Since an operator is usually associated with a matrix, in an operation such as $A:X\longrightarrow X$, giving $A\bf{x}=\bf{F}(x)$, I would give A=I as:
\begin{equation}A= \begin{pmatrix} 1&0 \\ \\ 0&1 \end{pmatrix} \end{equation}
But then, since we also have another type of operator, such as the differential operator, how would that look like as a matrix?
I know we can write it as:
$$Tx(t)=x'(t)$$
and satisfies the operation $$T:X\longrightarrow X$$
But can it be given as a matrix? I found this, but I am not sure I completely understand it.
Thanks for any response.
The differential operator acts on functions and spaces of functions are usually infinite dimensional so you would need an infinite matrix. However if you restrict to some sufficiently small space you can write the differential as a matrix.
For example consider the collection of all polynomials of degree at most 3. This is a 4-dimensional vector space where one possible basis is given by $1, x, x^2, x^3$. In this basis the polynomial $x^3+2x^2-x+5$ would correspond to the vector $(5,-1,2,1)$. Can you write out the matrix that corresponds to differention in this vector space?