Showing a list of polynomials is linearly dependent in $\mathcal{P}_m(\mathbb{F})$ Note: In case the notation isn't common, $\mathcal{P}_m(\mathbb{F})$ denotes the set of all polynomials $\mathbb{F} \to \mathbb{F}$ with at most degree $m$.
I'm having trouble with the following question:

Suppose $p_0, p_1, \dots, p_m$ are polynomials in $\mathcal{P}_m(\mathbb{F})$ such that $p_j(2)=0$ for each $j$. Prove that $p_0, p_1, \dots, p_m$ are not linearly independent.

Attempt: Assume the list of polynomials is linearly indepedent. Then the only choice of $a_i \in \mathbb{F}$ such that 
$$a_0p_0(z) + a_1p_1(z)+ \cdots + a_mp_m(z)=0$$
is $a_i=0$ for all $0\leq i \leq m$, for all $z\in \mathbb{F}$. But note
$$a_0p_0(2)+\cdots + a_mp_m(2)=a_0(0)+\cdots a_m(0)=0$$
thus...

I feel like I'm definitely misunderstanding polynomials here. I'm tempted to say something along the lines of, "since now we can pick any $a_i \in \mathbb{F}$, not all $0$, and achieve $a_0p_0(z)+\cdots + a_mp_m(z)=0$ for some particular $z \in \mathbb{F}$, then the list is linearly dependent". But to show linear dependence here, do I need $a_i \in \mathbb{F}$, not all $0$, such that $a_0p_0(z)+\cdots + a_mp_m(z)=0$ for all $z \in \mathbb{F}$ or for at least one $z \in \mathbb{F}$?
 A: Note that since you have $m+1$ polynomial, lineal independence would in particular imply that $<p_0,...,p_m>$ generates $\mathbb{F}_{\leq m}$. However, this is not so: clearly
$$
<p_0, \dots p_m> \subseteq \{p \in \mathbb{F}_{\leq m} : p(2) = 0\}
$$
which is a proper subspace of $\mathbb{F}_{\leq m}$ since, for example, it does not contain 1. In general you could have linearly independent polynomials that behave like this, as long as they are not 'too many'. Take for example $(X-2)X$ and $(X-2)X^2$ in $\mathbb{R}_{\leq 3}$, they both verify the former property but they are linearly independent, because  they have different degrees. We have then constructed a linearly independent set of polynomials that vanishes at $2$. 
In general, for any vector space, proving that $v_1, \dots v_n$ are linearly independent consist of showing the following implication
$$
\sum_{i=1}^na_iv_i = 0 \ \Rightarrow a_i = 0 \ (\forall i)
$$
In this particular scenario, the first equality to zero means the zero polynomial, that is, the one whose coefficients are all zero (this shouldn't be confused with $p(z) = 0 \ (\forall z)$, think for example of $X(X-1)$ in $\mathbb{Z}_2$).
A: I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
This book contains the following problem. (Exercise 2.A 16 on p.38)

Suppose $p_0, p_1, \dots, p_m$ are polynomials in $\mathcal{P}_m(\mathbb{F})$ such that $p_j(2)=0$ for each $j$. Prove that $p_0, p_1, \dots, p_m$ is not linearly independent in $\mathcal{P}_m(\mathbb{F})$.

My answer is here:

If $m=0$, $p_0$ is a constant function and $p_0(2)=0$.
So, $p_0$ is the zero function from $\mathbb{F}$ to $\mathbb{F}$.
So, $p_0$ is not linearly independent in $\mathcal{P}_0(\mathbb{F})$.
Suppose that $m\geq1$.
We can write $p_j(x)=(x-2)q_j(x)$ for any $x\in\mathbb{F}$, where $q_j\in\mathcal{P}_{m-1}(\mathbb{F})$ for each $j\in\{0,\dots,m\}$.
Since $\text{Span}(1,x,\dots,x^{m-1})=\mathcal{P}_{m-1}(\mathbb{F})$, $q_0,\dots,q_m$ is linearly dependent by 2.23 on p.35 in "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
So, there exist numbers $a_0,\dots,a_m\in\mathbb{F}$, not all $0$, such that $a_0 q_0(x)+\cdots+a_m q_m(x)=0$ for any $x\in\mathbb{F}$.
Obviously, $(x-2)(a_0 q_0(x)+\cdots+a_m q_m(x))=0$ for any $x\in\mathbb{F}$.
So, $a_0 (x-2) q_0(x)+\cdots+a_m (x-2) q_m(x)=0$ for any $x\in\mathbb{F}$.
So, $a_0 p_0(x)+\cdots+a_m p_m(x)=0$ for any $x\in\mathbb{F}$.
So, $p_0,p_1,\dots,p_m$ is not linearly independent in $\mathcal{P}_m(\mathbb{F})$.

