International Mathematics Competition
for University Students
2019

IMC2019: Day 2, Problem 9

Problem 9. Determine all positive integers \(\displaystyle n\) for which there exist \(\displaystyle n\times n\) real invertible matrices \(\displaystyle A\) and \(\displaystyle B\) that satisfy \(\displaystyle AB-BA=B^2A\).

Proposed by Karen Keryan, Yerevan State University & American University of Armenia, Yerevan

Solution. We prove that there exist such matrices \(\displaystyle A\) and \(\displaystyle B\) inf and only if \(\displaystyle n\) is even.

  

I. Assume that \(\displaystyle n\) is odd and some invertible \(\displaystyle n\times n\) matrices \(\displaystyle A,B\) satisfy \(\displaystyle {AB-BA=B^2A}\). Hence \(\displaystyle B=A^{-1}(B^2+B)A\), so the matrices \(\displaystyle B\) and \(\displaystyle B^2+B\) are similar and therefore have the same eigenvalues. Since \(\displaystyle n \) is odd, the matrix \(\displaystyle B\) has a real eigenvalue, denote it by \(\displaystyle \lambda_1\). Therefore \(\displaystyle \lambda_2:=\lambda_1^2+\lambda_1\) is an eigenvalue of \(\displaystyle B^2+B\), hence an eigenvalue of \(\displaystyle B\). Similarly, \(\displaystyle \lambda_3:=\lambda_2^2+\lambda_2\) is an eigenvalue of \(\displaystyle B^2+B\), hence an eigenvalue of \(\displaystyle B\). Repeating this process and taking into account that the number of eigenvalues of \(\displaystyle B\) is finite we will get there exist numbers \(\displaystyle k\leq l\) so that \(\displaystyle \lambda_{l+1}=\lambda_k\). Hence

\(\displaystyle \begin{align*} \lambda_{k+1}&=\lambda_k^2+\lambda_k\\ \lambda_{k+2}&=\lambda_{k+1}^2+\lambda_{k+1}\\ \ldots&\ldots\ldots\\ \lambda_{l}&=\lambda_{l-1}^2+\lambda_{l-1}\\ \lambda_{k}&=\lambda_{l}^2+\lambda_{l}. \end{align*} \)

Adding this equations we get \(\displaystyle \lambda_{k}^2+\lambda_{k+1}^2+\ldots+\lambda_{l}^2=0. \) Taking into account that all \(\displaystyle \lambda_i\)'s are real (as \(\displaystyle \lambda_1\) is real), we have \(\displaystyle \lambda_k=\ldots=\lambda_l=0,\) which implies that \(\displaystyle B\) is not invertible, contradiction.

  

II. Now we construct such matrices \(\displaystyle A,B\) for even \(\displaystyle n\). Let \(\displaystyle A_2=\begin{bmatrix} 0&1\\1&0 \end{bmatrix}\) and \(\displaystyle B_2=\begin{bmatrix} -1&1\\-1&-1 \end{bmatrix}\). It is easy to check that the matrices \(\displaystyle A_2,B_2\) are invertible and satisfy the condition. For \(\displaystyle n=2k\) the \(\displaystyle n\times n\) block matrices

\(\displaystyle A=\begin{bmatrix} A_2&0&\ldots&0\\ 0&A_2&\ldots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\ldots&A_2\\ \end{bmatrix}, \quad B=\begin{bmatrix} B_2&0&\ldots&0\\ 0&B_2&\ldots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\ldots&B_2\\ \end{bmatrix} \)

are also invertible and satisfy the condition.