Some of the comments seem to imply that the concept of a basis is somehow hard to define for infinite-dimensional spaces. That's not the case. The definition
If $V$ is a vector space over the scalar field $K$, then a family $B=(b_i)_{i\in d}$ of vectors is a basis of $V$ exactly if for every vector $x$ there is a unique family $(x_i)_{i\in d}$ of scalars with $x_i \neq 0$ for only finitely many $i$, such that $x = \sum_{i \in d} c_ix_i$. The family $(x_i)_{i\in d} =: \mathfrak{C}_B(x)$ is called the coordinatization of $x$ in basis $B$.
works for vector spaces of arbitrary dimension. Note that the crucial point is that every vector must be representable by a finite linear combination of basis vectors. Most property carry over from the finite-dimensional case - in particular, you still have that
If $T \,:\, V \to V$ is a linear map, and $(b_i)_{i\in d}$ a basis of $V$, then $T$ is fully determined by the images of the $b_i$ under $T$, i.e. by $(Tb_i)_{i\in d}$.
It follows that you can also generalize matrices, by allowing arbitrarily large index sets, and (just as in the finite-dimensional case), require the "columns" to be the images of the basis vectors.
If $T \,:\, V \to V$ is a linear map, and $B$ a basis of $V$, then the family of scalar $(a_{ij})_{i,j\in d} =: \mathfrak{C}_B(T)$ were $$
(a_{ij})_{i \in d} = \mathfrak{C}_B(Tb_j) \text{ for all $j \in d$,}
$$
i.e. where $(a_{i,j})_{i\in d}$ for a fixed $j$ is the coordinatization of the image of $b_j$ under $T$, is called the coordinatization of $T$ in $B$ or matrix of $T$ in $B$. One may call the family $(a_{ij})_{i \in d}$ for a fixed $j$ the $j$-th column (of $(a_{ij})_{i,j\in d}$), and the family $(a_{ij})_{j \in d}$ for a fixed $i$ the $i$-th row. Every coordinatization has the property that each column contains only finitely non-zero entries.
Just as in the finite-dimensional case, you can then define $A\cdot x$ for a pair of coordinatizations $A=(a_{ij})_{i,j\in d}$ and $x=(x_j)_{j\in d}$, by setting
$A\cdot x := (y_i)_{i\in d}$, where $y_i = \sum_{j \in d} a_{ij} x_j$.
Since $x_i \neq 0$ only finitely many times, it's clear that the sum always exists. So the question remains, is it alwas a valid coordinatization, i.e. is $y_i \neq 0$ also only finitely many times? One can restrict the attention to those $n$ columns of $A$ which correspond to non-zero $x_j$. Since each columns of a coordinatization $\mathfrak{C}_B(T)$ contains onyl finitely many non-zero entries, say $m_1,\ldots,m_n$ for the $n$ columns of interest, it follows that $y_i$ contains at most $m_1+\ldots+m_n$ non-zero entries.
You also get a product of matrices $A\cdot B$, by collecing the products of $A\cdot x$ as $x$ ranges over the columns of $B$, i.e. you have
$A\cdot B := (c_{ij})_{i,j\in d}$ where $(c_{i,j})_{i \in d} = A\cdot (b_{ij})_{i \in d}$
for all $j \in d$, i.e. the $j$-th column of $A\cdot B$ is $A$ times the $j$-th column of $B$.
