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Please note I have only little background im mathematics and I am working on formalizing theorems with theorem provers. This is very much a beginner question.

Suppose I have matrices, where the elements of the matrix are from a commutative ring with 1 element (commonly called ring I guess).

I looked at the wikipedia entries of matrix multiplication and semi-ring.

My guess is that matrix multiplication together with plus (+) is a semi-ring.

But does multiplication together with minus form a semi-ring? Is it called something else?

The following lemmas are true: A ** (B - C) = A ** B - A ** C and (A - B) ** C = A ** C - B ** C where ** is the matrix multiplication.

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up vote 2 down vote accepted

No. Subtraction is neither commutative nor associative. It's usually not considered as an operation in isolation without addition. With addition and subtraction, together with matrix multiplication, matrices form a ring.

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Let $R$ be a commutative ring with unit. Let $M_n(R)$ be the set of $n \times n$ matrices with entries taken from $R$. If $A = [A_{ij} \in M_n(R)$, then taking $-A = [-A_{ij}] \in M_n(R)$ yields a matrix $-A$ such that $A + (-A) = (-A) + A = 0 \in M_n(R)$. According to this wikipedia page, we have shown $M_n(R)$ to be a ring, since we have demonstrated an additive inverse for each element. Technically speaking, we should show $M_n(R)$ satisfies all the other ring axioms, but if one grants that $M_n(R)$ is a semi-ring, then that work is taken care of. In any event, the existence of additive inverses is the distinguishing fact, and verification from scratch that $M_n(R)$ satisfies the other ring axioms is not a difficult matter, so I will leave it for my audience to work out. Of course, if anyone needs more input in that matter, leaving a comment to that effect will beckon me to address this further, and I will be back to answer.

If the addtive operation is replaced by subtraction in any ring $S$, commutative or not, with or without a multiplative inverse, i.e. $a + b$ is replaced by $a - b =a + (-b)$ then we don't even have a semi-ring unless $a = -a$ for all $a \in S$ (or you may say $\text{char} S = 2$ if there exists $1_S \in S$ having the usual property $1_S a = a 1_S$). This is for the reason that, without $a = -a$, subtraction is neither commutative nor associative in general: taking $b = 0$ yields

$a - b = b - a \Rightarrow a = a - 0 = 0 - a = -a, \tag{1}$


$(a - b) - c = a - (b - c) \Rightarrow (a - b) -c = a + (-(b - c)) = a + (-b + c)$ $= (a - b) + c \Rightarrow c = -c \; \text{for all} \; c \in S. \tag{2}$

So in general, in the absence of the condition $a = -a$ for all $a \in S$, subtraction doesn't satisfy the necessary axioms, though $a(b - c) = ab - ac$ and $(a - b)c = ac - ab$ still apply.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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Thank you for the detailed answer! – mrsteve Feb 16 '14 at 8:38
My pleasure, sir! – Robert Lewis Feb 16 '14 at 8:40

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