Sylvester rank inequality: $\operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n$ If $A$ and $B$ are two matrices of the same order $n$, then
  $$ \operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n. $$


I don't know how to start proving this inequality. I would be very pleased if someone helps me. Thanks!

Edit I.  Rank of $A$ is the same of the equivalent matrix $A' =\begin{pmatrix}I_r &amp; 0 \\ 0 &amp; 0\end{pmatrix}$. Analogously for $B$, ranks of $A$ and $B$ are $r,s\leq n$. Hence, since $\operatorname{rank}AB = \min\{r,s\}$, then $r+s\leq \min\{r,s\} + n$. (This is not correct since $\operatorname{rank} AB \leq \min\{r,s\}$.

Edit II. A discussion on the rank of a product of $H_f(A)$ and $H_c(B)$ would correct this, but I don't know how to formalize that $\operatorname{rank}H_f(A) +\operatorname{rank}H_c(B) - n \leq \operatorname{rank}[H_f(A)H_c(B)]$.
Subtract $2n$ from both sides and then multiply by $-1$; this gives the equivalent inequality $\def\rk{\operatorname{rank}}(n-\rk A)+(n-\rk B)\geq n-\rk(AB)$. In other words (by the rank-nullity theorem) we must show that the sums of the dimensions of the kernels of $A$ and of $B$ gives at least the dimension of the kernel of $AB$. Intuititively vectors that get killed by $AB$ either get killed by $B$ or (later) by $A$, and the dimension of $\ker(AB)$ therefore cannot exceed $\dim\ker A+\dim\ker B$. Here is how to make that precise.

One has $\ker B\subseteq \ker(AB)$, so if one restricts (the linear map with matrix) $B$ to $\ker(AB)$, giving a map $\tilde B:\ker(AB)\to K^n$, one sees that $\ker(\tilde B)=\ker(B)$ (both inclusions are obvious). Since the image of $\tilde B$ is contained in $\ker A$ by the definition of $\ker(AB)$, one has $\rk\tilde B\leq\dim\ker A$. Now rank-nullity applied to $\tilde B$ gives $$\dim\ker(AB)=\dim\ker\tilde B+\rk\tilde B\leq\dim\ker B+\dim\ker A,$$ as desired.
In response to Mr. Albanese's correct critique. I submit the following. The rank of A,B is in the set of integers [0,n]. Assume we know the relatively straigtfoward proofs that rank(AB)&lt;= min(rank(A),rank(B)), -> rank(AB) = n if rank(A) and rank(B) have rank n. There are three rank possibilities. Let the rank of A = p, the rank of B = q with p,q integers. Then (p or q &lt; n) or (p and q &lt; n) or (p=q=n). The first two cases imply (rank(A) + rank(B) &lt; rank(AB) + n) or (p+q &lt; min(p,q) + n) as p+q &lt; 2n and (rank(A) + rank(B)  > rank(AB) + n) is a contradiction. This is a contrapositive proof of the strict inequality in the Sylvester theorem (rank(A) + rank(B) &lt; rank(AB) + n).  For the equality case p=q=n. Then (rank(A) + rank(B)  = rank(AB)) + n) or n + n = n + n which is true because rank(A) and rank(B) have full rank, n, implies rank(AB)=n.