Prove or give a counterexample: if \(A\) is an invertible complex \(n\times n\) matrix and some positive power of \(A\) is diagonal, then \(A\) is diagonalizable.
Solution
1. Let \(k\geq1\) be such that\[ D\coloneqq A^k \]is diagonal. Then every eigenspace of \(D\) is invariant under \(A\).
Since \(D=A^k\), one has \[ AD=A^{k+1}=DA. \] If \(v\) belongs to the eigenspace \[ V_\mu\coloneqq\ker(D-\mu I), \] then \[ D(Av) = A(Dv) = A(\mu v) = \mu Av. \] Thus \(Av\in V_\mu\).
2. For every eigenvalue \(\mu\) of \(D\), one has \(\mu\neq0\), and therestriction\[ A_\mu\coloneqq A|_{V_\mu} \]satisfies\[ A_\mu^k=\mu I_{V_\mu}. \]
Because \(A\) is invertible, so is \(D=A^k\); hence \(0\) is not an eigenvalue of \(D\). Step 1 makes \(A_\mu\) well-defined. On \(V_\mu\), \[ A_\mu^k = D|_{V_\mu} = \mu I_{V_\mu}. \]
3. Each restriction \(A_\mu\) is diagonalizable over \(\CC\).
By step 2, the minimal polynomial of \(A_\mu\) divides \[ p_\mu(x)=x^k-\mu. \] Since \(\mu\neq0\), \[ p_\mu'(x)=kx^{k-1} \] has no common root with \(p_\mu\): a common root would have to be \(0\), but \(p_\mu(0)=-\mu\neq0\). Thus \(p_\mu\) has distinct roots over \(\CC\). Any divisor of it also has distinct roots, so the minimal polynomial of \(A_\mu\) is squarefree and splits over \(\CC\). Hence \(A_\mu\) is diagonalizable.
4. The matrix \(A\) is diagonalizable.
Because \(D\) is diagonal, \[ \CC^n = \bigoplus_{\mu}V_\mu, \] where the sum ranges over the distinct diagonal values of \(D\). By step 1 these summands are \(A\)-invariant, and by step 3 each has a basis of eigenvectors of the corresponding restriction of \(A\). The union of these bases is a basis of \(\CC^n\) consisting of eigenvectors of \(A\).
5. Therefore the statement is\[ \boxed{\text{true}}. \]
Step 4 proves the asserted diagonalizability.
6. Q.E.D.
Step 5 answers the prove-or-counterexample question.