IMC 2019 · Day 2 · P9

We prove that there exist such matrices $A$ and $B$ inf and only if $n$ is even.

I. Assume that $n$ is odd and some invertible $n \times n$ matrices $A, B$ satisfy $AB - BA = B^2 A$. Hence $B = A^{-1} (B^2 + B) A$, so the matrices $B$ and $B^2 + B$ are similar and therefore have the same eigenvalues. Since $n$ is odd, the matrix $B$ has a real eigenvalue, denote it by $\lambda_1$. Therefore $\lambda_2 := \lambda_1^2 + \lambda_1$ is an eigenvalue of $B^2 + B$, hence an eigenvalue of $B$. Similarly, $\lambda_3 := \lambda_2^2 + \lambda_2$ is an eigenvalue of $B^2 + B$, hence an eigenvalue of $B$. Repeating this process and taking into account that the number of eigenvalues of $B$ is finite we will get there exist numbers $k \le l$ so that $\lambda_{l+1} = \lambda_k$. Hence $$\begin{gather*} \lambda_{k+1} = \lambda_k^2 + \lambda_k \\ \lambda_{k+2} = \lambda_{k+1}^2 + \lambda_{k+1} \\ \dots\dots\dots \\ \lambda_l = \lambda_{l-1}^2 + \lambda_{l-1} \\ \lambda_k = \lambda_l^2 + \lambda_l. \end{gather*}$$ Adding this equations we get $\lambda_k^2 + \lambda_{k+1}^2 + \dots + \lambda_l^2 = 0$. Taking into account that all $\lambda_i$'s are real (as $\lambda_1$ is real), we have $\lambda_k = \dots = \lambda_l = 0$, which implies that $B$ is not invertible, contradiction.

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