While solving a problem in calculus of variations, I have reached the following question:
Let $s>0$ be a positive number (in my application $s>\frac{1}{2}$, but I am not sure if it matters here). Let $M_2$ be the set of all real $2 \times 2$ matrices, and define the following sets:
$$X=\{ A \in M_2 \, |\, \det A=s^2 \,\text{and }\,\sigma_1(A)=\sigma_2(A)=s \}, \, \,\,\,\,Y=\{ A \in M_2 \, |\, \sigma_1(A)=0, \sigma_2(A)=1 \},$$
where $\sigma_1(A) \le \sigma_2(A)$ are the singular values of $A$.
(The point of the condition $\det A=s^2$ in the definition of $X$ is to make $\det A$ positive, i.e. to exclude $\det A=-s^2$).
Question: Do there exist $A \in X, B \in Y$ such that $A-B$ has rank $1$?
$A$ and $B$ exist iff $s\le1$. We have $X=\{sQ:Q\in SO_2\}$ and $Y=\{uv^T:\|u\|_2=\|v\|_2=1\}$. Hence $$ \det(sQ-uv^T) =\det(sQ)-v^T\operatorname{adj}(sQ)u =s^2-sv^TQ^Tu \ge s^2-s.\tag{1} $$ Therefore $sQ-uv^T$ is nonsingular when $s>1$.
When $0<s\le1$, pick any two unit vectors such that $v^Tu=s$. Let $A=sI$ and $B=uv^T$. Then $A-B$ is singular by $(1)$ but it is nonzero because $A$ has rank $2$ but $B$ has rank $1$. Hence $A-B$ is necessarily rank-one. Indeed, as $\operatorname{tr}(B)=v^Tu=s$, the eigenvalues of $B$ are $s$ and $0$. Therefore the eigenvalues of $A-B$ are $0$ and $s$.