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?
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.