Determine if subspace - complex numbers $$ \mathbb{F} := \mathbb{C} \\ V := \mathbb{C}_2[x] $$
$V$ is the set of all polynomials with degree less than or equal to $2$ with complex scalars.
$$ U := \left \{ p(x) \in \mathbb{C}_2[x] | p(x) = p(2x) \right \} $$
Given $ U \subseteq V$ I need to determine if $U$ is a linear subspace of $V$
My try:
$$ p(x) = ax^2 + bx + c \\ p(2x) = a(2x)^2 +2bx + c \\ p(x) = p(2x) \Rightarrow ax^2 + bx + c = 4ax^2 + 2bx + c \\ 3ax^2 + bx = 0 $$
Now, I know that I need to check $3$ things:

$\vec{0} \in U $
If $\vec{v_1} \in U$ and $\vec{v_2} \in U$ then $\vec{v_1} + \vec{v_2} \in U$
If $t \in \mathbb{C}$ then $t \cdot \vec{v_1} \in U$

But I only got the rule of thumbs that &quot; $ 3ax^2 + bx = 0$ &quot;
Of course the zero vector is in $U$ because we can substitude $x=0$.
But how do I check closure over addition? what is really the rule of the subspace? I would appreciate you help.. I am lost. Thank you!
Your analysis shows that $a\,x^2+b\,x+c\in U$ if and only if $3a\,x^2+b\,x$ is the zero polynomial. This is the case if and only if $a=0$ and $b=0$. Hence $U$ is exactly the set of constant polynomials,
$$
U = \{\,0\,x^2+0\,x+ c\mid c\in\mathbb C\,\} = \mathbb C_0[x] \subseteq \mathbb C_2[x].
$$
This is of course a subspace of $\mathbb C_2[x]$.
Your justification for the fact that $0\in U$ is wrong. We have $0\in U$ because where $0$ is the null polynomial and if $n(x)$ is the null polynomial then it is indeed true that $n(x)=n(2x)$.
If $p(x),q(x)\in U$, then $p(x)=p(2x)$ and $q(x)=q(2x)$. So\begin{align}(p+q)(x)&amp;=p(x)+q(x)\\&amp;=p(2x)+q(2x)\\&amp;=(p+q)(2x)\end{align}and therefore $p(x)+q(x)\in U$.
And if $p(x)\in U$ and $\alpha\in\Bbb C$, then $p(x)=p(2x)$ and \begin{align}(\alpha p)(x)&amp;=\alpha p(x)\\&amp;=\alpha p(2x)\\&amp;=(\alpha p)(2x)\end{align}and therefore $(\alpha p)(x)\in U$.