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 nth 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 nth 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.

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