[Math] Interpretation of Einstein notation for matrix multiplication

linear algebralinear-transformationsmatricestensors

Consider the matrix product $C = AB$ where $A \in \mathbb{R}^{m \times n}, B \in \mathbb{R}^{n \times p}$. The Einstein summation notation for this is
$$
c_{ik} = a_{ij}b_{jk}.
$$

Is there any example from math, physics, engineering, statistics etc. where each term in the sum $a_{ij}b_{jk}$ has a meaningful interpretation?

Equivalently, I guess: suppose we did not use Einstein summation notation, and instead defined $D \in \mathbb{R}^{m \times n \times p}$ as:
$$
D_{ijk} = A_{ij}B_{jk}.
$$

Would the elements of $D$ mean anything with respect to $A, B, C$ or the linear transformations represented by these matrices? Obviously $C_{ik} = \sum_{j} D_{ijk}$, but what does this mean?

Or suppose, instead of summing over the second index of $D$ to get $C$, we instead summed over the first index of $D$ to get $E \in \mathbb{R}^{n \times p}$:
$$
E_{jk} = \sum_{i} D_{ijk}.
$$

Obviously this cannot be represented in Einstein notation. So does this not have meaning as a tensor contraction, and unlikely to have a physical interpretation?

Edit: For the powers of adjacency matrices (eg $C = A^2 = AA$), the terms do have an interpretation. Iff $a_{ij}b_{jk} = 1$ (not using Einstein notation), then there is a path from node $i$ to node $k$ through node $j$.

Best Answer

Define a third order tensor whose components are equal to zero unless all three indices are equal $${\cal H}_{ijk} = \begin{cases} 1 \quad{\rm if}\; i\!=\!j=\!k \\ 0 \quad{\rm otherwise} \\ \end{cases} $$ Then you can use Einstein notation to write $${\cal D}_{ijk} = A_{ip}{\cal H}_{pjs}B_{sk}$$ This tensor is a useful addition to standard matrix algebra.

It can be used to generate a diagonal matrix $A$ from a vector $a$ (using a single-dot product) $$A = \operatorname{Diag}(a) = {\cal H}\cdot a \quad\implies A_{ij} = {\cal H}_{ijk}\,a_k$$ or to create a vector $b$ from the main diagonal of a matrix $B$ (using a double-dot product) $$b = \operatorname{diag}(B) = {\cal H}:B \quad\implies b_{i} = {\cal H}_{ijk}\,B_{jk}$$ or simply as a way to write ${\cal D}$ without resorting to index notation $${\cal D} = A\cdot{\cal H}\cdot B$$

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