Associated matrix to operator on infinite dimensional spaces. A linear operator on a vector space has a basis through which write its associated matrix. This is certainly true for finite dimensional spaces. But is it still true for infinite dimensional spaces? I can not imagine a matrix written with a base that contains infinite elements. How should it be this matrix? With endless columns and endless rows?
 A: Yes. A linear map between two $K$-vector spaces $V$ and $W$ with given bases can be represented as a matrix with respect to these bases. When the spaces are infinite-dimensional, this will indeed be a matrix with "endless" columns and "endless" rows. (But every column has only finitely many nonzero entries, because linear combinations are finite.) A matrix (in a sufficiently general meaning of this word) over the field $K$ is just a family $\left(a_{i,j}\right)_{\left(i,j\right)\in I\times J}$ of elements of $K$ indexed by a product $I \times J$ of two sets $I$ and $J$. The matrices usually seen in linear algebra books have $I = \left\{1,2,\ldots,n\right\}$ and $J = \left\{1,2,\ldots,m\right\}$, but there is nothing wrong with allowing arbitrary $I$ and $J$.
For instance, consider the ring $K\left[x\right]$ consisting of all polynomials in the variable $x$ over our ground field $K$. The matrix which represents the map $K\left[x\right] \to K\left[x\right], \ p \mapsto p'$ (where $p'$ means the derivative of $p$) with respect to the basis $\left(1,x,x^2,\ldots\right)$ of the $K$-vector space $K\left[x\right]$ has the form
$\left(\begin{array}{ccccc} 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & 2 & 0 & \cdots \\ 0 & 0 & 0 & 3 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots\end{array}\right)$.
