The rank after adding a rank-one matrix to a full rank matrix Suppose $A$ and $B$ are all $N$ by $N$ matrices. 
$$rank(A) = N, rank(B) = 1$$
What's the lower bound of:  $$rank(A+B)?$$ 
My guess in this specific case is $$rank(A+B) \geq N-1,$$ but I don't know if it's true, how to prove it, and under what condition we have $rank(A+B) = N-1$.
Can anyone help me with this? I know there is a nice result of the upper bound, $rank(A+B) \leq rank(A) + rank(B)$, but I didn't find anything about the lower bound online.

Think of $A,B$ as linear transformations. $rank(A)=N$ implies $A$ is one to one, hence maps every subspace of dimension $k$ to another subspace of the same dimension. $rank(B)=1$ implies $\dim\ker(B)=N-1$. Now $$\dim(A+B)(\ker(B))=\dim(A(\ker(B)))=\dim(\ker(B))=N-1$$implies $$\dim(Im(A+B))\geq N-1,$$and in other words $$rank(A+B)\geq N-1.$$ Assume now $rank(A+B)=N-1$, so there is $v\neq0$ such that $Av+Bv=0$. Obviously, $A(v)\in Im(B)$, but since $Im(B)$ is $1$-dimensional and $A$ is one to one, $A^{-1}(Im(B))$ is also $1$-dimensional, and actually $A^{-1}(Im(B))=span(v)$. It follows that for every $u\in A^{-1}(Im(B)),\;A(u)=-B(u)$. 
In conclusion, it is easy to check: Just pick $u\in A^{-1}(Im(B))$ and see whether $A(u)=-B(u)$.
