common invariant subspace

Now picture not one operator but a whole family of them — a set S of operators on V. A common invariant subspace is a subspace W that every member of S leaves invariant at once: A(W) is contained in W for every A in S. It is a single subspace that all the operators respect simultaneously.

Finding a common invariant subspace is the key that unlocks simultaneous reduction. If a family shares a nontrivial proper common invariant subspace, you can put all of them into block-triangular form in one shared basis; if no such subspace exists, the family is called irreducible. The contrast frames the central question: when must a family of operators have one?

Burnside's theorem gives the deepest answer over the complex numbers: a set of operators is irreducible (has NO common invariant subspace except {0} and V) if and only if it spans the entire algebra of all operators on V. So having a shared invariant subspace is the same as failing to generate everything — sharing structure means the family is, algebraically, small. For commuting families the story is even cleaner: they always share at least a common eigenvector, which seeds simultaneous triangularization.