I've stumbled upon this exercise on a linear algebra book that asks me to determine all the ordered pairs $(a,b)$ of real numbers to which there exists an unique symmetric matrix $A\in R^{2\times 2}$ so that $tr(A)=a$ and $Det(A)=b$. I don't even know how to tackle this problem, I've tried using Laplace expansion, and other properties of the trace and determinants.
Any help will be appreciated.
On
The characteristic polynomial of a $2 \times 2$ matrix can be written as: \begin{equation} p(\lambda) = \lambda^2 - \textrm{tr}(A)\lambda + \textrm{det}(A) \end{equation} (Check here). If a matrix $A$ is symmetric then it is diagonalizable such that: \begin{equation} A = Q \Lambda Q^T \quad \textrm{ where } \quad \Lambda= \textrm{diag}(\lambda_1,\lambda_2) \end{equation} (i.e. $\Lambda$ is the matrix with eigenvalues in its diagonal). Therefore we can assume the characteristic polynomial can be solved for real roots: \begin{equation} \lambda^2 - \underbrace{\textrm{tr}(A)}_{a}\lambda + \underbrace{\textrm{det}(A)}_{b} = 0 \end{equation}
The problem is therefore reduced to finding for which $(a,b)$ there is a solution to: \begin{equation} \lambda^2 - a\lambda + b = 0 \end{equation}
The roots of the equation are given by the usual formula: \begin{equation} \lambda = \frac{a \pm \sqrt{a^2 - 4b}}{2} \end{equation}
Real-valued solutions are the ones where $a^2 \geq 4b$. This solves the problem.
Let $A=\pmatrix{x&z\\z&y}$. Then you have the equations $a = x+y$ and $b=xy-z^2$. It shouldn’t be too hard to solve these. You’ll have one free parameter when you’re done.