Let \(A\) and \(B\) be commuting linear operators on a finite-dimensional vector space. What can be said about their eigenspaces? When can they be simultaneously diagonalized?
Solution
Suppose \[ AB=BA. \] Let \(E_\lambda(A)=\ker(A-\lambda I)\) be an eigenspace of \(A\). If \(v\in E_\lambda(A)\), then \[ A(Bv)=B(Av)=B(\lambda v)=\lambda Bv. \] Hence \[ B(E_\lambda(A))\subseteq E_\lambda(A). \] Thus every eigenspace of \(A\) is \(B\)-invariant. By symmetry, every eigenspace of \(B\) is \(A\)-invariant.
If, in addition, both \(A\) and \(B\) are diagonalizable over the base field, then they are simultaneously diagonalizable. Indeed, \[ V=\bigoplus_\lambda E_\lambda(A), \] and each \(E_\lambda(A)\) is \(B\)-invariant. The restriction of a diagonalizable operator to an invariant subspace is diagonalizable, so each \(E_\lambda(A)\) has a basis of \(B\)-eigenvectors. Combining these bases gives a basis of simultaneous eigenvectors for \(A\) and \(B\).