Prove that $\operatorname{Tr} (A^m) =\sum_{i} \lambda^m_i$

Question

The coefficients of the characteristic polynomial are expressed as $$ \chi_A(\lambda) = \lambda^n + D_1\lambda^{n-1} + \dots + D_n $$ Prove that $\operatorname{Tr} (A^m) = \sum_{i} \lambda^m_i$, where $\lambda_i$ are the eigenvalues of $A$ (a matrix) and from this prove the recurrent formula $$ m D_m + D_{m-1} \operatorname{Tr} (A) + D_{m-2} \operatorname{Tr} (A^2) + \dots + \operatorname{Tr} (A^m) = 0$$

My solution: Every diagonalisable matrix $A$ can be expressed as the product od the similar diagonal matrix $D$ with the eigenvalues on its diagonal and the bases containing the eigenvectors. Traces of those matrixes are same: $\operatorname{Tr}(A) = \operatorname{Tr} (D) = \sum_{i=1}^n a_{ii} = \sum_{i=i}^n \lambda_i$, so the same follows for $\operatorname{Tr} (A^m) = \operatorname{Tr} (D^m) = \sum_{i=i}^n \lambda^m_i$, for $1\le m \le n$.

I think I would prove that reccurent formula with Viet's relation, not sure how. Is there any proof of $\operatorname{Tr}(A) = \operatorname{Tr}(D)$?

I think in the recurrence formula the index should be $n$ not $m$. — Hyperplane, Mar 08 '17 at 14:42
More generally, a matrix is always unitarily upper triangularizable (Schur theorem), so $A^{m}\sim T^{m},$ and the diagonal entries of $T^{m}$ are the $m$th powers of the eigenvalues of $A.$ Then trace is preserved under similarity, which proves the first part. — RideTheWavelet, Mar 08 '17 at 14:51
Can't resist citing myself: The second part of your problem is Theorem 2.6 in The trace Cayley-Hamilton theorem. Of course, the proof I give is more general than necessary here (the whole point of that note is making do without eigenvalues). — darij grinberg, Mar 08 '17 at 15:24
Ah, here's a reference for your exact situation: Theorem 1 in arXiv:hep-th/0701116v1. Needless to say, the authors of this paper are hardly the first to invent it, but they seem to have written the proof quite nicely. — darij grinberg, Mar 08 '17 at 15:26

score 0 · Answer 1 · answered Mar 08 '17 at 14:42

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If you want to prove the recurrence formula, you would rather applies the Hamilton theorem which says that every matrix satisfies its own characteristic equation. As such, in the characteristic equation, substitute $\lambda$ by $A$. and then apply the equality on the trace operator given that: the trace of the sum of matrices is equal to sum of traces of each matrix.

Best

answered Mar 08 '17 at 14:42

Hmath

407

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I don't think the question follows from Cayley-Hamuilton. It does when $m \geq n$, but that's not the interesting case. – darij grinberg Mar 08 '17 at 15:27

score 0 · Answer 2 · answered Jun 30 '25 at 23:50

If you are willing to use a little topology, it is sufficient to consider only the diagonalizable case. More specifically, diagonalizable matrices are dense in the the set of all complex matrices, and the trace operator is continuous (as are polynomials), so if the result holds for diagonalizable $A$ then it holds for all $A$.

In the diagonalizable case, $$ A = PDP^{-1} $$ for invertible $P$ and where $D_{ij} = \lambda_{i} \delta_{ij}$. Therefore \begin{align*} tr(A^{n}) &= tr(PD^{n}P^{-1}) \\ &= tr(PP^{-1}D^{n}) \\ &= tr(D^{n}) \\ &= \sum \lambda_{i}^{m} \end{align*}

This approach also suffices to prove the Cayley-Hamilton Theorem, showing it holds for diagonalizable matrices and then using density. As Darij pointed out, the second claim is related to the trace version of Cayley-Hamilton, so it should work the same. Alternatively, you could just work directly from the result about $tr(A^{n})$, and note that $$ mD_{m} + D_{m-1}tr(A) + \dots + tr(A^{m}) = \sum_{i} \left( D_{m} + D_{m-1}\lambda_{i} + \dots + \lambda_{i}^{m} \right) $$ so you just need to prove that $$ \left( D_{m} + D_{m-1}\lambda_{i} + \dots + \lambda_{i}^{m} \right) = 0 $$ for each $\lambda_{i}$.

Prove that $\operatorname{Tr} (A^m) =\sum_{i} \lambda^m_i$

2 Answers2