Consider a three dimensional Lie algebra $L$ with basis $a_1, a_2$ and $a_3$ which satisfy following commutation relation. \begin{align} [a_1,a_2]&=0\\ [a_1,a_3]&=a_1\\ [a_2,a_3]&=-a_2 \end{align}
Is it possible to find a representation $\rho: L\to \mathrm{End}(V)$ with finite dimensional V such that $\rho(a_1)=(\rho(a_2))^{-1}$ as elements of $GL(V)$?
For context: this algebra describes a quantum circuit as described in this paper (https://journals.aps.org/prb/pdf/10.1103/PhysRevB.53.4027?casa_token=kelWUXKoPVoAAAAA%3AHF236bMnQ_GbDKWMdySqluoDE_9PZ3Cbxb-lN31u0FxcGnWrtwgqfaonQ-mvy17mZi2m3iioarigfA)
What I see so far is that there can not be two-dimensional matrix representation because, from the commutation relation, we see that $a_1$ and $a_2$ are traceless. Moreover, $a_1$ and $a_2$ are inverses of each other. But the inverse of a $2\times 2$ traceless matrix is just a constant multiplied by itself. Thus, we run into inconsistent commutation relations.
But for 3D, I do not see any ways to argue whether a faithful representation exists or not. And the same with higher dimensions.
Any hints on how to approach the problem for $3*3$ case?
Thanks for revising the question.
If we are working over a field of characteristic $0$, then the answer is no. Because e.g. equation 2 now forces
$$\rho(a_1) \rho(a_3) - \rho(a_3) \rho(a_1) = \rho(a_1)$$
but it is a very useful result of Jacobson's that for matrices over characteristic $0$ fields, $AB-BA = A$ forces $A$ to be nilpotent. (More generally, this is already true as soon as $A$ commutes with $AB-BA$.)
So if there were such a finite-dimensional representation, the matrix representing $a_1$ (and also the one representing $a_2$) would have to be nilpotent, and thus cannot even have an inverse.
I do not know if one can construct such a representation over a field of positive characteristic. I would not be surprised if one can. The above result certainly does not hold any more.Update: There is a somewhat "natural", three-dimensional representation over any field of characteristic $3$:
$\rho(a_1) = \pmatrix{0&1&0\\0&0&1\\1&0&0}$, $\rho(a_2) = \pmatrix{0&0&1\\1&0&0\\0&1&0}$, $\rho(a_3) = \pmatrix{0&0&0\\0&1&0\\0&0&-1}$.
And a four-dimensional representation of this Lie algebra over a field of characteristic $2$: Let the field be $\mathbb F_4 = \{0,1, \zeta, \zeta^2=\zeta^{-1}=\zeta+1 \}$ and set
$\rho(a_1) = \pmatrix{0&\zeta&0&0\\ 1&0&0&0\\0&0&0&1\\0&0&1&0}$, $\rho(a_2) = \pmatrix{0&1&0&0\\ \zeta^{-1} &0&0&0\\0&0&0&1\\0&0&1&0}$, $\rho(a_3) = \pmatrix{1&0&0&0\\ 0&0&0&0\\0&0&1&0\\0&0&0&0}$