I am given a symmetric $n \times n$ real matrix $A=A^\top$ and a vector $v\in\mathbb{R}^n$, and I know that $\det A = a$. Then I have the bordered matrix $F$ given by $$ F = \begin{bmatrix} A & v \\ v^\top & 0 \end{bmatrix}.$$ As far as I understand, it is not possible to find $\det F$ given only $a$ and $v$. However, is it possible to find some lower and upper bounds on $\det F$ given $a$ and $v$? E.g., $$f_{min}(a,v) \le \det F \le f_{max}(a,v),$$ where $f_{min}$ and $f_{max}$ are some $\mathbb{R}\times \mathbb{R}^n \to \mathbb{R}$ functions? If it helps, then $a \ge 0$.
I know also that $\det F =-v^\top \left(\operatorname{adj}A\right) v$, where $\operatorname{adj}A$ is the adjoint of $A$. But I do not see if it helps.
There are not any such bounds. E.g. when $b\ne0,\ A=\operatorname{diag}(\frac{a}{b}, b)$ and $v=(1,0)^T$, $$ \det(F)=-v^T\operatorname{adj}(A)v=-\pmatrix{1&0}\pmatrix{b&0\\ 0&\frac{a}{b}}\pmatrix{1\\ 0}=-b $$ is unbounded.