Let be $m,b\in\Bbb{N}$ and $A=(a_{ij})\in\Bbb{R}^{m \times n} $ It applies that $\sum_{j=1}^{n} a_{ij}= 0, \forall i\in \{ 1,\cdots,m\}$ Show that $Rank(A) < n$.
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I guess it's a crucial point that the $rank(A) < n$ and it's not $rank(A) = n$
Actually I have difficulties to understand this part $\sum_{j=1}^{n} a_{ij}= 0, \forall\in \{ 1,\cdots,m\}$.
Do we accumulate the elements $ a_{i1}+ \cdots a_{in}$? So is it the sum of every elements in the matrix? But it makes no sense to me. Thank you for your helps in advance.
Each and every column of the matrix sums to $0$. Hence, the $n$-dimensional vector of all $1$'s is mapped to the null vector. This gives a non-trivial nullspace, hence rank is $<n$.