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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!

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

1 Answer 1

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An easier approach: find the characteristic polynomial by examining the eigenvalue equation.

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 $\sum_{i=1}^n \alpha_{n-i} \lambda^{n-i} = \lambda^n$.

Then the last thing is to just work out the sign convention for the characteristic polynomial.

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.

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