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I have been contemplating extending the definition of cross product for matrices, and I wonder if this has been done before.

Basically my definition is, given two 3x3 matrices: $A=(a_{ij})_{i,j=1} ^ 3$ and $B=(b_{ij})$ then $A\times B=(A_i \times B_j)_{i,j=1}^3$ where $A_i=(a_{i1},a_{i2},a_{i3})$ the same with B (just replace A with B a with b and i with j).

Now I checked that it's linear with regard to simple matrix addition $$A\times (B+C)= A\times B + A\times C$$

And obviously it's antisymmetric, I am not sure if there's a nice connection with matrix multiplcation, I mean $A\times (BC)=?$ not sure if there's a simple identity here.

Has this been done already, where may I read on this?

Thanks.

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    I'm not sure I follow... Since each $A_i$ and $B_j$ are vectors, isn't each $A_i \times B_j$ also a vector? Does this mean that each entry of the matrix $A \times B$ is a vector?2011-04-14
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    I may be misunderstanding something, but it seems you are multiplying matrices with scalar entries to get a matrix with vector entries ($A_i\times B_j$ is a vector), so this wouldn't be an operation within the set of matrices with scalar entries; also it's not obvious how it could be extend to be an operation, say, within some algebra of matrices with tensors of arbitrary orders as entries; you'd need to define how to multiply the resulting matrices of vectors and so on for this to lead anywhere.2011-04-14
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    @Joriki, that's right The product gives a matrix of vectors entries instead of scalar entries.2011-04-14
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    Sorry for posting my reply too quick, I am just checking if there's any one who used this definition before and to what extent can it be applied.2011-04-14
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    @MathematicalPhysicist: So if this is a matrix of vectors, what happens then? Do you leave it at that, or can these in turn be multiplied? I suspect that for this to be interesting, you'd either have to also define how to multiply these, so you get some closed algebraic structure, or you'd have to say something about what they represent or how they relate to something else involving the original matrices -- as it stands, it looks like a somewhat arbitrary operation on the column vectors.2011-04-14
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    *The king's business requires haste*, as they say, but not math.SE questions nor replies.2011-04-14
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    This is just extension to $\mathbb{R}^3 \oplus \mathbb{R}^3 \oplus \mathbb{R}^3$, it is improper to call it generalization for matrices.2011-04-14

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