Columns of $AB$ are independent $\rightarrow$ Columns of B are independent Let $A\in M_{mxn}(F)$ and $B\in M_{nxk}(F)$

Columns of $AB$ are independent $\rightarrow$ Rank(AB)=k on the other hand $Rank(AB)\leq min(Rank(A),Rank(B))$ therefore $k \leq min(Rank(A),Rank(B))$ due to $B\in M_{nxk}(F)$ $k \leq Rank(B)=k$ and therefore $B$ columns are independent
is it a valid proof? Should I use the Rank?
 A: The proof is correct. But you could also have a different view on this. Let $B=(b_1,\dots, b_k)$ with $b_i$ the columns of $B$; then $AB=(Ab_1,\dots,Ab_k)$, that is, the columns of $AB$ are the vectors $Ab_i$. Now, if there is a non-trivial relation $\sum \lambda_ib_i=0$ (that is, some $\lambda_i\ne 0$) and the $b_i$ are linearely dependent then so are the vectors $Ab_i$ as they satisfy the same relation: $$0=A0=A(\sum \lambda_i b_i)=\sum A(\lambda_i b_i)=\sum \lambda_i Ab_i$$
and so the vectors $Ab_i$ are also linearely dependent.
A: There is a small glitch in your proof; you correctly state that
$$
k=\operatorname{rank}(AB)\le\min(\operatorname{rank}(A),\operatorname{rank}(B))
\le\operatorname{rank}(B),
$$
so, in particular, $k\le\operatorname{rank}(B)$.
On the other hand $\operatorname{rank}(B)\le\min(n,k)\le k$ (here is the glitch in your proof), so we can conclude that
$$
\operatorname{rank}(B)=k
$$
and so the columns of $B$ are linearly independent.
Actually, a more general result can be proved. Suppose $b_{i_1},b_{i_2},\dots,b_{i_r}$ are some columns of $B$ and that the corresponding columns of $AB$ are linearly independent. Then also $b_{i_1},b_{i_2},\dots,b_{i_r}$ are linearly independent.
Indeed the corresponding columns in $AB$ can be written $Ab_{i_1},Ab_{i_2},\dots,Ab_{i_r}$ and a zero linear combination
$$
0=\alpha_1b_{i_1}+\alpha_2b_{i_2}+\dots+\alpha_rb_{i_r}
$$
gives
$$
0=A0=\alpha_1Ab_{i_1}+\alpha_2Ab_{i_2}+\dots+\alpha_rAb_{i_r}
$$
so, by assumption, $\alpha_1=\alpha_2=\dots=\alpha_r=0$.
This is a particular case of an even more general theorem.
Suppose $f\colon V\to W$ is a linear map and that, for $v_1,v_2,\dots,v_r\in V$, the set
$$
\{f(v_1),f(v_2),\dots,f(v_r)\}
$$
is linearly independent. Then $\{v_1,v_2,\dots,v_r\}$ is linearly independent.
