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All of these are true. First note that the space of endomorphisms of $V$ is finite-dimensional, so even an infinite $S$ can just be replaced by finitely many matrices that have the same span (it's really more elegant to think about the span of $S$ as a Lie algebra, rather than $S$ itself). You actually may want to look at some discussion of abelian Lie algebras, since really your question is about the natural structure theorem for semi-simple representations of abelian Lie algebras (if you think in this language question 4) is obvious from the first 3, since any representation has a flag whose successive quotients are semi-simple).

The important point for proving 1) is that if A and B commute and are both diagonalizable, you should analyze the action of B on the eigenspaces of A. The spaces $E_\chi$ above are the eigenspaces of B's action on each eigenspace of A (and if there were a third matrix, you would take the eigenspaces of C acting on the eigenspaces of B in the eigenspaces of A, etc.). 2) is clear, and for 3) just pick any basis of these iterated eigenspaces.

For 3) => 1), you have associated to each basis vector a $\chi$, given by looking at how the elements of $S$ act on it. $E_\chi$ is just the span of all vectors associated to the particular map $\chi$.

For 4), there is a similar argument, but you have to use a flag (the $i$-th subspace being things killed by $(A-\lambda I)^i$ for some scalar $\lambda$) rather than a subspace decomposition. Still, matrices commuting means that B will preserve this flag, so just as one refined the eigenspaces, one can refine this flag.

3 added 206 characters in body; added 271 characters in body

All of these are true. First note that the space of endomorphisms of $V$ is finite-dimensional, so even an infinite $S$ can just be replaced by finitely many matrices that have the same span (it's really more elegant to think about the span of $S$ as a Lie algebra, rather than $S$ itself).

The important point for proving 1) is that if A and B commute and are both diagonalizable, you should analyze the action of B on the eigenspaces of A. The spaces $E_\chi$ above are the eigenspaces of B's action on each eigenspace of A (and if there were a third matrix, you would take the eigenspaces of C acting on the eigenspaces of B in the eigenspaces of A, etc.). 2) is clear, and for 3) just pick any basis of these iterated eigenspaces.

For 3) => 1), you have associated to each basis vector a $\chi$, given by looking at how the elements of $S$ act on it. $E_\chi$ is just the span of all vectors associated to the particular map $\chi$.

For 4), there is a similar argument, but you have to use a flag (the $i$-th subspace being things killed by $(A-\lambda I)^i$ for some scalar $\lambda$) rather than a subspace decomposition. Still, matrices commuting means that B will preserve this flag, so just as one refined the eigenspaces, one can refine this flag.

2 added 220 characters in body

All of these are true. The important point for proving 1) is that if A and B commute and are both diagonalizable, you should analyze the action of B on the eigenspaces of A. The spaces $E_\chi$ above are the eigenspaces of B's action on each eigenspace of A (and if there were a third matrix, you would take the eigenspaces of C acting on the eigenspaces of B in the eigenspaces of A, etc.). 2) is clear, and for 3) just pick any basis of these iterated eigenspaces.

For 4), there is a similar argument, but you have to use a flag (the $i$-th subspace being things killed by $(A-\lambda I)^i$ for some scalar $\lambda$) rather than a subspace decomposition. Still, matrices commuting means that B will preserve this flag, so just as one refined the eigenspaces, one can refine this flag.

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