Goal
I would like to show the following
$$\det\left(\begin{bmatrix} \alpha_{n-1} & \alpha_{n-2} & \dots & \alpha_{1} & \alpha_{0} \\ 1 & 0 & \dots & 0 & 0 \\ 0 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & 0 \end{bmatrix} - \lambda I\right) = (-1)^n \left[\lambda^n - \sum_{j=0}^{n-1} \alpha_j \lambda^j \right]$$
Attempt
We proceed by induction.
For the base case of $n=1$, the companion matrix is $[\alpha_0]$ and its characteristic polynomial is \begin{align} \det([\alpha_0] - \lambda I) & = \alpha_0 - \lambda \\ & = (-1) (\lambda - \alpha_0). \end{align} So, the base case satisfies the desired result. Next, for the inductive step, we suppose that the desired result is satisfied when $n=k$ for some $k>1$; i.e. \begin{equation} \det \left( \begin{bmatrix} \alpha_{k-1} & \alpha_{k-2} & \dots & \alpha_1 & \alpha_0 \\ 1 & 0 & \dots & 0 & 0 \\ 0 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & 0 \end{bmatrix} - \lambda I_{k \times k} \right) = (-1)^k \left[\lambda^k - \sum_{j=0}^{k-1} \alpha_j \lambda^j \right] \label{eq6} \end{equation} With this assumption, we wish to show that the desired result is also satisfied for $n=k+1$. Toward this end, we consider \begin{equation} \det \left( \begin{bmatrix} \alpha_{k} & \alpha_{k-1} & \alpha_{k-2} & \dots & \alpha_1 & \alpha_0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ 0 & 1 & 0 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 1 & 0 \end{bmatrix} - \lambda I_{(k+1) \times (k+1)} \right). \end{equation} Performing a cofactor expansion along the last column of the matrix above yields \begin{equation} (-1)^{k+2} \alpha_0 \det \left( \begin{bmatrix} 1 & -\lambda & 0 & \dots & 0 \\ 0 & 1 & -\lambda & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & -\lambda \\ 0 & 0 & 0 & \dots & 1 \end{bmatrix} \right) \\ + (-1)^{2(k+1)} (-\lambda) \det \left( \begin{bmatrix} \alpha_{k} & \alpha_{k-1} & \alpha_{k-2} & \dots & \alpha_1 \\ 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 0 \end{bmatrix} - \lambda I_{k \times k} \right). \end{equation} The first determinant above is simply $1$ -- since its argument is a triangular matrix with only $1$'s on the main diagonal -- while the second determinant can be evaluated with our inductive hypothesis. Also, note that $(-1)^{k+2} = (-1)^k$ and $(-1)^{2(k+1)} = 1$. With these observations, the quantity above simplifies to \begin{align} & (-1)^k \alpha_0 + (-\lambda) (-1)^k \left[\lambda^k - \sum_{j=1}^{k} \alpha_j \lambda^j \right] \qquad \qquad (1) \\[15pt] = & (-1)^k \left(\alpha_0 - \lambda \left[\lambda^k - \sum_{j=1}^{k} \alpha_j \lambda^j \right] \right) \\[15pt] = & (-1)^{k+1} \left(-\alpha_0 + \lambda \left[\lambda^k - \sum_{j=1}^{k} \alpha_j \lambda^j \right] \right) \\[15pt] = & (-1)^{k+1} \left(-\alpha_0 + \left[\lambda^{k+1} - \sum_{j=1}^{k} \alpha_j \lambda^{j+1} \right] \right) \\[15pt] = & (-1)^{k+1} \left[\lambda^{k+1} - \sum_{j=0}^{k} \alpha_j \lambda^{j+1} \right] \end{align}
...incomplete.
Problem
In the summation above, I need $\lambda^{j}$, not $\lambda^{j+1}$. I think there's a problem with the bounds on the summation in $(1)$, but it's not clear to me what that problem is.
If you could help me to finish this proof, I would very much appreciate it!