What is the smallest value $tr(A)$ can take if $tr(A^2) = 241$? ($tr(A)$ is the trace of the matrix $A$)

Question

I'm working on a problem involving a 2 × 2 matrix $A$ with different eigenvalues and a determinant $det(A) = 60$. Given that $tr(A^2) = 241$, I'm tasked with finding the smallest possible value for $tr(A)$. I've tried utilizing the relationship between the trace and determinant of a matrix with its eigenvalues, expressing $tr(A)$ and $det(A)$ in terms of the eigenvalues ${\lambda}_{1}$ and ${\lambda}_{2}$. However, I seem to be struggling to derive a conclusive relationship between $tr(A)$ and $det(A)$ that helps me solve for the minimum value of $tr(A)$ satisfying the given condition. Could someone provide guidance or a step-by-step approach to set up the equations necessary to solve for the smallest possible value of $tr(A)$ given the conditions provided?

Answer 1

Express the given values in terms of the eigenvalues $\{a,b\}$ $$\eqalign{ \def\tr{\operatorname{tr}} \det(A) &= ab\doteq 60,\qquad\tr(A^2) = a^2+b^2 \doteq 241 \\ }$$ Now expand the square of the trace $$\eqalign{ \Big(\tr(A)\Big)^2 &= (a+b)^2 \\ &= a^2+b^2+2ab \\ &= 241+2(60) \\ &= 361 \\ }$$ Therefore $\quad\tr(A) = \pm19$