Let \(A\) be an \(n \times n\) matrix with characteristic polynomial \[ f = (x - c_1)^{d_1} \cdots (x - c_k)^{d_k}. \] Show that \[ c_1 d_1 + \cdots + c_k d_k = \operatorname{trace}(A). \]
Solution
1. Coefficient of \(x^{n-1}\) from the determinant definition of the characteristic polynomial:
1.1. By definition, the characteristic polynomial of \(A = [a_{ij}] \in M_n(F)\) is \(f(x) = \det(xI - A)\).
definition of characteristic polynomial.
1.2. By the Leibniz formula for the determinant:\[ \det(xI - A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n (x \delta_{i, \sigma(i)} - a_{i, \sigma(i)}). \]
Leibniz formula for determinants.
1.3. For any non-identity permutation \(\sigma \neq \operatorname{id}\), \(\sigma\) fixes at most \(n - 2\) elements, so the corresponding product has degree at most \(n - 2\) in \(x\).
permutations with at least \(n - 1\) fixed points must be the identity.
1.4. Thus the terms of degree \(n\) and \(n - 1\) arise solely from the diagonal product (\(\sigma = \operatorname{id}\)):\[ \prod_{i=1}^n (x - a_{ii}) = x^n - \left(\sum_{i=1}^n a_{ii}\right) x^{n-1} + O(x^{n-2}) = x^n - \operatorname{trace}(A) x^{n-1} + O(x^{n-2}). \]
expanding \(\prod_{i=1}^n (x - a_{ii})\) and definition of trace.
1.5. Thus the coefficient of \(x^{n-1}\) in \(f(x)\) is \(-\operatorname{trace}(A)\).
3 and 4.
2. Coefficient of \(x^{n-1}\) from the factored form of \(f(x)\):
2.1. The factored characteristic polynomial is \(f(x) = (x - c_1)^{d_1} \cdots (x - c_k)^{d_k}\), with \(\sum_{j=1}^k d_j = n\).
given.
2.2. By Vieta’s formulas, the coefficient of \(x^{n-1}\) in a monic polynomial with roots (counted with multiplicity) is the negative of the sum of the roots:\[ f(x) = x^n - \left(\sum_{j=1}^k d_j c_j\right) x^{n-1} + \cdots + (-1)^n \prod_{j=1}^k c_j^{d_j}. \]
expanding \((x - \lambda_1) \cdots (x - \lambda_n)\) where \(\lambda_1, \dots, \lambda_n\) lists each \(c_j\) repeated \(d_j\) times.
2.3. Thus the coefficient of \(x^{n-1}\) in \(f(x)\) is \(-(c_1 d_1 + \cdots + c_k d_k)\).
2.
3. Equating the coefficients:
3.1. Equating the coefficients of \(x^{n-1}\) obtained in 1 and 2 gives:\[ -\operatorname{trace}(A) = -(c_1 d_1 + \cdots + c_k d_k). \]
uniqueness of polynomial coefficients.
3.2. Multiplying both sides by \(-1\) yields:\[ \operatorname{trace}(A) = c_1 d_1 + \cdots + c_k d_k. \]
scalar multiplication by \(-1\).
4. Conclusion:\(c_1 d_1 + \cdots + c_k d_k = \operatorname{trace}(A)\). Q.E.D.
3.