# Cross product for matrices

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.

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? –  Jesse Madnick Apr 14 '11 at 8:22
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. –  joriki Apr 14 '11 at 8:23
@Joriki, that's right The product gives a matrix of vectors entries instead of scalar entries. –  MathematicalPhysicist Apr 14 '11 at 8:27
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. –  MathematicalPhysicist Apr 14 '11 at 8:29
@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. –  joriki Apr 14 '11 at 8:31

Well, one way to think about it is this: instead of thinking about matrices, you should think about linear transformations. (The nomenclature below maybe a bit cukoo, since I am not really an algebraist...)

Let $V, W$ be two three-dimensional real (doesn't matter, any field will do; I choose the reals for simplicity of notation) vector spaces. Let $A:V\to W$ and $B:V\to W$ be linear transformations. Treat $V$ to be part of the (graded) tensor product algebra, that is, consider the (infinite) set of vector spaces $\{ \mathbb{R} = \otimes^0 V, V, V\otimes V, V\otimes V\otimes V, \ldots \}$, with the multiplication operation given by the tensor product. Treat $W$ as part of the (graded) exterior algebra, that is the set $\{ \mathbb{R}, W, W\wedge W, W\wedge W\wedge W\}$ (since $W$ has dimension 3, I stop at the third order exterior product; you can generalize this to dimension $n$ if you want) with the multiplication map given by the exterior product.

Your $A\times B$ is the map $V\otimes V \to W\wedge W$ induced by $A,B$ via the tensor product over $V$ and exterior product over $W$. In other words, your $\times$ is the induced bilinear map from $L(V,W)\times L(V,W) \to L(V\otimes V,W\wedge W)$ via the $(V,\otimes)$ and $(W,\wedge)$.

Seen this way it is pretty clear that there should not be, in general, a nice interpretation of $A \times (BC)$ in terms of $A\times B$ and $A\times C$. In general you have here three vector spaces $U,V,W$ with $A:U\to W$, $B:V\to W$ and $C:U\to V$, so that ($BC:U\to W$ and the "cross prodcut" against $A$ can be defined). But then since $B$ and $C$ each has domain/codomain different from $A$, you cannot "factor".

(You ask, but aren't all three-dimensional real vector spaces isomorphic? Yes, but since in the domain side we embed into the tensor algebra and on the codomain side we embed into the exterior algebra, you should think of them as distinct objects.)

I'm sure someone better versed in Category Theory can come along and give a cleaner presentation of this.

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Oh, right, forgot one minor detail. The above of course assumes you are familiar with the identification between the cross product of vectors on $\mathbb{R}^3$ with the exterior product. –  Willie Wong Apr 14 '11 at 11:07