I've tried to find the characteristic polynomial of the following matrix
$$A=\begin{pmatrix} 0 & 0 & \cdots & 0 & -a_n \\ 1 & \ddots & \ddots & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots& \ddots & 0 & -a_2 \\ 0 & \cdots & 0 & 1 & -a_1 \end{pmatrix} \in M_{n \times n}(\mathbb{K})$$
for $\mathbb K$ being some element of a field and $n\in\mathbb N$.
I tried for several hours but I just couldn't succeed! Maybe someone can find an easy solution? Appreciated!
Edit: Apparently this matrix is called a companion matrix.
Take $A-\lambda I$. Now
proceed like this, and in the end by doing these elementary matrix operations you end up with the following upper triangular matrix which has the same determinant of $A-\lambda I$: $$ \begin{pmatrix} -\lambda & & & -a_n \\ & -\lambda & & -a_{n-1}-\lambda^{-1}a_n \\ & & \ddots & \vdots \\ & & & -a_1-\lambda-\sum_{i=2}^na_i\lambda^{1-i} \\ \end{pmatrix} $$ The determinant is simply the product of the elements of the diagonal, hence yielding $$ {\rm det}(A-\lambda I) = (-\lambda)^{n-1} \left( -a_1-\lambda-\sum_{i=2}^na_i\lambda^{1-i} \right) = (-1)^n\sum_{i=1}^na_i\lambda^{n-i} $$