Is the characteristic equation in ODE the same characteristic equation in linear algebra? Can someone show me whether this "characteristic equation" thing in ODE is the same characteristic equation that we derive for a matrix?
For example, given $y'' + 2y = 0$, the characteristic equation is $\lambda^2 = -2$
How does this equation correspond to those of a matrix?
 A: Suppose you start with a polynomial
$$
                       p(w) = w^{n}+a_{n-1}w^{n-1}+a_{n-2}w^{n-2}+\cdots+a_{1}w+a_{0}.
$$
Consider the differential equation $p(D)f=0$ where $D=\frac{d}{dx}$ is the differentiation operator. That is, consider the differential equation
$$
             f^{(n)}+a_{n-1}f^{(n-1)}+a_{n-2}f^{(n-2)}+\cdots+a_{1}f'+a_{0}f=0.
$$
The solution space $M$ of this differential equation is a finite-dimensional linear space of dimension $n$, and this solution space is invariant under the operator $D$. So $D$ is represented by a matrix $[D]$ on $M$, and the characteristic polynomial of $[D]$ is $p$. In fact the minimal and characteristic polynomials of $[D]$ are both $p$, which is independently interesting.
For your case: The solution space of $(D^{2}+2I)f=0$ is a linear subspace $M$ which is invariant under $D$. When $D$ is represented as a matrix $[D]$ on the solution space $M$, it has characteristic polynomial and minimal $p(x)=x^{2}+2$. For example, one such basis is $\{ e^{i\sqrt{2}x},e^{-i\sqrt{2}x}\}$, and the matrix $[D]$ in this basis is
$$
        [D]=\left[\begin{array}{cc}i\sqrt{2} & 0 \\ 0 & -i\sqrt{2}\end{array}\right],
$$
which has minimal and characteristic polynomial $(\lambda-i\sqrt{2})(\lambda+i\sqrt{2})=\lambda^{2}+2$.
A: Let's re-write the ODE as a first-order system:
$$\begin{align*}
y_1 &= y\\
y_2 &= y' \\
y'_1 &= y_2 \\
y'_2 &= 2y_1.
\end{align*}$$
This is a linear system, so let's let $\mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}$. Then,
$$\mathbf{y}' = \underbrace{\begin{pmatrix} 0 & 1 \\ -2 & 0 \end{pmatrix}}_{=A} \mathbf{y}.$$
Now, we can use the matrix exponential to solve this ODE in the usual way, but rather let's explore the characteristic equation of the matrix $A$:
$$A-\lambda I = \begin{pmatrix} -\lambda & 1 \\ -2 & -\lambda\end{pmatrix}\\
\det A-\lambda I = (-\lambda)^2 + 2.$$
Setting this equal to zero, we get
$$\lambda^2 +2 = 0$$
which is precisely the characteristic equation obtained through the "traditional" method of converting derivatives of $y$ to powers of $\lambda$.
