Eigenvalues of the companion matrix

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. $$$$\det \left( \begin{bmatrix} \alpha_{k-1} & \alpha_{k-2} & \dots & \alpha_1 & \alpha_0 \\ 1 & 0 & \dots & 0 & 0 \\ 0 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & 0 \end{bmatrix} - \lambda I_{k \times k} \right) = (-1)^k \left[\lambda^k - \sum_{j=0}^{k-1} \alpha_j \lambda^j \right] \label{eq6}$$$$ With this assumption, we wish to show that the desired result is also satisfied for $$n=k+1$$. Toward this end, we consider $$$$\det \left( \begin{bmatrix} \alpha_{k} & \alpha_{k-1} & \alpha_{k-2} & \dots & \alpha_1 & \alpha_0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ 0 & 1 & 0 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 1 & 0 \end{bmatrix} - \lambda I_{(k+1) \times (k+1)} \right).$$$$ Performing a cofactor expansion along the last column of the matrix above yields $$$$(-1)^{k+2} \alpha_0 \det \left( \begin{bmatrix} 1 & -\lambda & 0 & \dots & 0 \\ 0 & 1 & -\lambda & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & -\lambda \\ 0 & 0 & 0 & \dots & 1 \end{bmatrix} \right) \\ + (-1)^{2(k+1)} (-\lambda) \det \left( \begin{bmatrix} \alpha_{k} & \alpha_{k-1} & \alpha_{k-2} & \dots & \alpha_1 \\ 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 0 \end{bmatrix} - \lambda I_{k \times k} \right).$$$$ The first determinant above is simply $$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!

• It is easier to find the form of possible eigenvectors instead.
– lhf
Commented Oct 22, 2019 at 16:52

Suppose $$Ax=\lambda x$$, then $$x_1=\lambda x_2,x_2=\lambda x_3,\dots,x_{n-1}=\lambda x_n$$. Thus $$x_k=\lambda^{n-k}x_n$$ for all $$k$$. One may assume that $$x_n \neq 0$$ because if it were then the eigenvector would be zero, so one may assume WLOG that it is $$1$$. So you have $$x_k=\lambda^{n-k}$$ for all $$k$$.
Now finally the top equation reads $$\sum_{i=1}^n \alpha_{n-i} x_i = \lambda x_1$$ which by substitution gives $$\sum_{i=1}^n \alpha_{n-i} \lambda^{n-i} = \lambda^n$$.
If you want to do it with cofactor expansion, you can, but it's easiest to do by expanding across the first row so that the $$\alpha$$'s are not involved in the minors. Doing that, with the first cofactor you get $$(\alpha_{n-1} - \lambda) M_1$$ and otherwise you get $$(-1)^{i-1} \alpha_{n-i} M_i$$, where in both cases $$M_i$$ is the determinant of the submatrix given by removing the first row and the $$i$$th column. Now these submatrices are triangular and $$M_i$$ has $$n-i$$ $$-\lambda$$'s and $$i-1$$ $$1$$'s on the diagonal, so their determinants are $$(-1)^{n-i} \lambda^{n-i}$$. Sum up the cofactors and you get the desired result.