## The matrix of coefficients of a system of first order differential equations arising from an nth order differential equation

Let $\sum_{k=0}^n a_ky^{(k)} = 0$ ($a_n = 0$) be an $n$th order differential equation with constant (real) coefficients. To this equation we can associate a linear system of first order differential equations by defining $y_i = y^{(i-1)}$ for $1 \leq i \leq n$, where $y^(0) = y$. A solution of this associated system of equations then gives a solution $y$ of the original $n$th order equation in the first entry.

Prove that the matrix of coefficients of the $n \times n$ system described above is the transpose of the companion matrix of $p(x) = \sum_k a_kx^k$.

Note that $\frac{d}{dt} y_i = \frac{d}{dt} y^{(i-1)} = y^{(i)} = y_{i+1}$ for $i < n$, and $\frac{d}{dt} y_n = y^{(n)} = - \sum_{k=0}^{n-1} a_k y^{(k)}$ $= -\sum_{k=1}^n a_{k-1}y_k$. So the corresponding matrix is $A = [\alpha_{i,j}]$, where $\alpha_{i,j} = 1$ if $j = i+1$, $a_{j-1}$ if $i = n$, and 0 otherwise. Then $A^\mathsf{T} = [\beta_{i,j}]$, where $\beta_{i,j} = 1$ if $i = j+1$, $a_{i-1}$ if $j = n$, and 0 otherwise. This is precisely the companion matrix of $p(x) = \sum a_kx^k$, as desired.