M_3) = M_1 M_2 + M_1 M_3\)
\(\land\)
\((M_2 + M_3) M_4 = M_2 M_4 + M_3 M_4\)
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2: Note
So, being distributive with respect to addition is not only for \(R\) real or complex or even any field but always for \(R\) any ring.
3: Proof
Whole Strategy: Step 1: see that \((M_1 (M_2 + M_3))^j_m = (M_1 M_2 + M_1 M_3)^j_m\); Step 2: see that \(((M_2 + M_3) M_4)^j_m = (M_2 M_4 + M_3 M_4)^j_m\).
Step 1:
\((M_1 (M_2 + M_3))^j_m = \sum_l ({M_1}^j_l ({M_2}^l_m + {M_3}^l_m)) = \sum_l ({M_1}^j_l {M_2}^l_m + {M_1}^j_l {M_3}^l_m)\), by the multiplication distributability with respect to addition of \(R\).
\((M_1 M_2 + M_1 M_3)^j_m = \sum_l ({M_1}^j_l {M_2}^l_m) + \sum_l ({M_1}^j_l {M_3}^l_m) = \sum_l ({M_1}^j_l {M_2}^l_m + {M_1}^j_l {M_3}^l_m)\), because addition of \(R\) is commutative.
So, \((M_1 (M_2 + M_3))^j_m = (M_1 M_2 + M_1 M_3)^j_m\).
So, \(M_1 (M_2 + M_3) = M_1 M_2 + M_1 M_3\).
Step 2:
\(((M_2 + M_3) M_4)^j_m = \sum_l (({M_2}^j_l + {M_3}^j_l) {M_4}^l_m) = \sum_l ({M_2}^j_l {M_4}^l_m + {M_3}^j_l {M_4}^l_m)\), by the multiplication distributability with respect to addition of \(R\).
\((M_2 M_4 + M_3 M_4)^j_m = \sum_l ({M_2}^j_l {M_4}^l_m) + \sum_l ({M_3}^j_l {M_4}^l_m) = \sum_l ({M_2}^j_l {M_4}^l_m + {M_3}^j_l {M_4}^l_m)\), because addition of \(R\) is commutative.
So, \(((M_2 + M_3) M_4)^j_m = (M_2 M_4 + M_3 M_4)^j_m\).
So, \((M_2 + M_3) M_4 = M_2 M_4 + M_3 M_4\).
