Let \(\phi, \psi\) be commuting endomorphisms of a finite-dimensional vector space \(E\) over a field \(K\) (so \(\phi \psi = \psi \phi\)). Show that if both \(\phi\) and \(\psi\) are diagonalizable (i.e. \(E\) has a basis of eigenvectors for \(\psi\) and a basis of eigenvectors for \(\phi\)), then \(E\) has a basis consisting of simultaneous eigenvectors for both \(\psi\) and \(\phi\).
Interpret the previous statement in terms of matrices similar to a diagonal matrix.
Solution
Let \(\lambda_1, \ldots, \lambda_k \in K\) be the distinct eigenvalues of \(\psi\) and \(E_{\lambda_i}(\psi) = \{v \in E : \psi(v) = \lambda_i v\}\). Since \(\psi\) is diagonalizable, \(E = \bigoplus_{i=1}^k E_{\lambda_i}(\psi)\).
1. (a) \(E\) has a basis of simultaneous eigenvectors of \(\phi\) and \(\psi\).
1.1. Each \(E_{\lambda_i}(\psi)\) is \(\phi\)-invariant.
For \(v \in E_{\lambda_i}(\psi)\), \(\psi(\phi(v)) = \phi(\psi(v)) = \lambda_i \phi(v)\).
1.2. The restriction \(\phi_i=\phi|_{E_{\lambda_i}(\psi)}\) is diagonalizable.
An endomorphism of a finite-dimensional space is diagonalizable if and only if its minimal polynomial is a product of distinct linear factors. The minimal polynomial of \(\phi\) has this form, and the minimal polynomial of \(\phi_i\) divides it, so it has this form as well.
1.3. Q.E.D.
By step 2, each \(E_{\lambda_i}(\psi)\) has a basis \(\mathcal{B}_i\) of eigenvectors of \(\phi\); each vector of \(\mathcal B_i\) is also a \(\lambda_i\)-eigenvector of \(\psi\). Since \(E = \bigoplus_i E_{\lambda_i}(\psi)\), the union \(\bigcup_i \mathcal{B}_i\) is a basis of \(E\) consisting of simultaneous eigenvectors.
2. (b) If \(A, B \in M_n(K)\) commute and are each similar to a diagonal matrix, then there is \(P \in \operatorname{GL}_n(K)\) with \(P^{-1} A P\) and \(P^{-1} B P\) both diagonal.
Apply step 1 to the endomorphisms \(x\mapsto Ax\) and \(x\mapsto Bx\) of \(K^n\) to obtain a basis \(v_1, \ldots, v_n\) of simultaneous eigenvectors, \(Av_j=\mu_jv_j\) and \(Bv_j=\lambda_jv_j\). The matrix \(P\) with columns \(v_1, \ldots, v_n\) is invertible, and \(P^{-1} A P = \operatorname{diag}(\mu_1, \ldots, \mu_n)\), \(P^{-1} B P = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)\).