Why the columns of $A$ are linearly dependent set? If $A_{m\times n}$ is a matrix such that $\sum_{j=1}^n a_{ij}=0$ for each $i=1,2,…,m,$ then why  the columns of $A$ are linearly dependent set, and hence $\operatorname {rank}(A)<n$?
 A: Hint: the equation that is given actually gives an explicit linear dependence relation between the columns.
A: I assume $m=n$ just to make it look good .
we have $\sum_{j=1}^m a_{ij}=0$ for all $1\leq i\leq m$
i..e, $\sum_{j=1}^m a_{1j}+\sum_{j=1}^m a_{2j}+\dots+\sum_{j=1}^m a_{mj}=0$
i.e., $(a_{11}+a_{12}+\dots+a_{1m})+(a_{21}+a_{22}+\dots+a_{2m})+\dots +(a_{m1}+a_{m2}+\dots+a_{mm})=0$
i.e., $(a_{11}+a_{21}+\dots+a_{m1})+(a_{12}+a_{22}+\dots+a_{m2})+\dots +(a_{1m}+a_{2m}+\dots+a_{mm})=0$
Do you see "the next step" would conclude that columns are linearly dependent. (??)
A: If $m < n$, then
$$\operatorname{rank} A \le \min \{m,n\} = m < n.$$
Let us now assume that $m \ge n$. Note that
$$A \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = 0_{m \times 1},$$
Aiming for a contradiction, we assume that $A$ if of a full rank (i.e., $\operatorname{rank} A = n$). That means that $A$ has a submatrix $B$ of order $n$ such that $B$ is of a full rank (made of the linearly independent rows in $A$), i.e., $B$ is invertible. We know that
$$B \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = 0_{n \times 1},$$
which is a contradiction with the invertibility of $B$, so $A$ is not of a full rank.
