I had been asked the following question:
Suppose $A$ and $B$ are two $10 \times 10$ matrices over $\mathbb{C}$ with $rank(AB)=6$. What is the maximum possible value of rank of $BA$?
First, I realized that both $rank(A)$ and $rank(B)$ are at least 6. Further, from Sylvester's inequality, we must have $$rank(A)+rank(B) \leq 16.$$ Now, counting down all the possibilities (For instance, one possibility is $rank(A)=6$ and $rank(B)=10$), I arrived at a possible maximum value of $rank(BA)$ which is 8. For this to happen we must have $rank(A) = rank(B) = 8$.
I tried, but could not get an answer to the question, `do there exist two matrices $A$ and $B$ each with rank $8$ and that $rank(AB)=6$, $rank(BA)=8$?
From Sylvester's inequality $$\operatorname{rank}(A)+\operatorname{rank}(B) \leq 6+10= 16.$$ Because $$\operatorname{rank}(BA)\leq\min\{\operatorname{rank}(A),\operatorname{rank}(B)\}$$ For $\operatorname{rank}(BA)$ to be as large as possible, the ranks of $A$ and $B$ must be as close as possible while satisfying Sylvester's inequality, thus $\operatorname{rank}(A)=\operatorname{rank}(B)=8$.
Following @user10354138's comment, we will utilize the matrices \begin{align*} X = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} && Y = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \end{align*} with $$XY=0\qquad YX = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $$
Let $$A=\operatorname{diag}(1,1,1,1,1,1,X,X),B=\operatorname{diag}(1,1,1,1,1,1,Y,Y)$$ We have $$\operatorname{rank}(A)=\operatorname{rank}(B)=8$$ and$$AB=\operatorname{diag}(1,1,1,1,1,1,0,0,0,0),BA=\operatorname{diag}(1,1,1,1,1,1,Y,Y)$$ Therefore, $$\operatorname{rank}(AB)=6,\operatorname{rank}(BA)=8$$