rank of a diagonal matrix + rank-one perturbation Let $D$ be a $n \times n$ diagonal matrix, and $A$ is a $n \times n$ rank-one matrix that can be rewritten as $A=a\cdot b^T$, where $a$ and $b$ are $n \times 1$ vectors. Now what is the lower bound for the rank of the matrix $D+A$ ? Any suggestions are welcome. 
 A: In general, one has $\def\rk{\mathop{\rm rank}}$ 
$$ \rk (A+B) \le \rk A + \rk B $$
for matrices $A, B \in \mathbb R^{n\times m}$ (note that the subspace generated by the union of the images of $A$ and $B$ contains the image of $A+B$). For your example, we have 
$$ \rk D \le \rk(D+A) + \rk(-A) $$
as $\rk(-A) = 1$, we have 
$$ \rk(D+A) \ge \rk D - \rk A = \rk D - 1. $$
A: A is of rank 1 then it only has one pivot. D is diagonal then it has (rank of D) pivots. If you add the two reducted matrix you will obtain a new diagonal matrix where there is between (rank of D - 1) pivots and (rank of D + 1) pivots.Then the lower bound of D+A is (rank of D - 1). Now i explain why rank of D - 1. In the reducted form of A you always can switch lines such as the pivot of A is placed on the diagonal. Then its place is (i,i). Now what can happen is that D(i,i)=-A(i,i) then we obtain a 0 which lower the rank of D from 1. For the upper bound the reasoning is approximately the same but it's if we have 0 on D(i,i).
