Let \(A\) and \(B\) be diagonalizable linear transformations of \(\mathbb R^n\) satisfying \(AB=BA\). Let \(E\) be an eigenspace of \(A\). Prove that the restriction \(B|_E\) is diagonalizable.
Solution
1. The eigenspace \(E\) is invariant under \(B\).
Let \(\lambda\) be the eigenvalue of \(A\) corresponding to \(E\), and let \(v\in E\). Then \[ Av=\lambda v. \] Using \(AB=BA\), \[ A(Bv) = B(Av) = \lambda Bv. \] Thus \(Bv\in E\), so \(B(E)\subseteq E\).
2. The minimal polynomial of \(B|_E\) divides the minimal polynomial of \(B\).
Let \(m_B(t)\) be the minimal polynomial of \(B\). Since \[ m_B(B)=0 \] on \(\RR^n\), its restriction to the \(B\)-invariant subspace \(E\) from step 1 is also zero: \[ m_B(B|_E)=0. \] By the defining divisibility property of the minimal polynomial, \[ m_{B|_E}(t)\mid m_B(t). \]
3. The polynomial \(m_{B|_E}\) splits over \(\RR\) into distinct linear factors.
Because \(B\) is diagonalizable over \(\RR\), its minimal polynomial splits into distinct linear factors: \[ m_B(t) = \prod_{j=1}^s(t-\mu_j) \] with the \(\mu_j\) distinct. By step 2, \(m_{B|_E}\) divides this polynomial, so it too is a product of distinct linear factors over \(\RR\).
4. Therefore\[ \boxed{B|_E\text{ is diagonalizable}}. \]
A linear transformation over \(\RR\) is diagonalizable exactly when its minimal polynomial splits into distinct linear factors. Step 3 verifies this criterion for \(B|_E\).
5. Q.E.D.
Step 4 is the required conclusion.