One can write a homogeneous linear differential equation with real constant coefficients $a_k \in \mathbb R$ of degree $n \in \mathbb N$ \begin{equation} \tag{1} 0 = \sum_{k = 0}^{n} a_{k} \cdot u^{(k)}(t) \qquad \text{with } a_n = 1 \end{equation} as a system of $n$ differential equations like so: \begin{align*} \begin{pmatrix} u \\ \vdots \\ u^{(n - 1)} \end{pmatrix}' + \underbrace{\begin{pmatrix} 0 & - 1 & 0 & \ldots & 0 \\ 0 & 0 & - 1 & 0 & 0 \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \ldots & \ldots & 0 & -1 \\ a_0 & a_1 & \ldots & \ldots & a_{n - 1} \end{pmatrix}}_{:= A \in \mathbb R^{n \times n}} \begin{pmatrix} u \\ \vdots \\ u^{(n - 1)} \end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix} \end{align*} The characteristic polynomial of $(1)$ is defined to be $\chi(\lambda) := \sum_{k = 0}^{n} a_{k} \cdot \lambda^{k}$ with $a_n = 1$. I want to show that the characterisitic polynomial of $A$, $\det(\lambda I - A)$, is equal to the characterisitic polynomial of the differential equation.
Here's is what I tried to prove it. If $A = \begin{pmatrix} B & C \\ D & E \end{pmatrix}$ is a block matrix where $E$ is a scalar and thus $C$ and $D$ are vectors, we have $\det(A) = (E - D B^{-1} C) \det(B)$, if $B$ is invertible.
Defining $$B := \begin{pmatrix} \lambda & -1 & 0 & \ldots & 0 \\ 0 & \lambda & -1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \ldots & 0 & \lambda & - 1 \\ 0 & \ldots & \ldots & 0 & \lambda \end{pmatrix} \in \mathbb R^{(n - 1) \times (n - 1)}, \qquad C := \begin{pmatrix} 0 \\ \vdots \\ 0 \\ - 1 \end{pmatrix} \in \mathbb R^{n - 1}, \qquad D :=\begin{pmatrix} a_0 & a_1 & \ldots & a_{n - 3} & a_{n - 2} \end{pmatrix} \in \mathbb R^{1, n - 1}$$ I obtain for $\lambda \ne 0$ (as $\det(B) = \lambda^{n - 1}$) \begin{align*} \det(\lambda I - A) & = \det\begin{pmatrix} \lambda & 1 & 0 & \ldots & 0 \\ 0 & \lambda & 1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \ldots & 0 & \lambda & 1 \\ a_0 & a_1 & \ldots & a_{n - 2} & a_{n - 1} + \lambda \end{pmatrix} \\ & = \lambda^{n - 1} \left[ (a_{n - 1} + \lambda) - \begin{pmatrix} a_0& \ldots & a_{n - 2} \end{pmatrix} \begin{pmatrix} \lambda^{-1} & \lambda^{-2} & \ldots & \lambda^{-(n - 1)} \\ 0 & \lambda^{-1} & \ddots & \vdots \\ \vdots & \ddots & \ddots & \vdots \\ 0 & \ldots & 0 & \lambda^{-1} \end{pmatrix} \begin{pmatrix} 0 \\ \vdots \\ 0 \\ - 1 \end{pmatrix} \right] \\ & = \lambda^{n - 1} \left[ (a_{n - 1} + \lambda) + \begin{pmatrix} a_0& \ldots & a_{n - 2} \end{pmatrix} \begin{pmatrix}\lambda^{-(n - 1)} \\ \vdots \\ \lambda^{-1} \end{pmatrix} \right] \\ & = \lambda^{n} + a_{n - 1} \lambda^{n - 1} + \begin{pmatrix} a_0 & \ldots & a_{n - 2} \end{pmatrix} \begin{pmatrix} 1 \\ \vdots \\ \lambda^{n-2} \end{pmatrix} \\ & = \lambda^{n} + a_{n - 1} \lambda^{n - 1} + \sum_{k = 0}^{n - 2} a_k \lambda^k. \end{align*} Edit I miscalculated $B^{-1}$, so now the calculation works. Is there a simpler way to show this result?