# Is there any thing like matrices for three-dimensional linear algebra?

I feel this must be an old question, but so far I have never seen it mentioned in any of the books I know. Most of modern treatment of advanced linear algebra is centered on tensor products and eigendecomposition, etc. So here is the question:

We know ordinary matrices provided a way of measuring a list of vectors $a_{i},i\in I_{n}$ such that for two different list of vectors we have the addition and multiplication. Further multiplication is associative, generates another list of vectors. Regardless of the specifics in row and column operations, I want to ask:

(1) Is it possible to define this for an array of vectors $a_{ij},i,j\in I_{n}$? I wish it to fulfill at least associativity, has an identity element, and makes some sense(such that respect to the individual $a_{ik}$ with $k$ fixed it resemble matrix multiplication?).

(2) Is it possible to define an exterior operation on it that resembling ordinary determinant?

(3) If the answer for above questions are both positive, is there any way to define a canonical decomposition of such $3$-dimensional matrices? I wish there be something resembling various decomposition theorems in ordinary linear algebra we have.

I had been thinking about this since I was a freshmen but I never got much progress (maybe I am too lazy). I will be happy to see if such an operation exists (or the reason why it should not exist).

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I don't understand your second paragraph. How do matrices "measure" lists of vectors? And how does the measuring somehow give addition and multiplication of vectors (what multiplication?)? – Arturo Magidin Mar 9 '11 at 1:04
oh, you need to think geometrically: a list of vectors created a subspace with dimension $k<n$. We associate it to some point in $R^{n^{2}}$. The respective scalar multiplication, addition, etc are preserved in this association, and multiplication two of them give you a new point this way. – Kerry Mar 9 '11 at 1:14
A list of vectors in $\mathbb{R^n}$ spans a subspace of dimension at most $n$ (it doesn't "create" subspaces). This gives you a point in the Grassmanian $\mathrm{Gr}(k,n)$, where $k$ is the dimension of the subspace; not sure how you are going all the way to $\mathbb{R}^{n^2}$. What you are talking about is a linear map (when you talk about "association" and "preservation"). And Still not clear what you mean when you talk about multiplying "two of them". How do you multiply two vectors in an arbitrary vector space? – Arturo Magidin Mar 9 '11 at 1:24
oh, multiplying two of the lists. I do not think you need to use Grassmanian in here. – Kerry Mar 9 '11 at 1:45
@user7887: How do you multiply two lists of vectors? In general, I do not know how to multiply two vectors to get a vector (except in the special case of one dimensional spaces, or in specific instances such as when I have algebras rather than vector spaces). What does it mean to "multiply two of the lists"? I think you probably are saying true things, just in a rather non-standard way, and this is getting in the way of my understanding what you are trying to say. – Arturo Magidin Mar 9 '11 at 17:26

## 2 Answers

There was a good discussion of this on mathoverflow.

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This answer is very helpful. – Kerry Mar 9 '11 at 1:20

I believe that three dimensional matrices are called Tensors. A study of them is practically a field in itself.

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