If $\operatorname{rk}(A) = 1$, when $B + A$ is invertible?

I should know how do this problem, but I have troubles with it.

Let $B$ be an invertible matrix and let $A$ be a matrix with $\operatorname{rk}(A) = 1$. Then $\exists \lambda$ such that $A^2 = \lambda A$ and the problem is for which values of $\lambda$ the matrix $B + A$ is invertible?

When $B = I$ then $B + A$ is invertible iff $\lambda \neq -1$ and in the general case I suppose is $\lambda \neq -\det(B)$.

I think it'll be better if I type my conclusion for $B = I$, if $I +A$ es invertible let $C$ such that $$(I + A)C = C(I + A) = I.$$ Then $AC = CA$ iff $C^{-1}AC = A$, but $C^{-1} = I + A$ and therefore we have $$A = (I + A)AC = (A +A^2)C=(A +\lambda A)C = (1 + \lambda)AC.$$ If $1 + \lambda = 0$ then $A = 0$ which has no rank 1, thereupon $\lambda \neq -1$ and if $\lambda \neq -1$ the inverse for $I+A$ is $I -\frac{1}{1+\lambda} A$.

Since $A$ is of rank 1, $A = uv^\top$ where $u,v\neq 0$. This implies $A^2 = (v^\top u)A$ and hence $\lambda = v^\top u.$
Now, consider first the case $B=I$, $I+A$ is singular iff there is a $x\neq 0$ with, $Ax = -x$. Clearly such an $x$ if it exists is of the form $\alpha u$. So $I+A$ is singular iff there exists a scalar $\alpha \neq 0$ such that $\alpha Au = -\alpha u$, or $\alpha ( (v^\top u) + 1) u = \alpha ( \lambda + 1) u = 0,$ this implies as $u \neq 0$ $A$ is singular iff $v^\top u = \lambda = -1.$
For the general case $B+A$ is singular iff $I + B^{-1}A$ is singular. Note $B^{-1}A = u_1v^\top$ where $u_1 = B^{-1}u$. Hence, $A+B$ is singular iff $v^{\top} B^{-1} u = -1.$
Suppose $A=\operatorname{diag}(\lambda,0,\dots,0)$. As you show $I+AB^{-1}$ invertible, so $(AB^{-1})_{11}\ne -1$, that is $\lambda (B^{-1})_{11}\ne -1$, or $\lambda C_{11}\ne \det B$, where $C_{11}$ matrix of cofactors. Considering permutation, $\lambda C_{ii}\ne \det B$ for any $i$.
• Wait a second, the $\lambda$ I mentioned isn't the $\lambda$ that you are talking about and what's the $\lambda$ for $AB^{-1}$? Another thing $A$ is not invertible and so $\det(A)=0$, it should be $B$ instead of $A$. – Diego Silvera Apr 27 '13 at 23:57