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Recently I have been doing some algebra of partitioned matrices, where product like below is common:

enter image description here

where $A_{11}, A_{12}, X, Y$ are all partitioned matrix blocks thus all are matrices in essence.

One of the Important features of product between matrices and/or vectors is that: it is not commutative, namely $ AB \neq BA $, so I have to remove the attributes of "Orderless" from Times[] to make sure the system does not change the order of product terms in result as it would often does.

To do this product in Mathematica, first I cleared the attributes:

ClearAttributes[Times, Orderless];

Then I tried to do this product several times, in order to simulate the long computation process I might need to do on this kind of algebra in the future:

({
    {Subscript[A, 11], 0},
    {0, Subscript[A, 22]}
   }) . ({
    {0, Y},
    {X, 0}
   }) // MatrixForm

However, it was to my surprise to find out that the result is unstable. For the same product, sometimes it will give out the result as:

enter image description here

While sometimes the result given is:

enter image description here

But only the second one is true, where in the relevant products $A_{11}, A_{12}$ go first and $X,Y$ go last.

The typical situation where the order/sequence of the product changes:

(1) After you do several other calculations and go back to do the same product;

(2) After you take a break for several minutes and come back to do the same product;

(3) When the name of a multiplier is too long, the system tends to adjust the order/sequence of the product to let the multiplier with long name goes last. For example, for matrix X, suppose the correct product is Transpose[X] X or Inverse[X] Xbut the system would tend to give it as X Transpose[X] or X Inverse[X]

So I was wondering: why the result would be so unstable even after the attributes of Orderless already being cleared? Is there anyway to clear the attributes stably and reliably so that when I do the algebra of partitioned matrices it will always gives the product in result in correct order/sequence?

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    $\begingroup$ First of all, the operator . is not Times but Dot. Secondly, I wouldn't mess with the properties of such a fundamental operator. I suggest defining a new operator or using something like NonCommutativeMultiply, for which you can find several examples on StackExchange. $\endgroup$
    – Domen
    Jan 10 at 9:44
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    $\begingroup$ Thanks Domen, but I would like to clarify: between the two large matrices outside, it is dot product, which does not have the attributes of Orderless by default, but to work out the product in entry level, Dot[] will compute sub-product as entry in result matrices by Times[], that is the why I have to modify Times furthermore to get the correct result. $\endgroup$
    – AlbertLew
    Jan 10 at 9:52
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    $\begingroup$ I would also recommend against changing a built-in, and maybe define your own version of Dot to use instead—consider using Inner, e.g. a = {{Indexed[A,{1,1}], 0}, {0, Indexed[A, {2,2}]}}; b = {{0,X},{Y,0}}; Inner[NonCommutativeMultiply, a, b, Plus]. Of course, you'd need to define NonCommutativeMultiply appropriately first, since it's nearly completely undefined by default, which is not a trivial task... $\endgroup$
    – thorimur
    Jan 10 at 10:05
  • $\begingroup$ @thorimur That's definitely true, it is indeed not a trivial task since I have been stuck here for nearly one week. For such a seemingly simple thing, the Partitioned Matrix Algebra, it is counter-intuitively hard to teach mathematica to do it correctly, how ironic it is! $\endgroup$
    – AlbertLew
    Jan 10 at 10:23
  • $\begingroup$ Changing attributes of the low level arithmetic functions Plus, Times and Power ranks high on the list of Things Not To Do. There is past discussion about this here. Generally one instead uses NonCommutativeMultiply, defining rules as needed. $\endgroup$ Jan 10 at 15:10

1 Answer 1

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First I assume that you want the outer "multiply to be "Dot" and the inner "Times" with order. (If the inner is also Dot, you must adapt the code inside "Sum")

Redefining Times is dangerous. Instead I would use NonCommutativeMultiply that has no definitions besides being non commutative.

First we define our multiplication operator:

Clear["Globals`*"]
Unprotect[NonCommutativeMultiply]
NonCommutativeMultiply[a_, b_] := 
 Table[Sum[a[[i1, i2]] b[[i2, i3]], {i2, n}], {i1, n}, {i3, n}]

Then we make a simple test example:

n = 2;
a=Table[RandomInteger[{0,10}],{i1,n},{i2,n},{i3,n},{i4,n}];
b=Table[RandomInteger[{0,10}],{i1,n},{i2,n},{i3,n},{i4,n}];
a // MatrixForm
b // MatrixForm
a ** b // MatrixForm

enter image description here

Update

To make it work with the case where the elements of a and b are symbolic matrices:

Clear["Globals`*"]
Unprotect[NonCommutativeMultiply]
NonCommutativeMultiply[a_?MatrixQ, b_?MatrixQ] := 
 Table[Sum[a[[i1, i2]] ** b[[i2, i3]], {i2, n}], {i1, n}, {i3, n}]

n = 2;
ma = Array[Subscript[a, #1, #2] &, {n, n}];
mb = Array[Subscript[b, #1, #2] &, {n, n}];

ma // MatrixForm
mb // MatrixForm
ma ** mb // MatrixForm

enter image description here

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  • $\begingroup$ Thanks Daniel. By default, mathematica identifies each matrix as one constituting of scalars. So when two matrices make dot produce, the Dot[] will employ Times[] to work out the result entry by entry, that is why when the matrices are made of symbolic matrix blocks or symbolic sub matrices, it is hard to teach mathematica to do the product correctly without commutation. $\endgroup$
    – AlbertLew
    Jan 10 at 11:35
  • $\begingroup$ Furthermore, Daniel, your solution looks impressive, but it does not work in my case, where the matrices are made of symbolic matrix blocks or symbolic sub matrices like X and Y in my post, and the specific values for each entry in those sub matrices is unknown, what I am doing is pure symbolic calculations. $\endgroup$
    – AlbertLew
    Jan 10 at 11:50
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    $\begingroup$ I made an update for this case. Note instead of using "Times" I use "**", that is "NonCommutatvve Multiply". This is necessary because otherwise MMA will reorder the terms. $\endgroup$ Jan 10 at 12:08

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