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Possible Duplicate:
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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    See [this post](http://math.stackexchange.com/a/24880/742) for discussion of objects satisfying all except for one of the axioms of a vector space.2012-03-07

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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, ...2012-03-07
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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.

See: http://en.wikipedia.org/wiki/Module_%28mathematics%29