Matrices and rank inequality Let $A \in K^{m\times n}$ and $B \in K^{n \times r}$
Prove that min$\{rk(A),rk(B)\}\geq rk(AB)\geq rk(A)+rk(B)-n$
My attempt at a solution:
$(1)$ $AB=(AB_1|...|AB_j|...|AB_r)$ ($B_j$ is the j-th column of $B$), I don't know if the following statement is correct: the columns of $AB$ are a linear combination of the columns of $B$, then $rk(AB) \leq rk(B)$. 
$(2)$In a similar way, $AB= \begin{bmatrix} —A_1B— \\ \vdots \\ —A_jB— \\ \vdots \\—A_mB—   \end{bmatrix}$ ($A_j$ denotes the j-th row of $A$), so the rows of $AB$ are a linear combination of the rows of $A$, from here one deduces $rk(AB)\leq rk(A)$.
From $(1)$ and $(2)$ it follows $rk(AB)\leq min\{rk(A),rk(B)\}$.
This is what I've done so far. I am having doubts with, for example (1), this statement I've conjectured: the columns of $AB$ are a linear combination of the columns of $B$, then $rk(AB) \leq rk(B)$, but wouldn't this be the case iff $AB=(\alpha_1B_1|...|\alpha_jB_j|...|\alpha_rB_r)$ with $\alpha_1,...,\alpha_n \in K$ instead of $(AB_1|...|AB_j|...|AB_r)$ ? This is a major doubt I have, the same goes for (2).
I need help to show the inequality $rk(AB)\geq rk(A)+rk(B)-n$

Use the dimension theorem on $A|_{\mathop{\rm Im}(B)}:\mathop{\rm Im}(B)\subseteq K^n\longrightarrow K^m$. Then $$\begin{align}\dim(\mathop{\rm Im}(B))&=\dim(\mathop{\rm Im}(A|_{\mathop{\rm Im}(B)}))+\dim(\mathop{\rm Ker}(A|_{\mathop{\rm Im}(B)}))\\&=\dim(\mathop{\rm Im}(AB))+\dim(\mathop{\rm Ker}(A)\cap\mathop{\rm Im}(B))\\&\leq\dim(\mathop{\rm Im}(AB))+\dim(\mathop{\rm Ker}(A)),\end{align}$$
and since $\dim(\mathop{\rm Ker}(A))=n-\dim(\mathop{\rm Im}(A))$, the claim follows.
