Prove Sylvester rank inequality: $\text{rank}(AB)\ge\text{rank}(A)+\text{rank}(B)-n$ 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.
 A: As noted in the other answer, it suffices to show $\dim\ \operatorname{Ker}(A)+\dim\ \operatorname{Ker}(B) \geq \dim\ \operatorname{Ker}(AB)$. This is equivalent to showing that $\dim\ \operatorname{Ker}(AB)/\operatorname{Ker}(B) \leq \dim\ \operatorname{Ker}(A)$. To do this, use the first isomorphism theorem for vector spaces on the linear map $\operatorname{Ker}(AB) \rightarrow \operatorname{Ker}(A)$ defined by $x \mapsto Bx$. This shows that $\operatorname{Ker}(AB)/\operatorname{Ker}(B)$ is isomorphic to a subspace of $\operatorname{Ker}(A)$, which proves the inequality.
A: 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.
Claim: $\{B(\alpha_{r+1}),\ldots,B(\alpha_n)\}$ is linearly independent.
Proof.
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 that $$B \left(\sum_{i=r+1}^{n}c_i\alpha_i\right) = 0,$$
or equivalently, we have $\sum_{i=r+1}^{n}c_i\alpha_i \in \ker B$. But we know $\beta$ is a basis for $\ker B$ (from @Babak Miraftab's proof). So we can write
$$\sum_{i=r+1}^{n}c_i\alpha_i = \sum_{i=1}^{r}d_i\alpha_i,$$
for some $d_i \in K$, or equivalently,
$$\sum_{i=r+1}^{n}c_i\alpha_i - \sum_{i=1}^{r}d_i\alpha_i = 0.$$
Expanding this sum, we can write it in the following way:
$$c_{r+1} \alpha_{r+1} + \ldots + c_{n} \alpha_{n} + (-d_1)\alpha_1 + \ldots + (-d_r)\alpha_r = 0.$$
Now let's be super explicit and let $e_i =-d_i$ for all $i$ which gives us:
$$c_{r+1} \alpha_{r+1} + \ldots + c_{n} \alpha_{n} + e_1\alpha_1 + \ldots + e_r\alpha_r = 0.$$
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, we have
$$c_{r+1} = \ldots = c_n = 0$$
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.
A: We claim $\dim \ker\,A+\dim\ker B \geq \dim\ker AB$.
Let $\beta=\{\alpha_1,\dots,\alpha_r \}$ be a basis for $\ker B$.
It is not hard to see that $\ker B\subseteq \ker AB$ so we can extend $\beta $ to a basis for $\ker AB$.
Suppose $\{\alpha_1,\dots,\alpha_r,\alpha_{r+1},\dots,\alpha_n \ \}$ be basis for $\ker AB$.  So $B(\alpha_{i})\neq 0$  for $i \in \{r<i<n+1\}$.
We show that $\{B(\alpha_{r+1}),\dots,B(\alpha_{n})\}$ is linear independent. We have $\dim\ker A\geq n-r$.
Assume that $\sum_{i=r+1}^n\gamma_iB(\alpha_i)=0$. In other words we have
$B(\sum_{i=r+1}^n\gamma_i\alpha_i)=0$ and as a result $\sum_{i=r+1}^n\gamma_i\alpha_i$ belongs to the kernel of $B$.
On other hand we already know that $\beta=\{\alpha_1,\dots,\alpha_r \}$ is a basis for the kernel B.
Next since the set $\{\alpha_1,\dots,\alpha_r,\alpha_{r+1},\dots,\alpha_n \ \}$ is an independent set, we infer that all $\gamma_i$ must be zero.
Now one can see that
$$\dim\ker A+\dim\ker B \geqslant n-r+r =n \Longrightarrow\dim\ker A+\dim\ker B \geqslant \dim\ker AB$$
A: Recall Linear Transformations Isomorphic to Matrix Space.
Using Rank–nullity theorem, $\operatorname{rank}(A)+\operatorname{nullity}(A)=n,\operatorname{rank}(B)+\operatorname{nullity}(B)=k$ and $\operatorname{rank}(AB)+\operatorname{nullity}(AB)=k.$ 
So, $\operatorname{rank}(A)+\operatorname{rank}(B)+\operatorname{nullity}(A)+\operatorname{nullity}(B)=n+\operatorname{rank}(AB)+\operatorname{nullity}(AB)$ 
$\implies \operatorname{rank}(AB)-\operatorname{rank}(A)-\operatorname{rank}(B)+n=\operatorname{nullity}(A)+\operatorname{nullity}(B)-\operatorname{nullity}(AB)$
$\geq \operatorname{nullity}(A)$[Since $Bv_2=0$ for $v_2\in Mat_{k\times 1}(F)\implies ABv_2=0$] $\geq 0.$ 
