Matrices and rank inequality exercise

[mahler1]

The problem statement

Let $A ∈ K^{m×n}$ and $B ∈ K^{n×r}$
Prove that min$\{rg(A),rg(B)\}≥rg(AB)≥rg(A)+rg(B)−n$

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 $rg(AB)≤rg(B)$.

(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 $rg(AB)≤rg(A)$.

From (1) and (2) it follows $rg(AB)≤min\{rg(A),rg(B)\}$.

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 $rg(AB)≤rg(B)$, but wouldn't this be the case iff $AB=(α_1B_1|...|α_jB_j|...|α_rB_r)$ with $α_1,...,α_n ∈ K$ 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 $rg(AB)≥rg(A)+rg(B)−n$

 

[LCKurtz]

The problem statement

Let $A ∈ K^{m×n}$ and $B ∈ K^{n×r}$
Prove that min$\{rg(A),rg(B)\}≥rg(AB)≥rg(A)+rg(B)−n$

Not sure I will answer your question, but I would suggest you might get more responses if you defined your terms. What is $K^{mxn}$? What does $rg(A)$ mean?