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.