Let $A$ be a $n \times n$ matrix with polynomial characteristic

$$f = (x-c_{1})^{d_{1}}...(x-c_{k})^{d_{k}}.$$

prove that

$c_{1}d_{1} + ...+c_{k}d_{k} = trace(A).$

i try this: $A$ defines e linear operator $T: V \rightarrow V$ so by C.H. Theorem $f(T) = 0$ for all $v \in V$ then exists a basis of $V$ of eigenvalues of $c_{i}$ and that show $A$ is diagonalizable, but i notice that its wrong because if this was true then all linear operator are diagonalizable. Any sugestion for the solution in the same line of idea?

Eduardo Silva
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  • Related:http://math.stackexchange.com/questions/546155/proof-that-the-trace-of-a-matrix-is-the-sum-of-its-eigenvalues – Arpit Kansal Jan 24 '16 at 17:39

2 Answers2


Hint: Recall that trace of similar matrices is same,and every matrix is similar to a Jordan block matrix.

Arpit Kansal
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  1. Note that the sum $\Sigma c_i d_i$ is the coefficient of the second highest term in the characteristic polynomial (think about multiplying it out by picking $n-1$ x's, and then one of the $c_i$). (This is true to sign depending on the degree anyway.)

  2. Now, try computing from the definition as $det (T - tI)$ the characteristic polynomial. Only keep track of the second highest degree term, because that is all you care about in this case. You will find that you compute the trace. Be careful about sign.

Elle Najt
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