I have a big question about the coeffecient of characteristic polynomials. Indeed I want to prove that for
$$p_f(X) = X^n + a_{n-1} X^{n-1} + \cdots + a_0$$
hence $a_n = (-1)^q tr(∧ ^qf)$.
I already know that we have
$$p_f(X) = X^n + (-1)^{n-1}tr(M)X^{n-1}+\cdots+ \det(M)$$
with $M$ the matrix of $f$. Hence concretly I have to prove that $tr(M) = tr(∧ ^qf)$ ? If yes how can I do it ?
$p_f$ is the characteristic polynomial of my function f ( endomorphism), $tr( ∧^qf)$ is the trace of the exterior product of f, M the matrix of my function.
Thank you in advance,
Question: "Hence concretly I have to prove that $tr(M)=tr(∧qf)$? If yes how can I do it ?"
Answer: If $k$ is a field and $V$ is an $n$-dimensional vector space over $k$ and $\phi: V \rightarrow V$ is a $k$-linear map, you may take the top exterior power of $\phi$ to get a $k$-linear map
$$\wedge^n \phi: \wedge^n V \rightarrow \wedge^n V$$
and $\wedge^n V$ is a one-dimensional vector space. The map $\wedge^n \phi$ is a $k$-linear map with the property that if you choose a basis $v$ for $\wedge^n V$ the map is as follows:
$$\wedge^n \phi(v)=det(A)v$$
where $A$ is a matrix defining $\phi$ with respect to a basis for $V$. Choosing a basis $e_1,..,e_n$ for $V$ it follows $v:=e_1 \wedge \cdots \wedge e_n$ is a basis for $\wedge^n V$.