Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it possible to consider a vector space over another vectorspace instead over a field as usual, where came into play that we need a field? And in such a vector space, the vector could be represented as tuples from $V^k$, where V is the vector space from which the other space is built from, and so the vectors are tuples of vectors (which could be considered as matrices). Is something wrong with such constructions?

share|improve this question
3  
Such a $V^k$ does not have any notion of multiplication by an element of $V$, only by elements of the field, so $V^k$ would still be a vector space over the original field. –  Thomas Andrews Jul 17 '12 at 19:46
1  
"Where came into play that we need a field?" The big advantage is the existence of bases and so. More generally you can look at modules over rings (not necessarily fields), but some of the important structure theorems of linear algebra are no longer true. –  Cocopuffs Jul 17 '12 at 19:56
    
I think the confusion starts with the implicit supposition that a vector space over a field $F$ is necessarily some $F^k$; it isn't. The field plays a role (only) for the notion of scalar multiplication. –  Marc van Leeuwen Jul 17 '12 at 23:38
add comment

2 Answers 2

up vote 3 down vote accepted

First of all, you can consider the space of matrices (of a particular size) to be a vector space over the scalar field directly, since you can add matrices and multiply a matrix by a scalar. There's a good chance the this will actually be sufficient for whatever concrete application (if any) you have in mind.

Another concept that may align somewhat with what you have in mind is that of a tensor product, which will allow you to view a matrix as a sum of various "row vectors" multiplied by each of a basis of "column vectors", or vice versa. That has more of the features you seem to be thinking of, but also a somewhat steeper learning curve.

However, if you want an actual "vector space over $V$", you run into the problem that the right-hand side of the vector space axiom $$ a\cdot(b\cdot v) = (ab) \cdot v $$ makes sense only if you can multiply scalars, and a general vector space $V$ does not have a concept that fits into that position.

Of course, if you can somehow define a multiplication operation on your vector space, you can plug it into the "scalar field" slot of the definition of vector space and see what happens. How such a multiplication should look depends on the vector space, and on your whims and desires. However, if you want the result to behave even slightly like a vector space, you'll want your multiplication to satisfy some reasonable rules -- at least associativity and distributivity over the vector addition, and possibly also commutativity. (Note in passing that requiring associativity rules out using the cross product in $\mathbb R^3$ here).

A vector space equipped with an associative and distributive multiplication is called an algebra, or a commutative algebra if the multiplication is commutative. You can speak of a "vector space" over an algebra, except that for historical reasons such as thing is called a module over the algebra rather than a "vector space". Since we're not assuming that things have multiplicative inverses or identities, modules can be a bit wilder than vector spaces in general.

If your multiplication is so nicely behaved that the original vector space happens to be a field (for example, $\mathbb C$ can be throught of as $\mathbb R^2$ equipped with a particular multiplication operation that makes it a field), you can of course have true vector spaces over that.

share|improve this answer
2  
A vector space (over $k$) equipped with the structure of a field is called a field extension of $k$. –  Colin McQuillan Jul 17 '12 at 20:31
add comment

Yes and no, depending on what you mean. The major reason we need a field (as opposed to, say, a ring or group) is to have multiplication and division operations. A vector space (underlying field aside) has none. That is, $a\vec{v}$ makes sense, but $\vec{u}\vec{v}$ does not. So a vector space cannot act as a replacement for a field, since it doesn't have all of the needed properties that a field does.

In terms of "a space of linear maps/matrices" there is the dual space, which is the space of all transformations from your original vector space to the underlying field. You could also consider a vector space of matrices (i.e. of all linear transformations) that map between two spaces, but this still uses the same underlying field that the original vector space did. So

  1. Can you make a vector space of linear transformations (essentially, of matrices)? Yes!
  2. Can you make a vector space with a field replaced by another vector space? No, because several of the vector space axioms no longer make sense.
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.