Lower bound for $\|A-B\|$ when $\operatorname{rank}(A)\neq \operatorname{rank}(B)$, both $A$ and $B$ are idempotent Let's first focus on $k$-by-$k$ matrices. We know that rank is a continuous function for idempotent matrices, so when we have, say, $\operatorname{rank}(A)>\operatorname{rank}(B)+1$, the two matrices cannot be close in norm topology.
But I wonder whether there is an explicit lower bound of the distance between two idempotent matrices in terms of their difference in their ranks. 
Thanks!

martini's example is the first that come to mind, and it shows that rank is not continuous. One possible criticism is that the norms are small, so maybe there's a "normalized" version that works. But it's not the case: consider
$$
A=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}, \ \ \ \
B=\begin{bmatrix}1&0\\ 0&\varepsilon\end{bmatrix}
$$
Then $\text{rank}(A)=1$, $\text{rank}(B)=2$, $\|A\|=1$, $\|B\|=1+\varepsilon$, and $\|A-B\|=\varepsilon$. 
