Prove that $\operatorname{null}(A)+\operatorname{null}(B)≥\operatorname{null}(AB)$ in matrix formulation I've already seen this question, but I'd like to prove the case in a matrix form as below; however, I have no justification for a particular part of my proof:
Let $A_{m*n}$, and $B_{n*k}$, so $AB_{m*k}$.
According to the rank-nullity theorem, we have:
$n(A) + r(A) = n$
$n(B) + r(B) = k$
$n(AB) + r(AB) = k$
Adding the first two equations and using the third one to get rid of $k$ yields:
$n(A) + n(B) + r(A) +r(B) = n + n(AB) + r(AB)$
Since $r(B) \ge r(AB)$, we have
$n(A) + n(B) + r(A) \le n + n(AB)$
Here is the problem: I know that $r(A) \le n$, so I can't justify how to eliminate $r(A)$ and $n$ to end up with
$n(A) + n(B) \ge n(AB)$ 
 A: What you are trying is not likely to work. You have numbers
\begin{align}
a+b&=n \\ 
c+d&=k \\
e+f&= k
\end{align}
with $d\geq f$, $b\geq f$, $a\leq e$, $c\leq e$. All those relations are not enough to obtain your implication. Consider for instance
\begin{align}
1+4&=5 \\ 
1+4&=5 \\
3+2&= 5
\end{align}
Then all relations are satisfied but $1+1<3$. You need more information to obtain your inequality. 
A: Not without linear maps, but different from the proof in the other question.
You had this:
$$n(A) + n(B) + r(A) +r(B) = n + n(AB) + r(AB)$$
Rearranging gives:
$$n(A) + n(B) - n(AB) = r(AB) - r(A) - r(B) + n$$
so to prove the desired inequality, it is sufficient to prove $r(AB) - r(A) - r(B) + n\ge 0$.
Consider the linear map $A|_{\operatorname{Im} B} : \operatorname{Im} B \to \operatorname{Im} A$, the restriction of $A$ to $\operatorname{Im} B$.
Using the rank-nullity theorem for $A|_{\operatorname{Im} B}$ we obtain:
$$r(B) = \dim\operatorname{Im} B = n\left(A|_{\operatorname{Im} B}\right) + r\left(A|_{\operatorname{Im} B}\right)$$
Notice that $n\left(A|_{\operatorname{Im} B}\right) \le n(A)$, and that $r\left(A|_{\operatorname{Im} B}\right) = r(AB)$.
$$r(B) =  n\left(A|_{\operatorname{Im} B}\right) + r\left(A|_{\operatorname{Im} B}\right) \le n(A) + r(AB) = n - r(A) + r(AB)$$
Rearranging gives $r(AB) - r(A) - r(B) + n\ge 0$, so we have $n(A) + n(B) - n(AB) \ge 0$.
