Let \(A\) be an \(n\times n\) complex matrix. Prove that if \[ A^k=I \] for some positive integer \(k\), then \(A\) is diagonalizable.
Solution
Let \(m_A(t)\) denote the minimal polynomial of \(A\).
1. One has\[ m_A(t)\mid t^k-1. \]
The hypothesis \(A^k=I\) says \[ (A^k-I)=0, \] so the polynomial \(t^k-1\) annihilates \(A\). By the defining property of the minimal polynomial, \(m_A(t)\) divides every polynomial that annihilates \(A\).
2. The polynomial\[ t^k-1 \]has no repeated roots in \(\CC\).
Its derivative is \[ kt^{k-1}. \] A repeated root would be a common root of \(t^k-1\) and \(kt^{k-1}\). Since the base field has characteristic zero, the only root of \(kt^{k-1}\) is \(0\), while \(0\) is not a root of \(t^k-1\). Hence no repeated root exists.
3. The minimal polynomial \(m_A(t)\) splits over \(\CC\) into distinct linearfactors.
The polynomial \(t^k-1\) splits completely over \(\CC\), and by step 2 its linear factors are distinct. Step 1 shows that \(m_A(t)\) is a divisor of that polynomial, so \(m_A(t)\) also splits and has no repeated root.
4. The matrix \(A\) is diagonalizable over \(\CC\).
A complex matrix is diagonalizable if and only if its minimal polynomial splits into distinct linear factors. Step 3 verifies this criterion for \(A\).
5. Q.E.D.
Step 4 is the required conclusion.