The effect of changing one of the matrix of matrix multiplication on the rank of the resulting new matrix? Consider matrix multiplication $C=AB$, where $A\in\mathbb{F}_q^{m\times n}$, $B\in\mathbb{F}_q^{n\times k}$, $q$ is the finite field size (the finite field constraint is actually not relevant to the problem). The ranks of matrices are denoted as $\text{rank}(A)$, $\text{rank}(B)$ and $\text{rank}(C)$. My question is, suppose that I change some rows of $B$ to obtain a new matrix $B'$ which has $\text{rank}(B')<\text{rank}(B)$ and denote $C'=AB'$. Is that safe to say that $\text{rank}(C')<\text{rank}(C)$ (or $\text{rank}(C')\leq \text{rank}(C)$)? If yes, how to rigorously prove?
What if another constraint is added to the question that $B'$ is obtained by replacing one/several row(s) of $B$ with linear combinations of existing rows of $B$?
Similarly, is that possible to have a similar argument on the change of rank if $C'=A'B$ where $\text{rank}(A')<\text{rank}(A)$? Thanks.
 A: There is no general statement. First, note that there is no hope to get strict inequality (just take $A$ as the zero matrix). Then take this example
$A=\left(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}\right)$
$B=\left(\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\\end{matrix}\right)$
$B^{'}=\left(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\\end{matrix}\right)$
where passing from $B$ to $B^{'}$ I changed the first three rows. The rank of $B^{'}$ is $2$, the one of $B$ is $1$, but $AB$ has rank $1$ and $AB^{'}$ rank $0$.
For the symmetric question you can argue with similar counterexamples.
For your further constraints:
$A=\left(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}\right)$
$B=\left(\begin{matrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\\end{matrix}\right)$
$B^{'}=\left(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\\end{matrix}\right)$
The constraints are respected, since the new first row is the old second one, the new second and third are equal to the old fourth one. As you can see, the result is the same as above.
A: No, that's not safe. Suppose $A = e_1e_1^T$, and $B = e_2e_2^T + e_3e_3^T$, where the $e$s are the standard basis vectors. Then, $\mathrm{rank}(AB) = 0$. Now, let $B' = e_1e_1^T$ so that $\mathrm{rank}(B') < \mathrm{rank}(B)$. Then, $\mathrm{rank}(AB') = 1 > \mathrm{rank}(AB)$.
