description/proof of that for 2 square matrices over commutative ring, trace of product of matrices does not depend on order of product
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of ring.
- The reader knows a definition of matrix over ring.
- The reader knows a definition of trace of matrix over ring.
Target Context
- The reader will have a description and a proof of the proposition that for any 2 same-dimensional square matrices over any commutative ring, the trace of each product of the matrices does not depend on the order of the product.
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:
\(R\): \(\in \{\text{ the commutative rings }\}\)
\(A\): \(\in \{\text{ the n } \times \text{ n matrices over } R\}\), \(= \begin{pmatrix} a^j_k \end{pmatrix}\)
\(B\): \(\in \{\text{ the n } \times \text{ n matrices over } R\}\), \(= \begin{pmatrix} b^j_k \end{pmatrix}\)
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Statements:
\(tr (A B) = tr (B A)\)
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2: Note
Typically, \(R\) is a field, and more typically, \(R\) is \(\mathbb{R}\) or \(\mathbb{C}\), but this proposition requires only that \(R\) is any commutative ring.
As an immediate corollary, for any more than 2 same-dimensional square matrices, the order of any product can be cyclically changed with the trace intact: \(tr (A_1 ... A_k) = tr (A_1 (A_2 ... A_k)) = tr ((A_2 ... A_k) A_1) = tr (A_2 ... A_k A_1)\), e.t.c..
3: Proof
Whole Strategy: Step 1: express \(tr (A B)\) and \(tr (B A)\) with the components of \(A\) and \(B\) and see that they are the same.
Step 1:
\(A B = \begin{pmatrix} \sum_{l \in \{1, ..., n\}} a^j_l b^l_k \end{pmatrix}\).
\(tr (A B) = \sum_{j \in \{1, ..., n\}} \sum_{l \in \{1, ..., n\}} a^j_l b^l_j\).
\(B A = \begin{pmatrix} \sum_{l \in \{1, ..., n\}} b^j_l a^l_k \end{pmatrix}\).
\(tr (B A) = \sum_{j \in \{1, ..., n\}} \sum_{l \in \{1, ..., n\}} b^j_l a^l_j = \sum_{j \in \{1, ..., n\}} \sum_{l \in \{1, ..., n\}} a^l_j b^j_l = \sum_{l \in \{1, ..., n\}} \sum_{j \in \{1, ..., n\}} a^l_j b^j_l = \sum_{j \in \{1, ..., n\}} \sum_{l \in \{1, ..., n\}} a^j_l b^l_j\), where the last equal is because the name of any dummy index can be arbitrarily changed, \(= tr (A B)\).
