Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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?
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.
