Oh yeah, sorry, silly mistake on my part. I suppose one could ask if, e.g., RA is the variety containing RRA and with a finite equational axiomatization where that finite equational axiomatization is as short as possible (by some metric - number of axioms, numbers of characters, Kolmogorov complexity, ...), but that seems much harder to establish, if true.

Is RA the smallest variety containing RRA and with a finite equational characterization? That wouldn't answer your question as stated, but, if true, could be taken as a different kind of evidence that these axioms are "the right ones".

An alternative if perhaps less standard way to see it is to not require \varphi nor A-modules to be unital, and then we have to allow the 0 representation as a simple module.

Oh, sorry, yes that's a potential issue but you already found the solution. Really what's happening in that case is that A is conjugate to matrices of the form $\begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix}$, and so is essentially a subalgebra of a smaller matrix algebra acting on a smaller vector space. And then the rest of what I wrote goes through in that smaller matrix algebra & smaller vector space.

@NoahSchweber: I admit I didn't quite follow all of that, but it sounds plausibly like what I was hoping for, so I would certainly appreciate reading more details about it!

@LSpice: I understood that. The results about representations I mentioned imply block-diagonalizability of any semisimple subalgebra of $M_d(F)$ into blocks each of which is a full matrix algebra on its own, though some blocks can be repeated. If you think there is a little new content I can spell out the details in an actual answer.

You can always put a semisimple matrix algebra into block-diagonal form over an algebraically closed field. This follows from the classification of the representations of a semisimple algebra: they are all direct sums of irreps, and each irrep is the standard d-dimensional module for $M_d(F)$, where $M_d(F)$ is one of the direct summands of the algebra.

I assumed that "algebraic group" meant G was finite-dimensional and that $G/G^0$ was finite. And then doesn't it follow that any irrep of $G^0$, and hence $G$, is finite-dimensional? But re: other fields, yes, I see there is potentially an issue for non-algebraically closed fields.

FYI the classification give in the Kac paper linked to by Sam Gunningham in an earlier comment is known to contain some errors, which were corrected in Dadok & Kac: doi.org/10.1016/0021-8693(85)90136-X

I would guess one might make some progress on this problem - perhaps in either direction - using Sergeichuk's normal form based on Belitskii's algorithm: Sergeichuk, "Canonical matrices for linear matrix problems", Lin. Alg. Appl. 2000, doi.org/10.1016/S0024-3795(00)00150-6