Connection between the sum of a matrix's columns and the largest eigenvalues 
Let's say I have a matrix $A \in \mathbb{R}^{n \times n}$. Matrix $A$ satisfies the condition that the absolute value sum of its columns is always less than or equal to $1$. Show that all eigenvalues are less than or equal to $1$.


What's the connection between the columns and the eigenvalues?
 A: $A$ has the same eigenvalues as its transpose, that I will denote $B$. For $B$, the hypothesis means that $\sum_{j=1}^n|a_{ij}|\leq 1$ for all $i$. If $x\neq 0$ is an eigenvector for $\lambda$, and $i$ is such that $x_i=\lVert x\rVert_{\infty}$, then we have $\sum_{j=1}^na_{ij}x_j=\lambda x_i$ hence $\sum_{j\neq i}a_{ij}x_j=(\lambda-a_{ii})x_i$ and 
$$\lVert x\rVert_{\infty}|\lambda-a_{ii}|\leq \sum_{j\neq i}|a_{ij}|\cdot |x_j|\leq \sum_{j\neq i}|a_{ij}|\cdot\lVert x\rVert_{\infty}\leq (1-|a_{ii}|)\cdot\lVert x\rVert_{\infty}.$$
As $x\neq 0$, we get $|\lambda-a_{ii}|\leq 1-|a_{ii}|$ hence $|\lambda|\leq 1$. 
(we proof the Gershgorin circle theorem)
A: More directly, we have:

for any (square) $B$ and any matrix norm $\|\cdot\|$ subordinate with
respect to some vector norm, we have $\rho(B)\leq\|B\|$. Hence, even
though it does not contradict or prove anything about the original
question, it hints that the statement is false from the start.

The norm you are referring to in the question is the infinity norm, so if the infinity norm is less than 1, so is the spectral radius.
You may check "Matrices with Applications in Statistics" by Graybill, Theorem 5.6.7.
