TITLE: The product of two zero-row-sum matrices
QUESTION [5 upvotes]: Suppose $A\in Mat_{m\times n}(\mathbb{R}), B\in Mat_{n\times r}(\mathbb{R}).$ and their row sums are $0$, i.e. 
$\displaystyle\sum_jA_{ij}=\sum_j B_{ij}=0.$ Suppose $AB=C$. Show that the row sum of $C$ is still $0$.

My attempt:
Suppose $A\in Mat_{m\times n}(\mathbb{R})$ is an arbitrary matrix and $B\in Mat_{n\times r}(\mathbb{R})$ is the matrix that the row sum of $B$ is $0$.
The $ith$ row sum is $$\begin{align}\displaystyle\sum_{j=1}^r (AB)_{ij} &=\sum_{j=1}^r\sum_{k=1}^n a_{ik}b_{kj} \\&=\sum_{k=1}^n \sum_{j=1}^r a_{ik}b_{kj} \\&=\sum_k^na_{ik}(\sum_j^r b_{kj}) \\& =\sum_k^n a_{ik}*0,~~~~\text{since the row sum of $B$ is $0$}.\end{align}$$
So, the row sum of product $AB$ is $0$. So, if the row sums of both two matrices are $0$, then the row sum of their product $C$ is also $0$ (because we can always have a zero-row-sum matrix on the right).

Is my proof correct? Can someone check it for me? Thank you very much.

REPLY [4 votes]: Indeed, your proof is correct. Well done.
There is another proof, however, which I find more intuitive since I am used to matrix analysis.  First, we can show that an $m \times n$ matrix $A$ will have a row-sum zero if and only if the product $AM$ is zero, where $M$ is the $n \times 1$ column-vector
$$
M = \pmatrix{1\\1\\ \vdots \\ 1}
$$
From there, we can use the associativity of the matrix product to show that the matrix product $(AB)M$ must be zero, where $M$ in this case has size $r \times 1$. You might find it interesting to observe how the movement of the parenthesis from $(AB)M$ to $A(BM)$ corresponds to the reordering of your sum.