MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we have two vectors $x\in \mathbb{R}^n$ and $y\in \mathbb{R}^m$. I could define the mapping $$ T: \mathbb{R}^n\times \mathbb{R}^m \rightarrow \mathbb{R}^{n\times m} $$ as follows $$ T(x,y) = ( x_i+y_j )_{i,j=1}^{n,m}, $$ i.e. $T(x,y)$ is $n\times m$ matrix with elements equal sum of corresponding vectors. If instead of $x_i+y_j$ one considers $x_i \cdot y_j$ then it is matrix multiplication of two vectors ($x\cdot y^T$, where $x,y$ are column vectors). But what about sum?

Is that operation known and studied somewhere?

share|cite|improve this question
It's not unlikely that this comes up in tropical geometry (, where the "product" is the original addition. – Tom De Medts Jun 17 '13 at 14:42
"Outer sum"? I think that's what the programming language APL called it. – Deane Yang Jun 17 '13 at 15:53
I've occasionally heard "tensor sum". – Noah Stein Jun 17 '13 at 17:28
Thank you Tom for direction. According to… It is indeed tropical product of two vectors $a \otimes b^T$. – Fedor Nikitin Jun 17 '13 at 17:52

I'd write your operation simply as $x e^T + ey^T$, where $e$ is the vector of all ones. I have never encountered this operation directly studied in a research context (my guess is that it's simply "not interesting enough", since it's a small variation to the common theme of a rank-2 matrix), but I have seen several times the related matrix with elements $$ T_{ij}= \frac{1}{x_i - y_j}, $$ which is known as Cauchy matrix.

share|cite|improve this answer
There are two vectors of ones $e_m$ and $e_n$ should be above because I assume that $x$ and $y$ are of dimensions. – Fedor Nikitin Jun 17 '13 at 20:11
Yes, you are correct. – Federico Poloni Jun 18 '13 at 6:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.