Find all ordered pairs of real numbers $(a,b)$ for which there is a unique $2\times 2$ real symmetric matrix $M$.

Question

Find all pair of real numbers (a,b) for which there is a unique 2x2 real symmetric Matrix $M$ with $\operatorname{tr}(M) = a$ and $\det(M) = b$

What I have already tried is forming a characteristic equation $$\lambda^2 - \operatorname{tr}(M)\lambda + \det(M)=0$$ then inputing values of $tr(M)$ and $\det(M)$ form $$\lambda^2 - a\lambda + b = 0$$ the discriminant of the characteristic equation using the quadratic formula is $a^2-4b$, as we have to find real pairs I can assume that $a^2-4b >= 0$

For simplicity, I assumed $a^2-4b = 0$ and I derived $$a = +2\sqrt{b},-2\sqrt{b}$$ Let $b=t$ and I get these two pairs $$(2\sqrt{t}, t), (-2\sqrt{t}, t)$$ therefore, for all real numbers (a,b), as $a^2-4b >= 0$ $$-2\sqrt{b}>=a >= 2\sqrt{b} $$

Does this seem valid, or is there a better way to do this, or maybe I'm doing it all wrong. Any help is appreciated. Thank you!

Answer 1

Here is a relatively quick approach.

Hint: For $d \neq 0$, the matrices $$ M_1 = \pmatrix{c&d\\d&c}, \quad M_2 = \pmatrix{c&-d\\-d&c} $$ satisfy $\det(M_1) = \det(M_2)$ and $\operatorname{tr}(M_1) = \operatorname{tr}(M_2)$.

Answer 2

Let $$M=\begin{pmatrix}p&q\\q& a-p\end{pmatrix}$$

where $$p(a-p)-q^2=b$$ $$-b+\frac14a^2=(p-\frac12 a)^2+q^2$$ You want to find values of $a$ and $b$ such that $p$ and $q$ are unique. If the LHS is negative there are no solutions, if it is positive there are infinitely many. So this requires that the above expression is zero and so $4b=a^2$.