### International Mathematics Competition for University Students

July 31 – August 6 2017, Blagoevgrad, Bulgaria

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### Problem 1

1. Determine all complex numbers $\lambda$ for which there exist a positive integer $n$ and a real $n\times n$ matrix $A$ such that $A^2=A^T$ and $\lambda$ is an eigenvalue of $A$.

Proposed by: Alexandr Bolbot, Novosibirsk State University

Solution. 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 $\frac{-1\pm\sqrt3i}2$. In order to verify that these values are possible, consider the matrices $$A_0 = \begin{pmatrix} 0 \end{pmatrix}, \quad A_1 = \begin{pmatrix} 1 \end{pmatrix}, \quad A_2 = \begin{pmatrix} -\frac12 & \frac{\sqrt3}2 \\ -\frac{\sqrt3}2 & -\frac12 \\ \end{pmatrix}, \quad A_4 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -\frac12 & \frac{\sqrt3}2 \\ 0 & 0 & -\frac{\sqrt3}2 & -\frac12 \\ \end{pmatrix}.$$ The numbers $0$ and $1$ are the eigenvalues of the $1\times1$ matrices $A_0$ and $A_1$, respectively. The numbers $\frac{-1\pm\sqrt3i}2$ 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 $2\pi/3$.