Let us consider $N \times N$ complex matrices, with $N >2$. Let D be a diagonal matrix, with $$D_{kk} = \sin \left(\frac{2\pi k}{N}\right), \space k = 0,..N - 1$$
I am looking for two hermitian matrices $H_1$ et $H_2$, such that :
$$ [H_1, H_2] = i\space D$$,where $i$ is the imaginary unit.
Specific answers for particular low values of $N$ (3, for instance) are also welcome, if no general solution can be found.
On
To complete the following answer, a possibility for a $N×N$ unitary matrix $U$ having $\frac{1}{\sqrt{N}}(1,1,\ldots,1)$ in its last row is :
for $\space k = 0,\ldots, N - 2$
$$U_{kk} = - \frac {(N - 1 - k)}{\sqrt{(N - k) (N - k - 1)}}$$
$$U_{kl} = \frac {1}{\sqrt{(N - k) (N - k - 1)}} for \space l > k$$
$$U_{kl} = 0 \space for \space l < k$$
with : $$U_{N - 1 l} = \frac {1}{\sqrt{N}}, for \space l = 0,\ldots, N - 1$$
If you can show that $\sum_{k=0}^{N-1}\sin\left(\frac{2k\pi}{N}\right)=0$, you may apply the result that every traceless skew-Hermitian matrix is the commutator of two Hermitian matrices (see here for a constructive proof).