3IR-Magic_Squares

A discussion of a 'matrix' vector space,
(no numbergenerating function, Regular Magic Condition)
(with thanks to an anonymous email contact,
who worked through the equations below,
and initiated the entire subject)

As a working square one can define:
 a b c d e f g h i
To define the magic sum one has to set pe. a+b+c=s
Ultimately the above leads to the following set of equations:
g = s - a - d = b + c - d
e = s - g - c = a + d - c
h = s - b - e = 2 c - d
f = s - e - d = b + 2 c - 2 d
i = s - c - f = a - 2 c + 2 d
i = s - e - a = b + 2c - a - d
because of the two different i-equations one derives d = (-2 a + b + 4 c) / 3
which means that the second and third row can be entirely 'hung-up' on the top row

'plugging in' the lineair independent IR3 vectors {(1,0,0),(0,1,0),(0,0,1)} in the above equations for (a,b,c) one gets:

 1.0 0.0 0.0 | 0.0 1.0 0.0 | 0.0 0.0 1.0 -2/3 1/3 4/3 | 1/3 1/3 1/3 | 4/3 1/3 -2/3 2/3 2/3 -1/3 | 2/3 -1/3 2/3 | -1/3 2/3 2/3

Which can serve as a base for the vector space 3IR-Magic_Squares which is even 'normalized' in the sense that the involved 'magic sums' are all 1.0
Note: B12 = I, B22 = (1/3) and B32 = (2/3) - I and thus: B32 + B12 = 2 B22 and so the basis forms a "Pseudo Pythagorian Matrix Triplet" (see: PseudoRings), also the transposed matrices B1t, B2t and B3t are expressable in this base

as a second suggestion for a base is:

 0 1 -1 | -1 1 0 | 1 1 1 -1 0 1 | 1 0 -1 | 1 1 1 1 -1 0 | 0 -1 1 | 1 1 1

of course since both are basis on the same space one can be put into the other, the coëfficients of the second base with respect to the first base are given in the top row of the second base. This second base however can also be used for the related
3IZ-Magic_Square
leaving only 3IN-Magic_Square 'baseless', since both bases has negative numbers in it, neither can be used in this case. Adding the third vector of the second base to he first two vectors gives a appropriate base in this instance:

 1 2 0 | 0 2 1 | 1 1 1 0 1 2 | 2 1 0 | 1 1 1 2 0 1 | 1 0 2 | 1 1 1

I was told that 0 was not in IN, however I use the definition of Peano which has 0 in IN.
The first two 'vectors' are the 'latin squares' which can be used to create a regular third order magic square (A * 3 + B + C; A the first and B the second vector of the base, C the base third vector needs to be added to have the numbers 1 .. n2 in the square)

'pure technical note: (Vector spaces)'
The exact definition I was given (by forementioned contact) was for a vector spaces over a field F which would mean that 'vector spaces' can only be defined in the cases IC, IR, and IQ (both bases above apply), however nothing in that definition prohibits one for defining a vector space over the ring IZ (second base is appropriate), besides that one can even define a vector space over IN. (only the third base approprite)
The given definition of a vector space is:
A vector space over (the field F) is a set V together with two binary operations addition +: V x V -> V and scalar multiplication *: F x V -> V which satisfy: a> V is commutative group under addition b> scalar multiplication satisfy: i) 1 * v = v for all v in V ii) a * (b * v) = (a * b) * v for all a,b in F and v in V c> scalar multiplication is distributive under addition i) (a + b) * v = a * v + b * v ii) a * (v + w) = a * v + a * w
There is nothing in the formulae above which prohibits one from substituting (the field F) by (the ring IZ) or even (the set IN).
A peak at mathworld told me that with IZ we have a Module and the set IN is neither because VectorSpace includes the need for an inverse under addittion which IN doesn't provide

In regular vector space (pe. IR3) one can define inner and outer product, I have never seen matrices treated as vector space 'vectors' so I don't know how one would define such concepts on matrices, For an inner product two candidates can be simply defined:

1. V . W = Vi,jWi,j = Aa + Bb + Cc + Dd + Ee + Ff + Gg + Hh + Ii
2. V . W = Vi,jWj,i = Aa + Bd + Cg + Db + Ee + Fh + Gc + Hf + Ii

(using the general matrix as a template (V in uppercast, W in lowercast letters))
both expressions are symmetrical, as is required for an inner product.
However the first one is definite positive when V == W while the second one can go negative, on the first set of base vectors above the first one would give lengths:

(39/9)1/2, (21/9)1/2 and (39/9)1/2
the second one
(27/9)1/2, (7/9)1/2 and (-7/9)1/2,

which would give an imaginairy length to the third vector above
(a strange but none the less definable concept)
Note due to the fact that the transposed base matrices are also within the space I suppose the first version is enough since (V.W)2 = (V.Wt)1

Some figuring in a puny spreadsheet suggest that:

(A*B)i,j = ((A(i+1)%3,(j+1)%3B(i+2)%3,(j+2)%3-A(i+2)%3,(j+1)%3B(i+1)%3,(j+2)%3) +
(B(i+1)%3,(j+1)%3A(i+2)%3,(j+2)%3-B(i+2)%3,(j+1)%3A(i+1)%3,(j+2)%3)) / 2

is expressable in the base so might be a candidate for a Crossproduct. Currently I don't know whether this is consistent with the features a outer (or Cross) product ought to have. Thus far seen tests on the base vectors suggest a decent vector within the space, however the expression A*B is symmetrical ie A*B = B*A. Unfortunately subtracting the two parts doesn't result in a vector in this space nor do the seperate parts. For the module 3IZ-Magic_Square one needs to multiply with 3 to stay within IZ with the numbers.
Some figuring show interestingly: B1t * B1t = - B1t

With the inner products one can define metric functions and since
V . W = |V||W|cos(<(V,W))
one can thus define the 'angle' between two 3IR-Magic_Square vectors,
the outer product ought to define a new vector which length ought to satisfy:
|V * W| = |V||W|sin(<(V,W))
The symmetrical nature of the above expression this doesn't seem to be the case!?

Note: I'm a bit out of touch with this, but as stated this is still a novel subject and as the second 'inner product' suggests imaginairy lengths, a symmetric 'outer product' might be possible though this isn't present in IR3 where the cross product is antisymmetrical (A*B = -B*A), but the above expressions seems possible in any case.

I implemented this vector space in the 3IR-Magic-Square.zip spreadsheet. For the expert mathematician the academic question whether one can call the above defined "crossproduct" a "crossproduct", I know that in IR3 the crossproduct is anti symmetric, allowing one to define right-handed and left-handed coordinate system. Also the above relation to angles are taken from IR3, future investigation might reveal what this all meaans in 3IR-Magic_Squares