# Compare full-rank probabilities of products of random matrices

Consider two matrices $C_1=A\times B_1$ and $C_2=A\times B_2$, where $A\in\mathbb{F}_q^{N\times K}$, $B_1\in\mathbb{F}_q^{K\times M}$ and $B_2\in\mathbb{F}_2^{K\times M}$; $M\leq N\leq K$.

It is known that $A$ is a random matrix with elements uniformly randomly chosen from the finite field $\mathbf{F}_q$, where $q$ is the field size; $B_1$ is in the form of $$B_1=\left[\begin{array}{c}I_{M\times M}\\P \end{array}\right],$$ where $I_{M\times M}$ is a $M$-size identity matrix and $P$ is of size $(K-M)\times M$ and consists of elements uniformly randomly chosen from $\mathbb{F}_q$; $B_2$ is a random matrix with all of its elements uniformly randomly chosen from $\mathbb{F}_q$. It is clear that $\mathrm{Pr}\{\text{rank}(B_1)=M\}=1$ while $\mathrm{Pr}\{\text{rank}(B_2)=M\}<1$ (it approaches $1$ as $q$ increases, but is not relevant here). My question is, is that possible to prove that $\mathrm{Pr}\{\text{rank}(AB_1)=M\}\geq \mathrm{Pr}\{\text{rank}(AB_2)=M\}$ under these conditions? What if $A$ is a random matrix with elements chosen from $\mathbb{F}_q$ with an unknown distribution?

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