Let $T$ be a linear transformation between vector spaces $$T: V \rightarrow W$$
Once we select a basis for $V$ and $W$, we can represent $T$ by a matrix.
Given a bilinear functional $$B:V \times W \rightarrow \mathbb{R}$$ Once we select a basis for $V$ and $W$ we can represent $B$ by a matrix.
But what if we had a third vector space equipped with a basis, say $Z$, and we had a bilinear function $$M: V \times W \rightarrow Z$$
I believe we could no longer represent this function by a matrix. But what if we had an array in 3 dimensional space? Then could we represent $M$ by a matrix?
What if we had an $n$-linear function $$V_1 \times ... \times V_n \rightarrow W$$ or perhaps a $n+1$-linear functional $$V_1 \times ... \times V_n \times V_{n+1} \rightarrow \mathbb{R}$$
Could you represent each of these as an $n+1$-dimensional array?
Thank you
Yes. To keep it simple, let's assume all the $V_i$ are the same, OK? Pick a basis $v_1, \ldots, v_k$ for $V$. Take your $n$-linear functional $f$, and compute $$ f(v_1, v_1, \ldots, v_1) = c_{1,1,1, \ldots, 1} \\ f(v_1, v_1, \ldots, v_2) = c_{1,1,1, \ldots, 2} \\ \vdots\\ f(v_k, v_k, \ldots, v_k) = c_{k,k,k, \ldots, k} $$ i.e., compute all $k^n$ coefficients by plugging in all possible combinations of subscripts. This is an $n$-index array of coefficients. For $n = 1$, this is a list; for $n = 2$ (the bilinear case) it's a matrix; for larger $n$ it's a cube or hypercube or whatever of coefficients.
And these completely determine your function, because of $n$-linearity.
In the event that your function has some degree of symmetry (e.g., for a 2-arg function, if you have $f(v_1, v_2) = f(v_2, v_1)$), then there are some relations among the coefficients. (In the 2-arg case, it means that the coefficient matrix is symmetric; arguably the anti-symmetric case is more interesting, at least for geometry, where my interests lie.)