Why missing the zero polynomial indicate subset is not closed under multiplication? My book has the following question:
Is this a subspace of
$P_{2}: \{ a_{0} + a_{1}x + a_{2}x^2 \mid a_{0} + 2a_{1} + a_{2} = 4\}$? If it is then parametrize its description.
My reasoning for it not being a subspace is that it is not a subspace of $P_{2}$ because the subspaces would need degrees less than 2. But after thinking about it, a vector space can be a subspace for itself as a trivial subspace. Then I picked a few numbers that made the equation not equal 4, thus I concluded it is not a subspace for not satisfying closure under addtion.
The book's reason is that it is missing the zero polynomial thus not closed under multiplication. I don't understand how the zero polynomial relates to it being a subspace?
Thanks!
 A: Recall the following fact:

Let $V$ be a vector space over a field $F$ and let $H$ be a subset of $V$. We will say that $H$ is a vector subspace of $V$ $\color{blue}{\text{if, and only if}}$, the following conditions are satisfied:

*

*$\vec{0}_{V}\in H$.

*For all $h_{1}$ and $ h_{2}$ in $H$, we have $ h_{1}+h_{2}\in H$.

*For all $\alpha$ in $F$ and for all $h$ in $H$, we have $\alpha\cdot  h\in H$.


Now, consider the set $$H=\{a_{0}+a_{1}x+a_{2}x^{2}\in P_{2}(\mathbb{R}): a_{0}+2a_{1}+a_{2}=4\}$$
Notice that $\color{red}{H\subseteq P_{2}(\mathbb{R})}$. Thus $H$ will be a vector subspace of $P_{2}(\mathbb{R})$ if, and only if, are satisfied the conditions $1),2)$ and $3)$.
Now, recall that $H$ is a subset of polynomials in $P_{2}(\mathbb{R})$ which satisfies one specific condition, that is, $a_{0}+2a_{1}+a_{2}=4$.
The vector zero of $P_{2}(\mathbb{R})$ under the usual operations is given by $$\vec{0}_{P_{2}(\mathbb{R})}=\color{red}{0}+\color{green}{0}x+\color{orange}{0}x^{2}$$
Now, the question is: is the vector zero of $P_{2}(\mathbb{R})$ in $H$? The answer is "no" because it doesn't satisfy the conditions in $H$. Indeed, notice that
$$\color{red}{0}+2\cdot\color{green}{0}+\color{orange}{0}=0\not=4$$
Hence $\vec{0}_{P_{2}(\mathbb{R})}$ is not in $H$ and then $H$ doesn't satisfy at least one conditions of fact on this first part. Therefore $H$ is not vector subspace of $P_{2}(\mathbb{R})$.

Also you can see the conditions $2)$ and $3)$ if you want to practice. You can continue following the similar way above and see if it holds or doesn't hold. Of course, this not change the conclusion above.
