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Let $C, R \in \mathbb{Z}^n$. If there is an $n \times n$-matrix $M$ with all entries being integers such that the sum of the entries of column $k$ equals $C(k)$, and the sum of the entries of row $k$ equals $R(k)$, for all $k \in \{1,...,n\}$, we say that the pair $(C, R)$ is representable.

For $(C, R)$ to be representable, it is easy to see that we need $\sum_{i=1}^n C(i) = \sum_{i=1}^n R(i)$.

Does the converse hold?

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    $\begingroup$ For non-negative integer vectors, this is a consequence of the RSK correspondence (Knuth, "Permutations, Matrices, and Generalized Young Tableaux", Pacific J. Math., 1970). Perhaps you already know that. $\endgroup$ Mar 20, 2018 at 16:21
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    $\begingroup$ Maybe I am misunderstanding the question, but given $(C,R)$ with $\sum C(i)=\sum R(i)$, can't they be realized as representable via the upper-triangular matrix $$M=\begin{bmatrix}C(1)&C(2)-R(2)&\ldots&C(n)-R(n)\\ &R(2)&&\\ &&\ddots&\\ &&&R(n)\end{bmatrix}$$ since the sum of the first row is $\sum C(i)-((\sum R(i))-R(1))=R(1)$? $\endgroup$
    – Sean Clark
    Mar 20, 2018 at 16:26

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