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I have a task with a few different questions 'if a given set is a vector space'. I have chosen one to show my way of solving such examples and ask you if it is a correct way. So

We can simply say that these conditions

  • $(p+q)v = pv + qv$
  • $p(u+v) = pu + pv$
  • and so on...

(where u,v belong to $\Bbb Q_n[x]$ and $p,q$ belong to $\Bbb Q$)

Are gonna be fulfilled as we operate on 'normal/common' rational numbers And now I check if Q x Qn[x] gives us a rational number as a result yes? If so I can easily state that operations in which we use only rational numbers give us a rational result.

If my way of thinking is wrong please correct me and if you are able to do that let help me find the correct way of solving tasks like this one. Thanks.

James Smith
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    The polynomials of degree $\le n$ form a subspace, since addition and scalar multiplication of two such polynomials again has degree $\le n$. – Dietrich Burde Dec 12 '17 at 13:32
  • @DietrichBurde So there is no need to prove in any other way (p+q)v = pv + qv p(u+v) = pu + pv conditions As there are obvious in this case. Am I correct? – James Smith Dec 12 '17 at 13:39
  • @JamesSmith, I've tried to clean up your post by adding markup to the math symbols, but I don't know what you intended by "Q -> Qn[x]". Is that an arrow pointing to the right? What is that notation about? – G Tony Jacobs Dec 12 '17 at 13:42
  • @GTonyJacobs It was to be a "x". I saw it in a few examples about this topic. However more clearly it should be like:
    1. u + v is in ℚn[x]
    2. qu is in ℚn[x] aswell
     – James Smith Dec 12 '17 at 13:47
  • @JamesSmith Assuming you have already shown that the set of all polynomials is a vector space, then, yes, you have already proved such things as "commutativity of addition" and "multiplication distributes over addition" for all polynomials so do not have to show that for a subset. – user247327 Dec 12 '17 at 13:48

1 Answers1

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This is correct, but there’s a faster way than checking all of the axioms. The space you’re interested, call it $S$, in lies inside of another vector space, namely $\mathbb Q[x]$. This means that almost all of the properties will automatically hold because they hold in $\mathbb Q[x]$. There are only three things you need to check to see that $S$ is a vector space once you know that $S\subseteq\mathbb Q[x]$:

  1. $\forall u,v\in S, u+v\in S$
  2. $\forall c\in\mathbb Q\forall v\in S, c\cdot v\in S$
  3. $0\in S$

This is similar to how you don’t have to go prove all the group axioms and can instead use the subgroup criterion. Analogously, this is called te vector subspace criterion. In general, the $\mathbb Q$ in line $2$ needs to be whatever the field you think $S$ is a vector space over is.

Stella Biderman
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