If $A$ is a $m \times n$ matrix and $B$ is a $n \times k$ matrix, where $k>n$, and $B$ is full rank, is $\text{rank}(AB)$ = $\text{rank}(A)$?
Edit: Hey apologies for the lack of information. Relatively new to stack exchange and linear algebra.
I've been trying to find similar questions, most are either square matrices or use Sylvester Rank Inequality, which I'm not supposed to use yet.
I was thinking since $B$ is full rank, $\text{rank}(B)$ = n. So ${rank}(AB)$ <= min($m, $n), hence using linear independent columns to find the rank, but not sure if it would be the correct way to go about it
Since "$\operatorname{rank}$" refers to the dimension of the image and $k>n$ (i.e., $B$ maps from a larger space to a smaller one), the fact that $B$ has full rank means that it is surjective. The image of $AB$ therefore equals the image of $A$. Hence, indeed, $\operatorname{rank}(AB) = \operatorname{rank}(A)$.