IMC 2017 · Day 1 · P1

By taking squares, $$A^4 = (A^2)^2 = (A^T)^2 = (A^2)^T = (A^T)^T = A,$$ so $$A^4 - A = 0;$$ it follows that all eigenvalues of $A$ are roots of the polynomial $X^4 - X$.

The roots of $X^4 - X = X (X^3 - 1)$ are $0$, $1$ and $\frac{-1 \pm \sqrt{3} i}{2}$. In order to verify that these values are possible, consider the matrices $$A_0 = \bigl( 0 \bigr), \quad A_1 = \bigl( 1 \bigr), \quad A_2 = \begin{pmatrix} -\frac12 & \frac{\sqrt3}{2} \\ -\frac{\sqrt3}{2} & -\frac12 \end{pmatrix}, \quad A_4 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -\frac12 & \frac{\sqrt3}{2} \\ 0 & 0 & -\frac{\sqrt3}{2} & -\frac12 \end{pmatrix}.$$ The numbers $0$ and $1$ are the eigenvalues of the $1 \times 1$ matrices $A_0$ and $A_1$, respectively. The numbers $\frac{-1 \pm \sqrt{3} i}{2}$ are the eigenvalues of $A_2$; it is easy to check that $$A_2^2 = \begin{pmatrix} -\frac12 & -\frac{\sqrt3}{2} \\ \frac{\sqrt3}{2} & -\frac12 \end{pmatrix} = A_2^T.$$ The matrix $A_4$ establishes all the four possible eigenvalues in a single matrix.

Remark. The matrix $A_2$ represents a rotation by $2\pi/3$.