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Is it possible to construct a quasi-vectorial space without an identity element?

I am looking for an example of a set and operations on this set that isn't quite a vector space. As in it meets some of the requirements, but not all of them.

For example it could meet all of the definitions except associativity and therefore not a vector space.

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marked as duplicate by Arturo Magidin, Asaf Karagila, Nate Eldredge, t.b., Jonas Teuwen Mar 8 '12 at 13:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See this post for discussion of objects satisfying all except for one of the axioms of a vector space. – Arturo Magidin Mar 7 '12 at 20:51

2 Answers 2

Invertible $n\times n$ matrices over some field will do it, I believe. Take the vectors to be $n \times n$ invertible matrices, but take vector addition to be matrix multiplication. It fails on two counts not one, though: 'addition' isn't commutative and scalar multiplication is not distributive.

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There is in fact nothing special about matrices in this example. You can take anything that you can multiply by the same species of objects and scalars, and the same idea still works :) Ideas include: * functions on some space $X$, possibly with some additional conditions, like continuous * special classes of matrices: invertible, diagonal, ... – Jakub Konieczny Mar 7 '12 at 20:50

You might be interested in the notion of a module. A module is something that resembles a vector space in that you have addition and scalar multiplication. The gist is that instead of scalars in some vector field, you have a ring.


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