Suppose $Ax = 0\Longrightarrow Bx=0$, then $A=PB$? Suppose $A$ and $B$ are not zero matrices. Suppose they have the same dimensions $n\times n$ and have real entries. Suppose that if $Ax = 0$ then $Bx = 0$. Is it true that there exists $C$ such that $A = CB$? Is is true that there exists $D$ such that $B = AD$?
I am not sure if this is the right way to think about it. I think of $B$ as a system of equations, and $A$ as the reduced form of $B$. Then, if $x$ solves $A$, it solve $B$. So we have $A = \text{[some row operations matrices]}B$. 
Is my argument correct? Is there any other argument? Thank you.
 A: This only answers one of your two questions, and that only incompletely.
$\newcommand{\rank}{\operatorname{rank}}$We have by assumption $$\ker A \subseteq \ker B $$ Now by the Rank-Nullity theorem (since the codomains of $A$ and $B$ have the same dimension): $$\rank(B) \le \rank(A) $$ If this inequality is strict, there might exist a matrix $C$ such that $B=CA$, however there cannot exist a matrix such that A = CB, because $$\rank(CB) \le \min\{\rank(C),\rank(B) \} < \rank(A)$$ see this question. If the inequality is not strict however then there is less we can say.
A: Let $E$ be a vector space of dimension n and $(e_1, ..., e_n)$ be a base of $E$.
Assume $a, b \in L(E)$ two endomorphisms of $E$.
If $a = id$ is the identity and $b = 0$, we clearly have $\ker a  = \{0\}\subset \ker b = E$ (ie $Ax= 0 \implies Bx= 0$ for any matrix $A$ and $B$ representing $a$ and $b$) however  $\forall c\in L(E)$ we have  $c\circ b =0\ne a$ thus the second property does not hold.
If $a(e_i) = e_1$ for all i and $b=id$, we have again $\ker a = \{0\} \subset \ker b = \{0\}$ however one can see that there exists no endomorphism $c\in L(E)$ such that $b = c\circ a$ (as long as $n\ge 1$ ...) thus the first property does not hold
