# Linear transformations in infinite dimensional vector spaces

If we look at an $n$ - dimensional vector space $V$ and a linear transformation $$T : V \to V, \quad x \mapsto Tx \quad \forall \, x \in V$$ then given a choice of basis for $V$ one can represent $T$ in terms of a $n \times n$ matrix $A_T = A_T(i,j)$

Is this also the case for linear transformations on infinite - dimensional vector spaces, where we replace the matrix $A_T$ by an integral ?

In particular, since differentiation is a linear map, that would mean differentiation can be written in the form of an integral ... I realize this is either a dumb question (because it is obviously wrong) or it is some classic result I haven't found yet. In both cases it would be great to get some reference where I can learn more about it, many thanks!

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This topic math.stackexchange.com/questions/4483/… may interest you. –  Davide Giraudo Feb 13 '12 at 22:16
In general, no. Not every linear transformation is described by an integral kernel. Many important linear transformations are, but also many important linear transformations are not. –  GEdgar Feb 13 '12 at 22:18
@DavideGiraudo thanks, exactly what I was asking, I searched for a similar question before posting this but didn't find it - I guess I have to refine the tags I use when I do a search :) –  harlekin Feb 13 '12 at 22:26
@GEdgar ok that sounds interesting, it shows I have to dig deeper. Many thanks for your comment! –  harlekin Feb 13 '12 at 22:26

The axiom of choice is equivalent to the assertion that any vector space (no matter how large) has a basis. Given a choice of basis of an arbitrary vector space $V$, one can represent a linear transformation $T : V \to V$ using an infinite row-finite matrix (this means that only finitely many entries in each row are nonzero).
However, in practice this is not useful, since most infinite-dimensional vector spaces (such as the space of smooth functions on $\mathbb{R}$, for example) are huge and don't have reasonable bases. To study these, we instead introduce a topology, and then we do functional analysis instead of linear algebra.
• For some $V$ one can talk about integral operators which on the one hand are a useful generalization of matrices but on the other hand do not represent all operators one might care about. In the examples I know about, integral operators are compact, for example, and differentiation usually is not; it's generally not even continuous!