Invertibility of Difference of Two Rank-One Matrices

Question

Suppose I have a two rank-one matrices, $A$ and $B$ with $A \ne B$. When is the difference $A-B$ invertible?

Answer 1

Suppose that $A,B$ are rank-1 matrices of shape $n \times n$. If $n \geq 3$, then $A - B$ cannot be invertible since its rank will be at most equal to $2$. For $n=1$, $A - B$ is "invertible" iff it is non-zero, which is to say that $A \neq B$. For $n = 2$, $A - B$ will be invertible if and only if it has rank $2$.

In general, the following statements are equivalent:

1. $A - B$ has rank $2$
2. $A$ and $B$ have distinct column-spaces and distinct row-spaces
3. $A$ and $B$ have distinct kernels and distinct cokernels.

Proof:

Because $A$ and $B$ have rank $1$, they can be written in the form $A = uv^T$ and $B = xy^T$ for non-zero column-vectors $u,v,x,y$ of size $n$. We can write $A - B$ as the product $A - B = PQ$, where $$ P = \pmatrix{u & x}, Q = \pmatrix{v^T\\ y^T}. $$ $1 \implies 2$: Suppose that $A - B$ has rank 2. We have $2 = \operatorname{rank}(PQ)\leq \min\{\operatorname{rank}(P),\operatorname{rank}(Q)\}$, which means that both $P$ and $Q$ are of rank $2$. From the fact that $P$ has independent columns, $A$ and $B$ must have distinct column spaces. Because $Q$ has independent rows, $A$ and $B$ must have distinct row spaces.

$2 \implies 1$: Suppose that $A$ and $B$ have distinct row spaces and distinct column spaces. Thus, $u,x$ must be independent and $v,y$ must be independent. Because $P$ has independent columns, we have $\operatorname{rank}(PM) = \operatorname{rank}(M)$ for all matrices $M$. So, $\operatorname{rank}(PQ) = \operatorname{rank}(Q)$. Because the rows of $Q$ are independent, it has rank $2$. So, $$ \operatorname{rank}(A - B) = \operatorname{rank}(PQ) = \operatorname{rank}(Q) = 2. $$

Statements 2 and 3 are equivalent because $\ker(M) = \operatorname{rowspace}(M)^\perp$.