If we look at an $n$ - dimensional vector space $V$ and a linear transformation \begin{equation} T : V \to V, \quad x \mapsto Tx \quad \forall \, x \in V \end{equation} 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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    $\begingroup$ This topic math.stackexchange.com/questions/4483/… may interest you. $\endgroup$ – Davide Giraudo Feb 13 '12 at 22:16
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    $\begingroup$ 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. $\endgroup$ – GEdgar Feb 13 '12 at 22:18
  • $\begingroup$ @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 :) $\endgroup$ – harlekin Feb 13 '12 at 22:26
  • $\begingroup$ @GEdgar ok that sounds interesting, it shows I have to dig deeper. Many thanks for your comment! $\endgroup$ – 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.

Three additional comments:

  • In most contexts, the easiest way to describe differentiation is as differentiation.
  • However, in complex analysis, differentiation actually is an integral, thanks to the Cauchy integral formula.
  • 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!
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    $\begingroup$ this is a really great answer, thanks very much for sharing your insight! $\endgroup$ – harlekin Feb 14 '12 at 1:01

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