I need help finding the inverse of the following operator. I am not sure about how to start. Any help would be hugely appreciated.
Operator: $( I + \frac{\partial^2}{\partial x^2})$
Edit:
I actually need the inverse of this whole thing:
$\begin{bmatrix} I + \beta\frac{\partial^2}{\partial x^2} & 0 \\ 0 & I \end{bmatrix}$.
I am assuming L2 space.
You haven't specified the domain and range of this function. If you mean the operator over the $C^\infty$ functions, note that it will not be invertible. Taking for example $f(x) = \sin(x)$, we have $$ \left(I + \frac{\partial^2}{\partial x^2}\right) f(x) = 0 $$ so your function is certainly not injective.
Suppose we are considering the operator over the space of functions with period $2$. We define the inner product $$ \langle f,g\rangle = \int_0^1 f(x) \overline{g(x)}\,dx $$ Then the family of functions $$ a(x) = 1\\ b_n(x) = \sin(n \pi x)\\ c_n(x) = \cos(n \pi x) $$ forms an orthonormal (Schauder) basis of this set. Relative to the basis $\{a,b_1,c_1,b_2,c_2,\dots\}$, we can represent the operator $(I + \beta\frac{\partial^2}{\partial x^2})$ as $$ \pmatrix{1\\ &1-\pi^2\beta\\ &&1-\pi^2\beta\\ &&&1 - 2^2 \pi^2 \beta\\ &&&&1 - 2^2 \pi^2 \beta\\ &&&&&1 - 3^2 \pi^2 \beta\\ &&&&&& \ddots} $$ We can find the inverse of this operator as we would find the inverse of any other diagonal operator.