TITLE: Is there an algorithm known to decompose quiver representation?
QUESTION [9 upvotes]: We have a finite dimensional representation of a finite quiver over, say, the rationals. Is there an algorithm known to decompose this representation into its irreducible components?
A related question: we have two (finite-dimensional) representations of a finite quiver. Is there an algorithm to check if one is a summand of the other?
Thanks a lot for hints and pointers!

REPLY [8 votes]: If you want to decompose a finite-dimensional representation over, say, $\mathbb{Q}$ into its indecomposables over, say, $\overline{\mathbb{Q}}$, there is not only an algorithm but an efficient one (at least theoretically efficient: polynomial-time). See 
Chistov, A., Ivanyos, G. and Karpkinski, M. Polynomial Time Algorithms for Modules over Finite Dimensional Algebras, ISSAC 1997.
It works more generally over finite fields, and the real or algebraic closures of number fields.
Not only that, but they also show how to test efficiently if two such representations are equivalent. These two pieces together answer your second question (when is one rep, say $R_1$, contained as a summand in another $R_2$): decompose both into indecomposables, then see if every indecomposable in $R_1$ appears in $R_2$ with greater or equal multiplicity (test each indecomposable of $R_1$ for equivalence to each indecomposable in $R_2$).
Update: Brooksbank and Luks (J. Algebra 2008, or freely available author's copy) provide another algorithm for these problems that is apparently efficient in both theory (polynomial-time) and practice (based on experiments in MAGMA). (This is all over finite-dimensional algebras; the OQ didn't specify whether the quiver could have directed cycles or not...)