Could you help me to solve this task? I am currently clueless and have no plan how I can solve this math question. The task was originally in German, but I tried my best to translate it into proper English.
Let $n \in \mathbb N$ and $V$ a $n$-dimensional Vector space over a field $\mathbb K$. Let $\phi: V\to V$ a endomorphism and $\chi_\phi(x) = \sum_{i=0}^n \lambda_i x^i$ the characteristic polynomial of $\phi$. Show that $$\lambda_n = (-1)^n, \lambda_{n-1} = (-1)^{n-1}.\text{Tr}(\phi) \text{ and } \lambda_0 = \det(\phi)$$
Thanks in advance!
Let $A = [a_{i,j}]$ be the matrix of $\varphi$ in the basis you want. Then $$\chi_{\varphi}(x) = \begin{vmatrix} a_{11} - x & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} -x & ... & a_{2n}\\ ... & ... & ... & ...\\ a_{n1} & a_{n2} & ... & a_{nn}-x\end{vmatrix}$$
Expand this determinant to see that:
the only way to obtain a term of degree $n$ is to make the product of the diagonal elements: $\displaystyle \prod_{i=1}^n (a_{ii}-x) = (-1)^nx^n + \sum_{i=1}^n a_{ii}\prod_{j\ne i} (a_{jj}-x)$ which gives $\lambda_n = (-1)^n$ and $\displaystyle \lambda_{n-1} = (-1)^{n-1}\sum_{i=1}^n a_{ii} = (-1)^{-1}\text{Tr}(\varphi)$
$\chi_{\varphi}(0) = \lambda_0 = \det(A-0.I) = \det(A) = \det(\varphi)$.