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I'm doing a question asking "A set of all polynomials with degree exactly 5. Does it form a vector space?" I'm a bit confusing showing multiplication part. If say the polynomial is ax^5+b, when the scalar is 0, then the polynomial is 0 degree. Does this prove this is not a vector space? Or can I argue that (0*a)x^5+(0*b) is included as it still has that form? So zero vector is in the space?

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Firstly, $0$ most definitely does not have degree 5. But say you include it anyway, i.e. try salvaging the problem statement.

You still have the right idea that closure is not satisfied. More specifically, you have the idea but applied it backwards. If $a$ and $b$ are in your candidate vector space, and $a+b$ does not fall into your set, then you just proved it is not a vector space.

This is something that can only happen with sub-sets of other vector spaces. Say you have some set $S$ (say, $\mathbb{R}^2$) and you want to define an operation $+$ on it that makes it a vector space. In this case, the very act of defining addition gives you closure for free, since that's what definitions mean. You certainly can't have a function on $\mathbb{R}$ where $f(3)$ is the color purple, right? You would definitely not call that a function from $\mathbb{R}$ to $\mathbb{R}$. However, if $V$ is a candidate vector space and $U \subset V$ as sets, it makes sense to talk about addition on $U$ by simply borrowing the operation on $V$. But then there is no reason whatsoever to think $+|_{U}$ is a function from $U$ to $U$. You have to prove that it lies in $U$. A priori, you only know that it falls in $V$. And if it doesn't, then it doesn't make sense to call the set $U$ a vector space using the addition and scalar multiplication operations on $V$.

I'll be even more formal. Say $F$ is a field and $V$ is a set together with a function $+: V \times V \rightarrow V$ and a scalar multiplication function $\cdot: F \times V \rightarrow V$, that satisfy the axioms for a vector space. Say $U \subset V$ with the property that $+(U \times U) \subset U$. Then there is another function, $+|_U : U \times U \rightarrow U$ you can define as $+|_U(u_1, u_2) = u_1 + u_2$, where the last $+$ refers to $V$'s plus operator. If scalar multiplication works out analogously, then $(U, +|_U, \cdot|_{F \times U})$ is a vector space. But it certainly isn't if $+(U \times U)$ is not a subset of $U$.

In conclusion, all you have to do is find two polynomials with degree exactly 5 whose sum is not exactly degree 5, to prove that restricting the $+$ operation does not give you a $+$ operation on your set since it isn't closed. Scalar multiplication by $0$ will also do in the problem as originally stated (not to mention the very lack of a zero vector).

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  • $\begingroup$ Thank you very much for your reply! I have showed that when scalar is zero, zero vector is not included in the set. But my friend keeps arguing that although scalar is zero,(0*a)x^5+(0*b)still has the form of ax^5+b, so zero vector is included. So I am just a bit confused by this expression. So in a word, zero is not in the set therefore the set is not vector space right? $\endgroup$ – Henry Ma Aug 30 '14 at 3:11
  • $\begingroup$ @HenryMa how do you define the degree of a polynomial? $\endgroup$ – djechlin Aug 30 '14 at 3:12
  • $\begingroup$ the highest degree of its terms $\endgroup$ – Henry Ma Aug 30 '14 at 3:15
  • $\begingroup$ @HenryMa what's the highest degree of ax^5+b when a is 0? $\endgroup$ – djechlin Aug 30 '14 at 3:16
  • $\begingroup$ 0, yeah i understand now. zero vector is not included $\endgroup$ – Henry Ma Aug 30 '14 at 3:17

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