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Given a matrix A =(aij) that sum of the entries on any row of A is equal to δ and the sum of the entries on any column of A is equal to δ. I was wondering if B=(bij) also has these properties, such that the sum of the entries of its rows and columns are equal to µ, where µ,δ∈ R. Whether the sum of entries for (A+B) columns and rows is (δ+µ)? And how you would prove this.

I thought maybe Induction could be used: - Base of induction, where A=(a11) B=(b11), therefore A=(δ) B=(µ) so A+B=(δ+µ), this is clear. - I am stuck for how to prove induction hypothesis.

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    do you **want** to prove this via **induction**? – 2017-01-09
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    Yeah I would like to see it done by induction as well, thanks. – 2017-01-09

2 Answers 2

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It seems you don't need induction, it is almost trivial:

Let $A =(a_{ij}), B = (b_{ij})$ then we assume

$$\sum_{i=1}^n a_{ij} = \sum_{i=1}^n a_{ji} = \delta\ \forall j$$

and

$$\sum_{i=1}^n b_{ij} = \sum_{i=1}^n b_{ji} = \mu\ \forall j$$

Then $(A+B)_{ij} = a_{ij} + a_{ij} \forall ij$. Therefore

$$\sum_{i=1}^n (A+B)_{ij} = \sum_{i=1}^n (a_{ij}+b_{ij}) = \sum_{i=1}^n a_{ij}+ \sum_{i=1}^n b_{ij} = \mu+\delta\ \forall j$$

and

$$\sum_{i=1}^n (A+B)_{ji} = \sum_{i=1}^n (a_{ji}+b_{ji}) = \sum_{i=1}^n a_{ij}+ \sum_{i=1}^n b_{ji} = \mu+\delta\ \forall j$$

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    Very helpful thanks, one typo (A+B)ij = aij +bij not (A+B)=aij + aij. – 2017-01-09
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One can prove that if some matrix $A$ has that property, then $A$ is the linear combination of some permutation matrices. Therefore it becomes clear that $A+B$ will have that property: given that both $A$ and $B$ are linear combinations of permutation matrices, $A+B$ also is and the sum of each row/column in $A+B$ is equal to the sum of each row in $A$ with each row in $B$.

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    Do you mean _linear combination_ of permutation matrices (rather than sum)? – 2017-01-09
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    @HenningMakholm maybe I should write that because the sum of each row/column need not be an integer. Thanks. – 2017-01-09
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    Even if the common sum _is_ an integer you still need linear combinations. Consider $\begin{pmatrix} 0.5 & 0.5 \\ 0.5 & 0.5\end{pmatrix}$ or $\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$. – 2017-01-09