If $A$ is a matrix with a row of zeros, then the product $AB$ also has a row of zeros I know there is a similar question posted on this site, but it does not have a proof, and I would rather have my proof for the problem criticized than look at a proof as an answer on the other post.
Suppose that $A$ is an $m \times r$ matrix with a row of zeros in the $i$-th row and let $B$ be an arbitrary $r \times n$ matrix.
Denote the $i$-th row vector of $A$ as $\mathbf{a}_i = \mathbf{0}$ where $\mathbf{0}$ is the zero row vector. Then the product $\mathbf{a}_i B$ is the $i$-th row vector of the matrix $AB$ denoted $(\mathbf{AB})_i$ By definition of matrix multiplication, we have
$$((\mathbf{AB})_i)_{ij} = (\mathbf{a}_i B)_{ij} = \sum_{k=1}^r a_{ik}b_{kj} = 0$$
for each $j = 1, \dotsc, n$.
Therefore,  
$$(\mathbf{AB})_i = \mathbf{0}$$
so the matrix $AB$ has a row of zeros in the $i$-th row.
 A: It seems fine. A minor comment is rather than using $((AB)_i)_{ij}$, why not just use $((AB)_i)_{j}$
A: If you transpose the problem, your question becomes the following : if the matrix $B$ has a column of zeros, then the matrix $AB$ as a column of zeros. If you translate this problem in terms of linear maps, the question becomes : if $f : V_1 \to V_2$ and $g : V_2 \to V_3$ are linear maps and $v \in V_1$ satisfies $f(v) = 0$, then $g(f(v)) = 0$. This is super mega trivial. All you have to do now is translate! 
(Hint : you can do the "translation" by letting $V_1 = \mathbb R^n$, $V_2 = \mathbb R^r$, $V_3 = \mathbb R^m$ and let $v$ be one of the standard basis vectors of $\mathbb R^n$.)
P.S. Your proof sounds good to me. However, I hope that you are aware that writing $_{ij}$ as an index is not just an indicator that we are interested in a coefficient of the matrix but rather the coefficient in row $i$ and column $j$. Therefore writing $( (AB)_i)_{ij}$ should be technically written $((AB)_i)_{1j}$ if you consider it as a $1 \times n$ matrix, or $((AB)_i)_j$ if you consider it as a vector (i.e. not a $2$-dimensional array, but a one-dimensional one). My preference would be to fix $i$ (since we are studying the $i^{\text{th}}$ row and write $(AB)_{ij}$, which is much clearer.
Hope that helps,
