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For an $m \times n$ fat matrix ($m<n$)

$$A = \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\ a_{21} & a_{22} & \ddots & \cdots & a_{2n} \\ a_{31} & \cdots & \ddots& \ddots & \vdots \\ \vdots & \cdots & \cdots & \ddots & \vdots \\ a_{m1} & \cdots & \cdots & \cdots & a_{mn} \end{pmatrix}$$

with $a_{ij} \geq 0$ and at least one positive entry in every column, why is the smallest column sum always smaller than the largest row sum?

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The $n$ column sums add up to $\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$, so the smallest column sum is at most the average column sum, which is $\dfrac{1}{n}\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$. The $m$ row sums add up to $\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$, so the largest column sum is at least the average row sum, which is $\dfrac{1}{m}\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$.

Hence, the smallest column sum is less than or equal to $\dfrac{1}{n}\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$, which is less than $\dfrac{1}{m}\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$, which is less than or equal to the largest row sum.

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  • $\begingroup$ Thank you very much. Is it also the case that the largest row sum is always greater than the largest column sum? $\endgroup$
    – Analysis
    Commented Aug 6, 2020 at 14:19
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    $\begingroup$ No, that isn't true. Consider $\begin{bmatrix} 10 & 1 & 1 \\ 10 & 1 & 1\end{bmatrix}$. $\endgroup$
    – JimmyK4542
    Commented Aug 6, 2020 at 18:22

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