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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.

Val
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    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 Sep 06 '13 at 03:41
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    Can you represent the example in my form and repeat the argument? – Val Sep 06 '13 at 03:42
  • What else can I say? – Pedro Sep 06 '13 at 03:44
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    If the sum is finite and $A$ and $B_i$ are numbers than what you have is regular old distributivity. – Devin Murray Sep 06 '13 at 03:45
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    Do you mean that this is not commutativity? A and ∑ do not commute in my example? – Val Sep 06 '13 at 03:46
  • @Val Is $A$ a function of some sort? In such case -- which seems to the case here -- make it explicit. – Pedro Sep 06 '13 at 03:46
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    I think of it like operator – Val Sep 06 '13 at 03:47
  • Then it is neither. – copper.hat Sep 06 '13 at 03:47
  • $A$ looks like a homomorphism of an additive group. – Andrey Sokolov Sep 06 '13 at 03:48
  • Well, in this case one says that "$A$ commutes with sums" if you may, but that is a less formal way to say things. – Pedro Sep 06 '13 at 03:48
  • A is distributive = A commutes with sums? – Val Sep 06 '13 at 03:50
  • It's the multiplication of a ring that is distributive. This "multiplication" can even be composition in a ring of operators. – Karl Kroningfeld Sep 06 '13 at 04:11
  • @KarlKronenfeld And how does it clarify the indicated relationship with commutativity? – Val Sep 06 '13 at 04:13
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    The elements of a ring of operators can commute, but summation of operators is never an element of that ring, so you cannot say that an operator commutes with summation. – Karl Kroningfeld Sep 06 '13 at 04:14
  • But I clearly see that it is. Isn't it? What is commutativity if not A∑ = ∑A? How is it never commutative if it is? – Val Sep 06 '13 at 04:16
  • Val: Suppose the domain of $A$ is $D$. The domain of $A\sum_{i=1}^n$ is the set of $n$-tuples consisting of elements of $D$. Since $A$ and $\sum_{i=1}^n$ commute as operators, the domain of $A$ must be the set of all $n$-tuples of elements of $D$ union $D$. Unless $D$ is empty, we get a contradiction to the axiom of foundation. – Karl Kroningfeld Sep 06 '13 at 04:35
  • Good point. Actually, I concocted my example using differentiation, as a model for A. You know the identity that sum of direvatives is a direvative of the sum. So, $B_n$ must also be functions. Operators operate on functions. Don't they? And, not all functions are constants. I have renamed $B_n$ to $f_n$ to remove the confusion. – Val Sep 06 '13 at 05:22
  • @KarlKronenfeld What if we take the domain of $A$ to be the union of $D^n$ for all $n$ and let $A$ act componentwise on a tuple of elements of $D$. True we are defining a new $\hat{A}$ on this union of tuples, but $A$ and $\hat{A}$ encode the same information. Then for a tuple $T=(T_1,\ldots,T_N)$ we have $\hat{A}\Sigma(T)=\hat{A}(T_1+\ldots+T_n)=A(T_1+\ldots+T_n)=A(T_1)+\ldots+A(T_n)=\Sigma A(T_i)=\Sigma\hat{A}(T)$. So $\hat{A}$ and $\Sigma$ commute. Hence we can say that it is true $A$ commutes with $\Sigma$ to the same extend as we can identify $A$ with $\hat{A}$. – OR. Sep 06 '13 at 05:24
  • @ABC Then the domain of operation of $\Sigma$ is different from that of $\hat A$. – Karl Kroningfeld Sep 06 '13 at 05:27
  • $\Sigma$ is defined on tuples. You input a tuple, it returns the sum of its entries. This last you can see then as a $1$-tuple and then is an element of the set of tuple again. – OR. Sep 06 '13 at 05:29
  • @ABC What, then, is $\Sigma\Sigma$, and does that compare in anyway with how the rest of the world defines it? – Karl Kroningfeld Sep 06 '13 at 05:37
  • Ok, I must figure out what the sum is and how can we differentiate the sum first. Since Σ produces a constant, derivative of sum is always 0 and I do not understand how the sum of direvatives is derivative of sums then. – Val Sep 06 '13 at 05:38
  • If we interpret $A$ as a map and $\sum$ as sum, then $A(\sum f_i) = \sum(A(f_i))$ kind of looks like half of the statement "$A$ is linear", doesn't it? – kahen Sep 06 '13 at 05:40
  • @kahen Yes, this is exactly how I came here. I started to think that linear operator is one that commutes with the sum operator. This also reminds me that ∑ is a matrix and the result of $\sum {f_i}$ is a function rather than a constant! – Val Sep 06 '13 at 05:41
  • @KarlKronenfeld Let $T=(T_1,\ldots,T_n)$ be a tuple. $\Sigma(T)$ is the $1$-tuple $(T_1+\ldots+T_n)$. Then $\Sigma\Sigma(T)$ is again the same $1$-tuple $(T_1+\ldots+T_n)$ as $\Sigma$ adds the elements of a tuple. The problem is not how people see it. Structures and formalism in math are made to aid thinking, not to limit it. Val is seeing a similarity between commutativity and associativity, because there is one. It is not that $A$ per se commutes with $\Sigma$ as you pointed out, but that $\hat{A}$ commutes with $\Sigma$. $\hat{A}$ and $A$ are not the same thing when written in the common... – OR. Sep 06 '13 at 05:45
  • formalism in mathematics. But that formalism can't limit the fact that $\hat{A}$ and $A$ contain pretty much the same kind of information. So, there is a way of 'seeing' that commutativity=distributivity that Val is seeing, which is by 'seeing' $\hat{A}$ and $A$ as the same thing. They are not, but they 'can be seen' as the same. – OR. Sep 06 '13 at 05:46
  • Do you mean that A applies to $F=[\vec f_1 \vec f_2 \cdots]$, the matrix of column-vectors $\vec f_i$ whereas matrix ∑, which consists of all 1s, applies to the $F^T = [\vec f_1 \vec f_2 \ldots]^T$, the matrix of row-vectors $\vec f_i$ so that the first column of $F^T$ is a vector of first components of $\vec f_i$, which, applied to ∑, produces the first component of their sum? This way, I see why the informational content ($F=F^T$) admits the commutation of A and ∑, yet, not the blind syntactic switch. I wonder how matematicians always see how objects should be transposed to fit each other. – Val Sep 06 '13 at 06:38
  • Please define the domain and range of your operators A and $\sum$. Without that, talking about commutativity is pointless. – TenaliRaman Sep 06 '13 at 07:11
  • @TenaliRaman Why do people teach the pointless things like "vector space", "matrix", "function" and derivative of sum is the sum of derivatives? – Val Sep 06 '13 at 07:15
  • Do you understand my question? Given two operators A and B, AB = A(B(x)). Unless B's range is A's domain, A(B(x)) does not make sense. Similarly, BA = B(A(x)) therefore, A's range has to be B's domain. Finally, for AB = BA to be true, A's range has to be same as B's range.

    Thereby, unless you define A and B's domain and range, you cannot talk about commutativity. That is why, I asked you, please write the domain and range of the two operators that you speak. Then we can think about commutativity.

     – TenaliRaman Sep 06 '13 at 12:37
  • @TenaliRaman I have got why you needed the ranges, thanks. This tides up everything very nicely. But, I think that Hurkyl has already answered my question. Do you have anything to add? – Val Sep 06 '13 at 16:19
  • No, nothing to add. Hurkyl (+1), pretty much has said everything I wanted to convey. – TenaliRaman Sep 06 '13 at 16:30

3 Answers3

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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

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

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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.

Marc van Leeuwen
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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".

Did
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  • 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 06 '13 at 07:18
  • 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 06 '13 at 07:47
  • How is different? I've asked about the case where A commutes with ∑. Are you responding to something different? – Val Sep 06 '13 at 07:54
  • To be commutative $\ne$ To commute with. – Did Sep 06 '13 at 08:02