Some review in Lie Algebra

Proof. For two commuting matrices A and B, each eigenspace of A is invariant under B. Indeed, if (A−λIn)v = 0 then (A−λIn)Bv = B(A−λIn)v = B(0) = 0. For n = 1, each element of F is a complex number; thus, the statement of the lemma is true. Suppose the statement of the lemma is true for all n < m for some m ≥ 2. Let F be a set of mutually commuting diagonalizable matrices in Mm(C). Take A ∈ F . Let E(λ1), E(λ2),...,E(λk) be the eigenspaces of A. E(λj) = {v ∈ C : (A− λjIn)v = 0} , C = E(λ1)⊕ E(λ2)⊕ ...⊕ E(λk). Each B ∈ F can be viewed as a linear operator on E(λj). Since dimE(λj) < m, by the induction hypothesis, E(λj) has a basis Bj that diagonalizes every member of F . Then the basis of C obtained by concatenating B1, B2,...,Bk diagonalizes every member of F .