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Here's an image showing what I am trying to do:

3*3*3 multiplication

The two hypermatrices are multiplied together by taking appropriate slices from each hypermatrix, and realised into a vector by an associated vector, given by (a,b,c) in this case.

But I can't figure out how to do the simple matrix multiplication. For example red times green times vector equals blue, but there doesn't seem to be an obvious orientation for the matrices.

JMP
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    What do you want your multiplication operation to be useful for? For example ordinary matrix multiplication can be motivated as composing the linear transformations that the operand matrices represent -- do you have a similar interpretation for your hypermatrices that can be used to guide your choice of multiplication operation? – hmakholm left over Monica Oct 12 '15 at 00:34
  • What does this look like in terms of sums and indices? – Chappers Oct 12 '15 at 00:36
  • @HenningMakholm; i'm just chuffed it seems to work at the moment. if i bring both colors to the front, we get the usual presentation – JMP Oct 12 '15 at 00:37
  • I believe this is a case where it is easier to define the resulting "hypermatrix" by using summation notation and indexes. A preferred orientation of the matrices may be hard to impute by the pictorial method. Presumably you want the horizontal $3\times 3$ slice and the vertical $3\times 3$ slice to produce one-ninth of the result, but I don't see a criterion that would give use the three blue dots in your image. – hardmath Oct 12 '15 at 00:39
  • I have tried embedding a hypercomplex number system into the matrices in an analogous way that complex numbers sit in 2x2 matrices (or 3x3 matrices) and then use the hypercomplex multiplication to motivate a generalization for 3x3x3 matrices. I'm still working on it. – Paul Sundheim Oct 12 '15 at 00:40
  • @hardmath; red$\cap$green? – JMP Oct 12 '15 at 00:41
  • @JonMarkPerry: But before you can conclude that it "seems to work", you need to come up with a criterion for what "work" means in this context. Simply generalizing the rank-2 case when one of the dimensions is $1$ doesn't mean much unless you have something that actually generalizes to other dimensions.. – hmakholm left over Monica Oct 12 '15 at 00:41
  • Yes, but I suspect it would be clearer to write it out so you produce one dot at a time. The front three red dots "dotted" with the front three green dots give us the front blue dot, etc. – hardmath Oct 12 '15 at 00:43
  • "Work" could mean that the multiplication is associative and that the resulting sytem produces a determinant that is multiplicative. – Paul Sundheim Oct 12 '15 at 00:43
  • @hardmath: The problem with using the kind of summations usually considered in multilinear algebra is that we have 6 indices to the left and 3 to the right, so we have to get rid of an odd number of indices by contracting, which is a bit unconventional. When the dimension is 3 we could throw in an $\epsilon_{ijk}$, corresponding to multiplying matrices with 3d vectors as elements and using cross products in place of products in the ordinary formula -- but that doesn't immediately generalize to dimensions other than 3. – hmakholm left over Monica Oct 12 '15 at 00:44
  • @hardmath; i see your point, but i want to know how to do it my way – JMP Oct 12 '15 at 00:46
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    A multi-dimensional matrix or vector is called tensor, so maybe you should check out tensor product. – Lucian Oct 12 '15 at 05:14
  • Note that currently you're diagram simply describes 2-d matrix multiplication 3 times "deep". I.e. ignore all but the front face and you see normal 2-d matrix multiplication, and then you can see it again in the middle slice and finally in the back face.

    You need to specify what you want the result to be or do; i.e how it is to behave, and/or at least mention how much of the final result should vary when varying each of the points in the multiplicands.

     – Mark Hurd Oct 13 '15 at 12:21
  • I think the main problem is that we ought to expect that, regardless of the orientation of the multiplication, the result should always be the same, which I imagine will prove to be difficult. – JMP Jul 06 '18 at 09:25
  • Is there a continuous version of hypermatrices that allows for a similar product? – Wakem Mar 11 '24 at 17:37

1 Answers1

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Think of two 2x2 matricies. There's no 'intrinsic' orientation either. Depending on your SPECIFIC matrix definitions you INTERPRET a secondary concept of orientation based on you application, but structurally, all that matters is 'associative order' (which matrix is on the left vs right).

Beyond that question, this can be seen as a basic abstraction where all operational structures increase by 1 dimension. So, in 2d, have 1d horizontal slice (row) x 1d verical slice (column) = 0d scalar (element in resulting 2d matrix). Here its 2d horizontal slice (row sheet/plane) x 2d vertical slice (column sheet) = 1d element in resulting 3d matrix).

Similarly, you'd expect absraction to nd via n-1 horizontal slice x n-1 verical slice = n-2 element in resulting nd matrix. Tensorially it'd be like a tensor product followed by a contraction (for tensors, contraction reduces dimension (rank) by 2)

user570621
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  • Welcome to Math.SE. You will find the ability to post with mathematical expressions useful as you contribute Questions and Answers. I don't get much illumination out of what you wrote, but then the Question itself was largely pictorial in presentation and lacked the clarification requested as to what the intended "multiplication" would apply to. Certainly there is a central issue of how two $3\times 3$ "slices" should produce a $3\times 1$ result, a concern highlighted in the Question itself. A good Answer would shed some light on that issue. – hardmath Jun 18 '18 at 15:40