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While dealing with another problem I saw that I need to classify subspaces $V$ of $\mathbb{C}^{n \times n}$ that are multiplicatively closed.

So to get an idea of the nature of the subspaces I noticed that even for $\mathbb{C}^{2 \times 2}$ the question does not have an obvious answer to me.

Clearly, the $0$ and $4$ dimensional space are closed ( under multiplication), in the one-dimensional space, the only closed one appears to be the one generated by the identity. But for $2$ or $3$ dimensional subspaces this question is really a mess (at least if you want to answer it "by hand"). So I was wondering whether there is a more abstract view on this whole topic.

Since this question is rather accessible, I suspect that it is well-studied, but I could not find the right words to get a "google-match" as the results are always messed up with scalar multiplicatively closedness.

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    $\begingroup$ If $V$ is multiplicatively closed, then so is the subspace $V+\text{span}(\text{Id}_{n\times n})$. So you might as well assume that $\text{Id}_{n\times n}$ is in $V$. In that case, you are asking about the classification of subalgebras of $\mathbf{Mat}_{n\times n}(\mathbb{C})$, i.e., complex algebras together with a faithful $n$-dimensional representation. There is no hope to classify these for arbitrary $n$. $\endgroup$
    – Jason Starr
    Jan 13, 2016 at 14:03
  • $\begingroup$ Agreeing with Jason in general, I would still mention that one can describe subalgebras of $2\times 2$ matrices. Any 2-dimensional one has the form $\mathbb C[A]$ for some matrix $A$, and is generated so by almost any its matrix. Next, it sems that any 3-dimensional subalgebra is the set of matrces sharing a common eigenvector. $\endgroup$
    – Ilya Bogdanov
    Jan 13, 2016 at 15:55

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By no means an answer to the general question, but for the $2 \times 2$ case, check out @coudy's answer to this question.

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