Characteristic Polynomial of a Higher-order system Transform the nth-order equation $y^{(n)}=a_0y+a_1y'+\cdots+a_{n-1}y^{(n-1)}$ into a system of first-order equations by setting $y_1=y$ and $y_j=y'_{j-1}$ for $j=2, \ldots, n$. Determine the characteristic polynomial of the coefficient matrix of this system.
 A: I'll show you how to do the transformation. We need to set this up as first order ode in terms of $$Y=\begin{bmatrix}y\\y^{(1)}\\y^{(2)}\\\vdots\\y^{(n-3)}\\y^{(n-2)}\\y^{(n-1)}\end{bmatrix}\text{ and }Y'=\begin{bmatrix}y^{(1)}\\y^{(2)}\\y^{(3)}\\\vdots\\y^{(n-2)}\\y^{(n-1)}\\y^{(n)}\end{bmatrix}.$$ Lets think about a linear equation relating $Y'_1=y^{(1)}$ to the elements of $Y$. We notice rather quickly that $y^{(1)}=Y_2$, so we can write $$Y'_1=\sum_{j=1}^nm_{1j}Y_j$$ where $m_{12}=1$ and $m_{1j}=0$ in all other cases. We can find similar equations for $Y'_2,\dots, Y'_{n-1}$. For $Y'_n$, we just use the ode $$y^{(n)}=a_0y+\dots+a_{n-1}y^{(n-1)},$$ so we can write $$Y'_n=\sum_{j=1}^nm_{nj}Y_j$$ with $m_{nj}=a_{j-1}$. We can summarize this in the matrix equation $$\begin{bmatrix}y^{(1)}\\y^{(2)}\\y^{(3)}\\\vdots\\y^{(n-2)}\\y^{(n-1)}\\y^{(n)}\end{bmatrix}=
\begin{bmatrix}0&1&0&\dots&0&0&0\\
0&0&1&\ddots&0&0&0\\
0&0&0&\ddots&\ddots&0&0\\
\vdots&\vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\
0&0&0&\dots&0&1&0\\
0&0&0&\dots&0&0&1\\
a_0&a_1&a_2&\dots&a_{n-3}&a_{n-2}&a_{n-1}\end{bmatrix}
\begin{bmatrix}y\\y^{(1)}\\y^{(2)}\\\vdots\\y^{(n-3)}\\y^{(n-2)}\\y^{(n-1)}\end{bmatrix},$$ or more simply $$Y'=MY.$$ To find the characteristic polynomial, you should calculate $$\text{det}(M-\lambda I_n).$$ To take the determinant, consider using row/column operations, or the Laplace expansion https://en.wikipedia.org/wiki/Laplace_expansion
A: For anyone who encounters this same problem, I'll comment that after you get to the matrix equation Y' = MY where M is the sparse matrix shown by hellowworld112358 in their answer, you should consider separately the cases where n is even and n is odd, and perform a Laplace expansion on the bottom row.
