Consider the case of the $1 \times 1$ matrix $A = [1]$. Its only eigenvalue is $1$. In this case, both $v$ and $c$ are numbers, so we're looking at eigenvalues of
$$
\pmatrix{
c & v \\
v & 1
}
$$
i.e., roots of
$$
(c-x)(1-x) - v^2 = 0,
$$
which is just
$$
x^2 - (c+1)x - v^2 = 0,
$$
whose roots are
$$
\frac{(c+1) \pm \sqrt{(c+1)^2 + 4v^2}}{2}
$$
Letting $u = (c+1)/2$, this becomes
$$
x_1, x_2 = u \pm \sqrt{u^2 + v^2}
$$
Pick any two numbers $a$ and $b$. If you set $u = \frac{a+b}{2}$, then for any $v$ you get
$$
x_1 + x_2 = a + b \\
x_1 - x_2 = \sqrt{\left(\frac{a+b}{2}\right)^2 + v^2}
$$
Setting this second thing to be $a-b$ gives you
\begin{align}
(a - b) &= \sqrt{\left(\frac{a+b}{2}\right)^2 + v^2} \\
(a - b)^2 &= \left(\frac{a+b}{2}\right)^2 + v^2 \\
(a - b)^2 - \left(\frac{a+b}{2}\right)^2 &= v^2 \\
a^2 -2ab + b^2 + \frac{a^2}{2} + ab + \frac{b^2}{2} &= v^2\\
\frac{3a^2}{2} -ab + \frac{3b^2}{2} &= v^2\\
\end{align}
so we can let
$$
v = \sqrt{\frac{3}{2} (a^2 + b^2) - ab}
$$
and we'll have arranged for the two eigenvalues of our matrix to be $a$ and $b$.
In short: Knowing that the only eigenvalue of $A$ is $1$ tells you nothing about the potential eigenvalues of $$
\pmatrix{
c & v \\
v & 1
}
$$
