The Char. Poly. of a Linear Homo. nth Order ODE with Constant Coef.

# The Characteristic Polynomial of a Linear Homogeneous nth Order ODE with Constant Coefficients

Theorem 1: Let $y^{(n)} + a_{n-1}y^{(n-1)} + ... + a_1y' + a_0y = 0$ be a linear homogeneous $n^{\mathrm{th}}$ order ODE with constant coefficients. Then the characteristic polynomial $p(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + ... + a_1\lambda + a_0$ is the same as the characteristic polynomial of the companion matrix of the ODE. That is, $\det (\lambda I - A) = p(\lambda)$. |

**Proof:**We prove this result by induction on submatrices from the bottom right on the companion matrix. Consider the companion matrix for this linear homogeneous $n^{\mathrm{th}}$ order ODE with constant coefficients:

\begin{align} \quad A = \begin{bmatrix} 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 \\ -a_0(t) & -a_1(t) & -a_2(t) & -a_3(t) & \cdots & a_{n-1}(t) \end{bmatrix} \end{align}

- For each $k \in \{ 1, 2, ..., n \}$ we let $A_k$ be the lower right $k \times k$ submatrix of $A$.

- For the base step, we have that:

\begin{align} \quad \det (\lambda I_1 - A_1) = \lambda + a_{n-1} \end{align}

- Assume that $\det (\lambda I_k - A_k) = \lambda^k + a_{k-1}\lambda^{k-1} + ... + a_{n-k}$. Then we compute $\det (\lambda I_{k+1} - A_{k+1})$ by expanding along the first column of $\lambda I_k - A_k$ to get:

\begin{align} \quad \det (\lambda I_{k+1} - A_{k+1}) &= \begin{vmatrix} \lambda & -1 & 0 & \cdots & 0 & 0\\ 0 & \lambda & -1 & \cdots & 0 & 0\\ 0 & 0 & \lambda & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & \lambda & -1 \\ -a_{n-k-1} & -a_{n-k} & -a_{n-k+1} & \cdots & -a_{n-2} & -a_{n-1} \end{vmatrix} \\ &= \lambda \det (\lambda I_k - A_k) + (-1)^{k+1} a_{n-1} (-1)^{k-1} \\ & \overset{I.H.} = \lambda (\lambda^k + a_{k-1} \lambda^{k-1} + ... + a_{n-k} + a_{n - k+1} \\ &= \lambda^{k+1} + a_{k-1} \lambda k + ... + a_{n-k}\lambda + a_{n-k+1} \end{align}

- So by the principle of mathematical induction, the result holds for all $k \in \{ 1, 2, ..., n-1 \}$. $\blacksquare$