Home > Olympiad, Problem Solving > IMC 2014 Day 1 Problem 1

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.

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