# What is the difference between commutatitivity and distributivity?

Inspired by linearity property, I see that operator A distributes over sum: $A(\sum {f_i}) = \sum {(A(f_i))}\;\,$. What do I have between A and ∑: commutativity or distributivity? The fact that I can switch the order, $A\sum = \sum A$, implies commutativity. Meantime, we see that A distributes over sums in this case. Can I say that commutative and distributive are the same things since A distributes over ∑ iff A and ∑ are commutative?

update I consider A and ∑ as operators (matrices) over functions (aka vectors) $f_i\,$. Might be I just confuse multiple definitions of the sum (if there are any). For instance, sum of vector elements is not the same as sum of multiple vectors.

• Commutativity is $ab=ba$. Distributivity is $a(b+c)=ab+ac$, or $(a+b)c=ac+bc$, or both. One relates on operation with itself, the other two operations. I guess this should help clear out your confusion. – Pedro Tamaroff Sep 6 '13 at 3:41
• Can you represent the example in my form and repeat the argument? – Val Sep 6 '13 at 3:42
• If the sum is finite and $A$ and $B_i$ are numbers than what you have is regular old distributivity. – Devin Murray Sep 6 '13 at 3:45
• Do you mean that this is not commutativity? A and ∑ do not commute in my example? – Val Sep 6 '13 at 3:46
• I think of it like operator – Val Sep 6 '13 at 3:47

The equation $A \Sigma = \Sigma A$ doesn't quite make sense here, because you're using $A$ to mean two different things: on the left, you mean an operator that takes one value $x$ and returns $A(x)$, and on the right you mean an operator that takes many values $(x_i)$ and returns the sequence $(A(x_i))$.

For addition specifically, we would use the term "additive", or possibly in certain contexts "linear". However, there is a generalization of the notion of commutativity that applies to this situation.

Suppose you have an $m \times n$ array of numbers $x_{ij}$, an $m$-ary operation $f$, and an $n$-ary operation $g$.

Then you could apply $f$ on the columns to get an $n$-long sequence which you can plug into $g$

$$g(f(x_{11}, x_{21}, \ldots, x_{m1}), \cdots, f(x_{1n}, x_{2n}, \ldots, x_{mn}) )$$

or, you could apply $g$ to the rows to get an $m$-long sequence which you can plug into $f$:

$$f(g(x_{11}, x_{12}, \ldots, x_{1n}), \cdots, g(x_{m1}, x_{m2}, \ldots, x_{mn}) )$$

If you get the same value either way, then $f$ and $g$ are said to commute.

As an example, incidentally, the distributive law is precisely the statement that, for each $c$, the binary addition operator commutes with the unary operation "multiply by $c$". Given a 1x2 array of numbers

$$(a,b)$$

If we apply $+$ across rows, and "multiply (on the right) by $c$" down columns, the two ways of calculating are

$$(a+b) \cdot c = (a \cdot c) + (b \cdot c)$$

Neither commutativity nor distributivity involve one operation of one argument like (the application of) $A$. Commutativity involves one binary operation ($a\star b=b\star a$), and distributivity involves two binary operations $a\star(b\circ c)=(a\star b)\circ(a\star c))$; this makes the two laws rather uncomparable. The verb "commute" can also be applied to individual elements $a,b$ in case a generally non-commutative operation (often function composition) happens to give the same result when applied to $a,b$ or to $b,a$.

The property of $A$ you state could be described as saying $A$ is compatible with addition, a morphisms of additive groups. There is no other operator of the same kind here to say it commutes with $A$. However one could informally say that $A$ commutes with addition; this would be using a rather vague generalisation of "commuting" since the operations involved are not of the same nature.

The property $A\sum=\sum A$ is probably best described as additivity.

Neither commutativity nor distributivity are used in this sense (by mathematicians), as demonstrated ad nauseam in the comments.

Finally, the mention in your post that "A distributes over ∑ iff A and ∑ are commutative" is rather mysterious. Note that $A\sum=\sum A$ might hold even if the underlying "additions" are not commutative (and if they are not, it is true one should probably not be calling them additions) and that I do not even know what is meant by "A is commutative".

• I do not say that A is commutative (alone). I always say with respect to what it is commutative, despite it is always commutative with ∑ in the case that I consider. – Val Sep 6 '13 at 7:18
• Indeed, a composition law is commutative alone and nobody is "commutative with respect to" anybody else. Some transformations "commute" with others--but this is different. – Did Sep 6 '13 at 7:47
• How is different? I've asked about the case where A commutes with ∑. Are you responding to something different? – Val Sep 6 '13 at 7:54
• To be commutative $\ne$ To commute with. – Did Sep 6 '13 at 8:02