Trace Finder

The solution I had in mind is equivalent to Guillermo's, but using the language of eigenvalues:

If $A\mathbf{v}=\lambda \mathbf{v}$ , then $A^3\mathbf{v}=\lambda^3 \mathbf{v}=\mathbf{v}$ , so that $\lambda^3=1$ . Now $\lambda_1=\lambda_2=1$ or $\lambda_1=\omega=e^{2\pi i/3}, \lambda_2=\omega^2$ . The trace is the sum of the eigenvalues, so that the minimum is $\omega+\omega^2=\boxed{-1}$ .

One solution to this problem is the matrix $\left(\begin{array} {c c} 0 & -1 \\ 1 & -1 \end{array} \right)$ . As $A^3 = I_2 \Rightarrow A^3 - I_2 = 0 = (A - I_2)(A^2 + A + I_2) \Rightarrow$ the minimum polynomial $p_{m}(A)$ associated to A is $A - I_2$ or $A^2 + A + I_2$ because its degree is less or equal than 2 and it belongs to $\mathbb{R}[X]$ . In the first case, $A = I_ 2$ and a + d = 2, but this case leads to a larger value, which we ignore, so we take $p_{m}(A) = A^2 + A + I_ 2$ and then $p_{m}(A)$ has to coincide with the characteristical polynomial $p_c(A)$ of A, which implies that trace(A) = a + d = -1 due to $p_c(A) = A^2 - trace(A) \cdot A + det(A)$