Row rank = Column rank for infinite-dimensional matrix Let $V$ and $W$ be vector spaces over field $F$. Let $T: V \to W$ be a linear transformation.
Using the axiom of choice, choose a basis $\{e_i\}$ for $V$ and $\{f_j\}$ for $W$, indexed by $I$ and $J$ respectively.
The columns are $\{T(e_i) \mid i \in I\} \subseteq W$.
I'm going to have to brute-force the rows because the adjoints wouldn't make sense.
Decompose $T(e_i)$ into $\displaystyle \sum_{j \in J} a_{ij} f_j$. Then, the rows are $\displaystyle\left\{\sum_{i \in I} a_{ij} e_i ~\middle|~ j \in J \right\} \subseteq V$.



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*Is the notion of rank (i.e. maximum number of linearly independent elements) well-defined for infinite sets?

*Must the column rank equal the row rank?

 A: The notion of rank / column rank continues to make perfect sense: it's the dimension of the image of $T$, where dimension is still defined as the cardinality of a basis (which is still independent of the choice of basis).
Unfortunately your attempt at defining the rows does not work; the problem is that the sums $\sum_{i \in I} a_{ij} e_i$ may be infinite, and so don't obviously make sense. For an explicit example, we might have $T(e_i) = f_1$ for all $i$, so that $a_{i1} = 1$ for all $i$. More generally, the matrix of a linear transformation between infinite-dimensional vector spaces has finitely many entries in each column but need not have finitely many entries in each row; among other things this means that the transpose of such a matrix is not another such matrix. 
In finite dimensions, the row rank of a linear transformation $T : V \to W$ can be defined as the rank of its adjoint $T^{\ast} : W^{\ast} \to V^{\ast}$. This definition continues to make sense in infinite dimensions, although since the dual vector spaces of infinite-dimensional vector spaces are larger than them, it's no longer true that row rank is equal to column rank with this definition. 
For example, take $T$ to be the identity linear operator from a countable-dimensional vector space $V$ to itself. Then $T^{\ast}$ is the identity linear operator from $V^{\ast}$ to itself, which has uncountable dimension.
For a positive result, I think everything is still fine if the rank of $T$ is finite.  
