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Let vector space $W$ be $\{p(z)\in P_{3}(\mathbb{C}):p(a)=0 \Rightarrow a\in \mathbb{R}\}$ under usual addition and usual scalar multiplication.

How to show that W is not a vector space over $\mathbb{C}$?

It should violate one or more of the 10 axioms of complex vector space, but I am stuck here, I tried and it all seems that 10 axioms holds, how can W not be an complex vector space?

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Both $p(x)=x^2$ and $q(x)=1$ belong to $W$. However, $p(x)+q(x)=x^2+1\notin W$, since its roots are $\pm i$, none of which is real.

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  • $\begingroup$ oh, okay, thank you!!! $\endgroup$ – Logan Nov 23 '20 at 20:00
  • $\begingroup$ $q(x)=1$ is in W ? how is $q(x)=1$ in W? $\endgroup$ – Logan Nov 23 '20 at 20:08
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    $\begingroup$ Sure. The set of roots of $q(x)$ is $\emptyset$ and $\emptyset\subset\Bbb R$. $\endgroup$ – José Carlos Santos Nov 23 '20 at 20:09
  • $\begingroup$ wait, the set of roots of q(x) is 0? so q(x)=1=0, how can the root be 0? I don't think the root exist $\endgroup$ – Logan Nov 23 '20 at 20:20
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    $\begingroup$ What I wrote whas that the set of roots of $q(x)$ is $\emptyset$, which is the symbol for the empty set. In other words, I wrote that $q(x)$ has no roots. $\endgroup$ – José Carlos Santos Nov 23 '20 at 20:23
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Jose already picked out an answer, but it’d also suffice to observe that the zero polynomial isn’t in $W$.

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  • $\begingroup$ hi, 0 polynomial isn't in W? So the root of the polynomial em... let's say $p(x)=0$ is all value of x(that include complex numbers), so 0 polynomial is not in W. Is this the right way to think about it? $\endgroup$ – Logan Nov 23 '20 at 23:02
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    $\begingroup$ Yeah. The polynomial $p(z)=0$ isn’t in $W$, but a vector space always has a zero element. Alternatively, you can see this as a case of $W$ not being closed under scalar multiplication, since $0q \not \in W$ for all $q \in W$. $\endgroup$ – AJY Nov 23 '20 at 23:04
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    $\begingroup$ I can't believe I missed that! (+1) $\endgroup$ – José Carlos Santos Nov 24 '20 at 7:10
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You mention $10$ axioms of a complex vector space, but you only need one for the vast majority of cases. Assuming scalar multiplication and usual addition we have that $$W \text{ is VS } \iff p, q\in W \implies \alpha p+\beta q \in W$$ for every $\alpha, \beta \in \Bbb C$. You simply need to find two polynomials with real zeros whose combination has a non-real zero, and Jose's answer does exactly this.

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    $\begingroup$ That’s necessary and sufficient to show a given set $W$ is a subspace of an already known vector space $V$, but there are additional axioms that establish the algebraic properties of the vector space operations (e.g. that addition is associative and commutative, that $1v=v$, that addition and multiplication are distributive, etc.). $\endgroup$ – AJY Nov 23 '20 at 22:48
  • $\begingroup$ I know, and I figured that's what the ten were, but my point is that for the vast majority of sets that students are asked about, if they fail the criteria of vector space, it is usually because they fail at least this property. (I.e. there are relatively few sets $S$ where $p,q\in S \implies \alpha p+\beta q\in S$ but where $S$ is not a vector space), to the point where it is sensible to check if it fails this before checking the other axioms if it does hold. $\endgroup$ – Rhys Hughes Nov 24 '20 at 4:37
  • $\begingroup$ Fair, but (1) the distinction is still significant and (2) I find anecdotally that most linear algebra texts are careful to distinguish between showing a set is a space and showing a set is a subspace. $\endgroup$ – AJY Nov 24 '20 at 4:39
  • $\begingroup$ @RhysHughes: I think that the fact that it is often enough to show that the set is a subspace of a well-known vector space (your comment) should appear directly in the question. That's really the key point of this answer. $\endgroup$ – Taladris Nov 24 '20 at 5:43

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