An example for almost commuting matrices

Question

I'm looking for examples of the following kind:

2 sequences of unitary matrices ($A_n,B_n$) (where these are $n\times n$ matrices) that don't commute and satisfy $$ ||A_nB_n-B_nA_n||_1\longrightarrow_{n\rightarrow\infty}0 $$ Where the $||\cdot||_1$ norm is the sum of the absolute values of entries.

I know that for other norms (actually for any $p>1$, there exists an example through a permutation matrix and a sign matrix - for example the right shift and the ordered roots of unity. This unfortunately doesn't work for the absolute value norm so I'm stuck)

I'd appreciate any help, thank you.

Answer 1

You could take rotation matrices in $\mathbb{R}^3$ which usually do not commute. If you define the matrices in such way that the angle of rotation becomes smaller when $n$ gets larger, the matrices will approach the identity matrix for increasing $n$ and the norm of the commutator will approach $0$, independently of the choice of the particular norm. Example: $$ A_n = \begin{pmatrix} \cos\frac{\pi}{n} & -\sin\frac{\pi}{n} & 0 \\ \sin\frac{\pi}{n} & \cos\frac{\pi}{n} & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ and $$ B_n = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\frac{\pi}{n} & -\sin\frac{\pi}{n} \\ 0 & \sin\frac{\pi}{n} & \cos\frac{\pi}{n} \\ \end{pmatrix} $$

Answer 2

Take $$ A_n=\begin{bmatrix}\cos\tfrac1n&\sin\tfrac1n\\ \sin\tfrac1n&-\cos\tfrac1n\\ &&1\\ &&&\ddots&\\ &&&&1 \end{bmatrix}, \ \ \ \ \ B_n=\begin{bmatrix}-1&\\ &1\\ &&1\\ &&&\ddots&\\ &&&&1 \end{bmatrix}. $$ Then $$ \|A_nB_n-B_nA_n\|_1=\left\|\begin{bmatrix}0&2\sin\tfrac1n\\-2\sin\tfrac1n&0\end{bmatrix}\right\|_1=4\,\left|\sin\tfrac1n\right|\leq\tfrac4n $$