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In a linear algebra book I've found, on the chapter about characteristic polynomials it says somewhere that if $A \in \Bbb F^{n \times n},c_{A}(x)=(-1)^{n}x^n+a_{n-1}x^{n-1}+...+a_0$

$\implies a_0=detA, a_{n-1}=(-1)^{n-1}TrA$

I'm wondering if $a_{n-1}=(-1)^{n-1}TrA$ only true for the term $a_{n-1}x^{n-1}$ or it holds also for $a_{n-2}x^{n-2},a_{n-3}x^{n-3}$ ... ?

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    Why would it, and why would it stop holding for $a_0$ ? – 2017-02-08

2 Answers 2

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For the sake of the counterexample, consider a $3\times3$ diagonal matrix with coefficients $1,2,4$. By Vieta's formulas, its characteristic polynomial is

$$(1-x)(2-x)(4-x)=-x^3+7x^2-14x+8.$$

(Of course, $\text{Tr}A=7,\det A=8$.)

Side note:

Should the property hold, any characteristic polynomial would have the form

$$(-1)^{n}x^n+Tx^{n-1}-Tx^{n-2}+Tx^{n-3}-...+D$$

which has only two independent coefficients. That would make the matrices quite "information-poor".

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    To the upvoters: I have changed the answer, giving a numerical counter-example instead, for simplicity. If you don't like it, feel free to un-upvote. – 2017-02-08
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    Thank you. (also to the power of three^) – 2017-02-08
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    @Michael: ooops, fixed :) – 2017-02-08
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Let $A\in \mathbb{R}^{n\times n}$. Then the characteristic polynomial $P_A(X)$ is defined as $P_A(X)=\det(X\cdot Id-A)$. Clearly $P_A(X)$ is a monic polynomial of degree $n$. By the fundamental theorem of algebra, we can write $$P_A(X)=\prod_{i=1}^n(X-\lambda_i)$$ where the roots $\lambda_i$ are complex numbers in general.

Now one can expand this product to discover what the coefficients are in terms of the roots. Let's take an easy example first. Suppose $n=3$, then \begin{eqnarray}P_A(X)&=&(X-\lambda_1)(X-\lambda_2)(X-\lambda_3)\\&=&X^3-(\lambda_1+\lambda_2+\lambda_3)X^2+(\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3)X-\lambda_1\lambda_2\lambda_3.\end{eqnarray}

It is well-known that $\text{Tr}(A)=\sum_{i=1}^n\lambda_i$ and $\det(A)=\prod_{i=1}^n\lambda_i$. Hence this accounts for the second and last coefficient in $P_A(X)$.

In general you can find by expanding $P_A(X)=\prod_{i=1}^n(X-\lambda_i)$ that the coefficients are given by symmetric polynomial expressions in the roots/eigenvalues $\lambda_i$ (up to sign). In fact this last observation is somehow crucial to the idea of Galois theory and already indicates how the symmetric polynomials enter this theory naturally.

Remark: Often the characteristic polynomial is defined by $P_A(X)=\det(A-X\cdot Id)$. There is nothing wrong with this definition, but the polynomial is no longer monic as the highest degree term might be $-1$ and I don't like that.