Linear Algebra

The characteristic equation of $A$ is: $\det(A - \lambda I) = (2 - \lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0$ $(\lambda - 3)(\lambda - 1) = 0 \implies \lambda_1 = 3, \quad \lambda_2 = 1$ The eigenvalues of $A^3$ are $\lambda_1^3$ and $\lambda_2^3$: $\lambda_1^3 = 3^3 = 27$ $\lambda_2^3 = 1^3 = 1$ The trace of $A^3$ is the sum of its eigenvalues: $\text{Trace}(A^3) = 27 + 1 = 28$