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Question:

is there a name for the following operation
$$\sum_{i=1}^n\sum_{j=1}^mx_iy_j^T,\ x_i,y_j\in \mathbb{R}^k$$ i.e. for generating a square matrix that is the sum of the cartesian product of a set $X$ of column vectors with a set $Y$ of row vectors?

The reason for asking is that it came up in the solution of a problem.

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  • $\begingroup$ I haven't seen anything more fine-grained than just saying that it's the dyad formed from $\sum_{i=1}^{n} x_i $ and $\sum_{j=1}^{m} y_j $. $\endgroup$ Commented Jul 18, 2023 at 13:13
  • $\begingroup$ this is one way to write the singular value decomposition of a matrix $\endgroup$ Commented Jul 18, 2023 at 13:39
  • $\begingroup$ @CarloBeenakker do you have a reference, where that is explained? I think you have noticed that the sum of dyadic products is invariant under a permutation of the column vectors and also of the row vectors; I just want to be sure that my question didn't give rise to misinterpretation. $\endgroup$ Commented Jul 18, 2023 at 14:55
  • $\begingroup$ ah indeed, that's different from the SVD. $\endgroup$ Commented Jul 18, 2023 at 15:07

1 Answer 1

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The name for the above operation is 2nd rank tensor composition / matrix multiplication.

I will explain in a matrix language, since it is easier to grasp. Assume there is an orthonormal basis in which all vectors $ \mathbf{x}_{k},\mathbf{y}_{k} $ are expressed

If you organize the $ \mathbf{x}_{k}$ vectors as columns of an $n\times n$ matrix, say $\mathbf{A}$, and the $ \mathbf{y}_{k}$ vectors as the rows of another $n\times n$ matrix, say $\mathbf{B}$, then the expression above is exactly the product $\mathbf{AB}$.

This is based on the fact that the product of any two matrices $\mathbf{A,B}$ can be seen as: $$\mathbf{AB} = \sum_{k=1}^{N} \mathbf{c}_k^{\mathbf{A}} \otimes \mathbf{l}_k^{\mathbf{B}},$$ where $\mathbf{c}_k^{\mathbf{A}}$ are the columns of $\mathbf{A}$ and $\mathbf{l}_k^{\mathbf{B}}$ are the rows of $\mathbf{B}$, while $N$ denotes the number of columns of $A$ (the same as the number of rows of $B$, such that the multiplication could be performed)

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  • $\begingroup$ Did you mean the rows of $B$ are $y_k^T$? $\endgroup$ Commented Dec 3, 2023 at 12:19
  • $\begingroup$ You seem to be assuming $k=n=m$ (in the nomenclature of the OP)? I don't see that in the OP. $\endgroup$ Commented Dec 3, 2023 at 15:02

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