IMC 1999 · Day 1 · P1

a) The diagonal matrix $$A = \lambda I = \begin{pmatrix} \lambda & & 0 \\ & \ddots & \\ 0 & & \lambda \end{pmatrix}$$ is a solution for equation $A^3 = A + I$ if and only if $\lambda^3 = \lambda + 1$, because $A^3 - A - I = (\lambda^3 - \lambda - 1) I$. This equation, being cubic, has real solution.

b) It is easy to check that the polynomial $p(x) = x^3 - x - 1$ has a positive real root $\lambda_1$ (because $p(0) < 0$) and two conjugated complex roots $\lambda_2$ and $\lambda_3$ (one can check the discriminant of the polynomial, which is $\left( \frac{-1}{3} \right)^3 + \left( \frac{-1}{2} \right)^2 = \frac{23}{108} > 0$, or the local minimum and maximum of the polynomial).

If a matrix $A$ satisfies equation $A^3 = A + I$, then its eigenvalues can be only $\lambda_1$, $\lambda_2$ and $\lambda_3$. The multiplicity of $\lambda_2$ and $\lambda_3$ must be the same, because $A$ is a real matrix and its characteristic polynomial has only real coefficients. Denoting the multiplicity of $\lambda_1$ by $\alpha$ and the common multiplicity of $\lambda_2$ and $\lambda_3$ by $\beta$, $$\det A = \lambda_1^{\alpha} \lambda_2^{\beta} \lambda_3^{\beta} = \lambda_1^{\alpha} \cdot (\lambda_2 \lambda_3)^{\beta}.$$ Because $\lambda_1$ and $\lambda_2 \lambda_3 = |\lambda_2|^2$ are positive, the product on the right side has only positive factors.