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This is a beginner question (and not any homework). I want to get a feeling for Lie group/algebra generators. Do the three matrices $$A=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0& 1&0 \end{pmatrix}$$, $$B=\begin{pmatrix} 0& 0& 1 \\ 0& 0& 0 \\ 1&0&0 \end{pmatrix}$$ and $$C=\begin{pmatrix} 0&1& 0 \\ 1& 0& 0 \\ 0&0& 0 \end{pmatrix}$$ generate a Lie group/ Lie algebra? I think they do not, because the commutators are outside of the set. Is this correct? It seems to me that they also do not generate a Lie group/algebra if each matrix is multiplied by the imaginary unit i. True?

Only the anticommutators are inside the set; but that does not define a Lie group/algebra. Do anticommutators also generate some algebraic structure?

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    $\begingroup$ Since these matrices are not invertible and you're talking about commutators, I think you mean to ask which Lie algebra they generate (not Lie group). $\endgroup$ – Eric O. Korman Jul 15 '13 at 19:18
  • $\begingroup$ Anticommutators (by which I imagine you mean the operation $\{a,b\}=\tfrac12(ab+ba)$) give you a structure of a Jordan algebra —notice I added a $\tfrac12$ in there. $\endgroup$ – Mariano Suárez-Álvarez Jul 15 '13 at 19:47
  • $\begingroup$ They do not form a Lie algebra on their own. (They don't even form a vector space.) But this isn't what the word "generate" means to my mind. "What Lie algebra do A, B and C generate?" means "what is the smallest Lie algebra containing A, B and C?". There is definitely a Lie algebra containing A, B and C (e.g. $\mathfrak{gl}_n$), and so there is definitely a smallest Lie algebra (take all the Lie algebras containing A, B and C, and intersect them). Consider an analogy: three vectors don't necessarily form a vector space, but they can generate (span) one. Likewise three generators of a group. $\endgroup$ – Billy Jul 15 '13 at 20:13
  • $\begingroup$ You are correct that the set does not form a Lie algebra because the set does not close under commutation, nor does it if you multiply by $i$, however it does close if you swap the number $1$ furthest to the bottom left for $-1$ in all matrices. $\endgroup$ – Graham Hesketh Jul 15 '13 at 21:25
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You are writing down three real Gell-Mann matrices, namely $\lambda_6,\lambda_4,\lambda_1$, respectively.

Their commutators and commutators thereof, and on and on, all close into the Lie algebra of SU(3), spanned by 8 generators. That is, successive commutations of these three will generate the remaining five λ matrices which will then all close into that Lie algebra--the most famous one in physics, arguably.

Specifically, the commutators of your set will generate $\lambda_7,\lambda_5,\lambda_2$, which, then, commuted with the original set will also generate $\lambda_3,\lambda_8$, as well.

But your set does not close to a sub algebra of SU(3) under commutation.

As you say, however, they plainly close into the set of 3 under anticommutation. I am not familiar with this structure. Some type of Jordan algebra as @Mariano Suárez-Álvarez remarks.

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