Eigenvalues of the subtraction of two symmetric matrices $A \in \mathbb R^{2 \times 2}$ is a real symmetric positive-definite matrix, and
$$
S(x)=\begin{bmatrix}
2x & -2x \\
-2 x & 4 x-x^2
\end{bmatrix}
$$
where $0<x$. Let $C(x)=A-S(x)$.
It can be verified that as $x$ increases, one eigenvalue of $S(x)$ goes to positive infinity, while the other one tends to negative infinity. Therefore, there exists an $x_m>0$ such that $C(x_m)$ is positive semi-definite and singular. My question is that whether the $x_m$ is unique, i.e., for any $0<x<x_m$, $C(x)$ is positive-definite, while for any $x>x_m$, $C(x)$ has at least one negative eigenvalue.
Can anyone provide some ideas? Thanks.
 A: The point $x_m$ is not unique:
Let
$$A=\begin{bmatrix}a&b\\b&c\end{bmatrix}.$$
We know that $A$ is positive definite iff $\text{tr}A>0$ and $\text{det}A>0$. This is equivalent to
$$a>0,\ ac>b^2.$$
Now,
$$C(x)=\begin{bmatrix}a-2x&b+2x\\b+2x&c+x^2-4x\end{bmatrix}$$
Analogously, we know $C(x)$ is positive semidefinite and singular iff $\text{tr}\,C(x)\geq0$ and $\text{det}\,C(x)=0$, that is
$$a\geq2x,\ \big(a-2x\big)\big(c+x^2-4x\big)=\big(b+2x\big)^2.$$
So to prove that $x_m$ is not unique we have to show that the system
$$\begin{cases}x>0\\ac>b^2\\a\geq2x\\\big(a-2x\big)\big(c+x^2-4x\big)=\big(b+2x\big)^2\end{cases},$$
has more than one real solution for $x$, for some values of $a,b,c$.
EDIT: I misread Mathematica results. The point $x_m$ is not unique:
Putting $a=100$, $b=0$ and $c=1$ we get three real solutions for the above system of equations. You can see that one eigenvalue of $C(x)$ becomes negative near $x=0$ then becomes positive near $x=4$ and finally becomes negative near $x=50$.
