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?
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 $A$:
$$A-\lambda I = \begin{pmatrix} -\lambda & 1 \\ -2 & -\lambda\end{pmatrix}$$ and $$ \det (A-\lambda I) = (-\lambda)^2 + 2.$$
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 $\lambda$.
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 $(\lambda-i\sqrt{2})(\lambda+i\sqrt{2})=\lambda^{2}+2$.
