Let \(A\) be an \(n\times n\) complex matrix. Suppose \(m\ge1\) and \(A^m\) is diagonalizable. Prove that \(A^{m+1}\) is diagonalizable.
Solution
Let \(V\coloneqq\CC^n\). For each eigenvalue \(\lambda\) of \(A^m\), set \[ E_\lambda\coloneqq\ker(A^m-\lambda I). \]
1. The space \(V\) decomposes as\[ V=\bigoplus_\lambda E_\lambda, \]and every \(E_\lambda\) is invariant under \(A\).
The direct-sum decomposition follows from the diagonalizability of \(A^m\). Since \(A\) commutes with \(A^m\), if \(v\in E_\lambda\), then \[ A^m(Av)=A(A^m v)=A(\lambda v)=\lambda Av. \] Thus \(Av\in E_\lambda\), so each \(E_\lambda\) is \(A\)-invariant.
2. On \(E_0\), the operator \(A^{m+1}\) is zero and hence diagonalizable.
If \(v\in E_0\), then \(A^m v=0\), so \[ A^{m+1}v=A(A^m v)=0. \] Thus the restriction of \(A^{m+1}\) to \(E_0\) is the zero operator.
3. If \(\lambda\ne0\), then the restriction\(B\coloneqq A|_{E_\lambda}\) is diagonalizable.
For every \(v\in E_\lambda\), \[ B^m v=A^m v=\lambda v, \] so \(B\) is annihilated by \[ p(x)=x^m-\lambda. \] Because \(\lambda\ne0\) and the ground field is \(\CC\), the polynomial \(p\) has no repeated root: a common root of \(p\) and \(p'(x)=mx^{m-1}\) would have to be both nonzero and zero. Hence the minimal polynomial of \(B\) divides a polynomial with distinct linear factors. Therefore \(B\) is diagonalizable.
4. For every eigenvalue \(\lambda\) of \(A^m\), the restriction of\(A^{m+1}\) to \(E_\lambda\) is diagonalizable.
2. If \(\lambda\ne0\), step 3shows that \(B=A|_{E_\lambda}\) is diagonalizable. Therefore its power\[ B^{m+1}=A^{m+1}|_{E_\lambda} \]is diagonalizable as well.
5. The matrix \(A^{m+1}\) is diagonalizable.
By step 1, \(V\) is the direct sum of the \(A\)-invariant spaces \(E_\lambda\). Step 4 gives a basis of each \(E_\lambda\) consisting of eigenvectors of \(A^{m+1}\). The union of these bases is therefore an eigenbasis of \(V\) for \(A^{m+1}\).
6. Q.E.D.
Step 5 is the required conclusion.