Let \(A\) be an \(n\times n\) complex matrix such that \[ \operatorname{tr}(A^k)=0 \qquad(k=1,\ldots,n). \] Prove that \(A\) is nilpotent.
Solution
Let \(\lambda_1,\ldots,\lambda_n\) be the eigenvalues of \(A\), counted with algebraic multiplicity, and set \[ p_k=\sum_{j=1}^n\lambda_j^k. \]
1. For \(1\leq k\leq n\),\[ p_k=0. \]
Over \(\CC\), the matrix \(A\) is triangularizable. In an upper-triangular form its diagonal entries are \(\lambda_1,\ldots,\lambda_n\), and the diagonal entries of its \(k\)th power are \(\lambda_1^k,\ldots,\lambda_n^k\). Hence \[ p_k = \operatorname{tr}(A^k) =0 \] by hypothesis.
2. Every elementary symmetric polynomial\[ e_r=e_r(\lambda_1,\ldots,\lambda_n), \qquad 1\leq r\leq n, \]vanishes.
1. Thus \(r e_r=0\). Since the ground field has characteristic zero,\(e_r=0\).
3. The characteristic polynomial of \(A\) is\[ \chi_A(t)=t^n. \]
The characteristic polynomial is \[ \chi_A(t) = \prod_{j=1}^n(t-\lambda_j) = t^n-e_1t^{n-1}+e_2t^{n-2}-\cdots+(-1)^ne_n. \] All \(e_r\) vanish by step 2.
4.\[ A^n=0. \]
By the Cayley–Hamilton theorem, \[ \chi_A(A)=0. \] Using step 3 gives \(A^n=0\).
5. The matrix \(A\) is nilpotent.
This is immediate from step 4.
6. Q.E.D.
Step 5 is the required conclusion.