Invertability of sum of matrices

Question

Consider two different n$\times$n matrix A and B generated by vector outer product. Therefore, both A and B are symmetric positive semidefinite but not invertible. In this case, is there a condition (necessary/sufficient/necessary and sufficient) for the sum of two matrices A+B to be invertible?

Answer 1

The outer product of two vectors $u$ and $v$ can also be written as a matrix multiplication as $u\otimes v = uv^T$, where $\cdot^T$ is the transpose operation. Therefore, let $A = uv^T$ and $B=xy^T$. Then both $A$ and $B$ are rank-1 and their respective images are the spans of $u$ and $x$, respectively. Therefore, $A+B = uv^T + xy^T$ can be at most rank-2. This shows that one necessary condition of the invertability of $A+B$ is that $n\leq 2$.

If $n=1$, then the sufficient invertibility condition is $A+B\neq 0$. When $n=2$, $A+B$ is invertible iff it has full rank. This is achieved iff $u\neq \alpha x$ for some $\alpha\in\mathbb{R}$ and $v\neq \alpha y$ for some $\alpha\in\mathbb{R}$. The first condition guarantees that $A$ and $B$ have linearly independent column spaces, so $A+B$ is surjective (only in $\mathbb{R}^2$!). The second condition guarantees that $A+B$ is injective, as if this condition were not satisfied, then then $(A+B)z = 0$ for any $z\perp v$.