# Column-sum and row-sum for fat matrices

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

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

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?

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.
• No, that isn't true. Consider $\begin{bmatrix} 10 & 1 & 1 \\ 10 & 1 & 1\end{bmatrix}$. Aug 6, 2020 at 18:22