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## IMC 2014 Day 1 Problem 1

Determine all pairs ${(a,b)}$ of real numbers for which there exists a unique symmetric ${2 \times 2}$ matrix ${M}$ with real entries satisfying ${tr(M)=a}$ and ${\det(M)=b}$.

IMC 2014 Day 1 Problem 1

Solution: First note that the matrix ${M}$ must be diagonal, since

$\displaystyle \begin{pmatrix} x& y \\ y & z \end{pmatrix} \text{ and } \begin{pmatrix} x& -y \\ -y & z \end{pmatrix}$

Have the same trace and determinant.

If we have a diagonal matrix ${\begin{pmatrix} x& 0 \\ 0 & z \end{pmatrix}}$ then ${\begin{pmatrix} z& 0 \\ 0 & x \end{pmatrix}}$ has the same trace and determinant. For it to be unique we would need that ${x=z}$ and therefore ${M = \lambda I_2}$. This means that ${a = 2\lambda}$ and ${b = \lambda^2}$ and the characteristic polynomial of ${M}$ is ${P_M = X^2 - 2\lambda X +\lambda^2=(X-\lambda)^2}$.

Conversely, if ${a=2\lambda, b=\lambda^2}$ then ${M}$ has both eigenvalues equal to ${\lambda}$. Since ${M}$ is symmetric, ${M}$ is diagonalizable and there exists a matrix ${P}$ such that ${\lambda I = P^{-1}MP}$. This last relation implies that ${M=\lambda I}$ and the matrix ${M}$ is unique.