How to find trace of adj$A$ from the characteristic polynomial of $A$?

Question

Let the characteristic polynomial for $A$ be $t^n+c_1 t^{n-1}+c_2t^{n-2}+\cdots+c_{n-1}t+c_n$. From it, is it possible to find the trace of adj$(A)$ ?

Ewan Delanoy · Accepted Answer · 2014-09-08T11:49:13.753

3

If $A$ is diagonalizable with eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n$,

$$ {\sf trace}(adj(A))={\sf trace}(\det(A)A^{-1})= \lambda_1\lambda_2\ldots \lambda_n (\frac{1}{\lambda_1}+\ldots+\frac{1}{\lambda_n})= (-1)^{n-1}c_{n-1} $$

This stays true for nondiagonalizable $A$’s because the diagonalizable matrices form a dense subset of $M_n({\mathbb R})$.

edited Sep 08 '14 at 11:49

answered Sep 08 '14 at 11:42

Ewan Delanoy

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+1. However, if I understand correctly, I believe you mean $c_{n-1}$ instead of $c_2$. – user2097 Sep 08 '14 at 11:44
@user2097 : corrected, thanks. – Ewan Delanoy Sep 08 '14 at 11:49
@user2097 but anyway, the solution is really nice and beautiful. thank you so much – KON3 Sep 08 '14 at 11:56
Isn't this also true trigonalizable? – Bman72 Sep 08 '14 at 15:08
@Ale It is true for all matrices. – Ewan Delanoy Sep 08 '14 at 15:11

How to find trace of adj$A$ from the characteristic polynomial of $A$?

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