TITLE: Prove Sylvester rank inequality: $\text{rank}(AB)\ge\text{rank}(A)+\text{rank}(B)-n$
QUESTION [11 upvotes]: If $A$ is a $m \times n$ matrix and $B$ a $n \times k$ matrix, prove that
$$\text{rank}(AB)\ge\text{rank}(A)+\text{rank}(B)-n.$$
Also show when equality occurs.

REPLY [3 votes]: Because I can't comment (yet), I will answer @user185640's question in a comment to @Babak Miraftab's answer which is "Why  $\{B(\alpha_{r+1}),\ldots,B(\alpha_n)\}$ is linear independent" in this post. 
Now, suppose $\sum_{i=r+1}^{n}c_i B(\alpha_i) = 0$ for some $i \in K$ where $K$ is the scalar field.
This would imply: $$B \left(\sum_{i=r+1}^{n}c_i\alpha_i\right) = 0$$ 
Hence, it is immediate to see that $\sum_{i=r+1}^{n}c_i\alpha_i \in \ker B$. But in his proof we know $\beta$ is a basis for $\ker B$. So we can write $\sum_{i=r+1}^{n}c_i\alpha_i$ as $\sum_{i=1}^{r}d_i\alpha_i$ for some $d_i \in K$. In particular we can write it as the following:
$$\sum_{i=r+1}^{n}c_i\alpha_i = \sum_{i=1}^{r}d_i\alpha_i \implies \sum_{i = r+1}^{n}c_i\alpha_i - \sum_{i=1}^{r}d_i\alpha_i = 0$$
To see this even more clearer, we can expand the sum to see this:
$$c_{r+1} \alpha_{r+1} + \ldots + c_{n} \alpha_{n} - d_1\alpha_1 - \ldots - d_r\alpha_r = 0$$
This may look unusual but it is not. We can further rewrite it as:
$$c_{r+1} \alpha_{r+1} + \ldots + c_{n} \alpha_{n} + (-d_1)\alpha_1 + \ldots + (-d_r)\alpha_r = 0$$
Let $e_i =-d_i$ for $i = 1,\ldots,n$ to see things even more clearer. We can then rewrite the above as:
$$c_{r+1} \alpha_{r+1} + \ldots + c_{n} \alpha_{n} + e_1\alpha_1 + \ldots + e_r\alpha_r = 0$$
Now since $\{\alpha_1, \ldots, \alpha_n\}$ is linearly independent (as it is a basis for $\ker AB$), we must have that
$$c_{r+1} = \ldots = c_n = e_1 = \ldots = e_r = 0$$
In particular, 
$$c_{r+1} = \ldots = c_n = 0$$
which directly implies that indeed $B(\alpha_{r+1}),\ldots,B(\alpha_n)$ are linearly independent and hence the set containing these vectors is a linearly independent set. Done.