Problem
Let $p(T) := T^n-\alpha_{n-1}T^{n-1}-\alpha_{n-2}T^{n-2}-\cdots-\alpha_0 \in K[T]$. Additionally we have the companion matrix of $p$ $$A:= \begin{bmatrix} 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0\\ 0 & \ddots & \ddots & \ddots & \ddots & \vdots \\\vdots & \ddots & \ddots & \ddots & 1 & 0 \\0 & 0 & \ddots & \ddots & 0 & 1 \\\alpha_0 & \alpha_1 & \alpha_2 & \cdots & \alpha_{n-2} & \alpha_{n-1}\end{bmatrix}$$
Furthermore, the characteristic polynomial of $p$ is denoted by $\chi_A(T)$
I need to show the following:$(a)$ $\;\chi_A(T) = (-1)^n p(T)$
$(b)$ $\;$ If $\lambda$ is a root of $p(T)$, then $\left( 1, \lambda, \ldots, \lambda^{n-1} \right)^T$ is an eigenvector of $A$ for eigenvalue $\lambda$
$(c)$ $\;$ Suppose $p(T)$ has $n$ distinct roots $\lambda_1, \ldots, \lambda_n \in K$. Find a matrix $S$ such that $S^{-1}AS$ is a diagonal matrix.
Approach
The given companion matrix $C(p) = A$ is transposed. The finite degree $n$ determines the number of columns of $A$, and because $A$ is a square matrix $A \in K^{n \times n}$. We can calculate $\chi_A(T)$ by solving $det(A - \lambda I) = 0$ and applying Laplace expansion. Picking the first row, by expansion we would get something like this
$$-\begin{vmatrix} 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & \vdots & \vdots & 1 & 0 \\ 0 & 0 & \cdots & 0 & 1 \\ \alpha_0 & \alpha_2 & \cdots & \alpha_{n-2} & \alpha_{n-1} \end{vmatrix} \in K^{n-1 \times n-1}$$
So if you repeat this the coefficients $\alpha_i$ would be eliminated. How would this be equal to $\pm p(T)$?