885: For 2 Square Matrices over Commutative Ring, Trace of Product of Matrices Does Not Depend on Order of Product

<The previous article in this series | The table of contents of this series | The next article in this series>

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
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

---

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:
$R$: $\in \{\text{ the commutative rings }\}$
$A$: $\in \{\text{ the n }\times \text{ n matrices over }R\}$, $=\left(\begin{array}{c}{a}_k^j\end{array}\right)$
$B$: $\in \{\text{ the n }\times \text{ n matrices over }R\}$, $=\left(\begin{array}{c}{b}_k^j\end{array}\right)$
//

Statements:
$tr(AB)=tr(BA)$
//

---

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 $tr(AB)$ and $tr(BA)$ with the components of $A$ and $B$ and see that they are the same.

Step 1:

$AB=\left(\begin{array}{c}\sum _{l\in \{1,...,n\}}{a}_l^j{b}_k^l\end{array}\right)$.

$tr(AB)=\sum _{j\in \{1,...,n\}}\sum _{l\in \{1,...,n\}}{a}_l^j{b}_j^l$.

$BA=\left(\begin{array}{c}\sum _{l\in \{1,...,n\}}{b}_l^j{a}_k^l\end{array}\right)$.

$tr(BA)=\sum _{j\in \{1,...,n\}}\sum _{l\in \{1,...,n\}}{b}_l^j{a}_j^l=\sum _{j\in \{1,...,n\}}\sum _{l\in \{1,...,n\}}{a}_j^l{b}_l^j=\sum _{l\in \{1,...,n\}}\sum _{j\in \{1,...,n\}}{a}_j^l{b}_l^j=\sum _{j\in \{1,...,n\}}\sum _{l\in \{1,...,n\}}{a}_l^j{b}_j^l$, where the last equal is because the name of any dummy index can be arbitrarily changed, $=tr(AB)$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>