Linear Independence in polynomial space.

Question

Here $P_m(F)$ is the space of polynomials of degree at most m.

My reasoning thus far has been to write each $p_j$ as $$p_j=(x-2)\times q_j$$ where $q_j$ is a polynomial of degree at most $(m-1)$.

If we consider the expression $$a_0p_0 + a_1p_1 +\cdots+ a_mp_m = (x-2)\times [a_0q_0 + a_1q_1 +\cdots+ a_mq_m] = 0$$ then at $x=2$ any collection of $a_j$ would satisfy the expression. Hence the list $(p_0,p_1,\cdots,p_m)$ is not linearly independent in $P_m(F)$.

I'm not sure about this reasoning though. In my understanding, shouldn't the $a_j$ be independent of the $x$, i.e. the expression $a_0p_0 + a_1p_1 +\cdots+ a_mp_m$ equals zero because the coefficients of the ${x^j}$ cancel each other out, but not because $x=2$ makes it so?

Answer 1

Hint:

Let $Z$ be the subspace of $\mathcal P_m$ which vanish at $x=2$. Using your notations, observe the linear mapping \begin{align} Z&\longrightarrow\mathcal P_{m-1}\\ p&\longmapsto q=\frac p{x-2} \end{align} is injective, hence the dimension of the image is equal to $\dim Z$. What is the dimension of $\mathcal P_{m-1}$?

Answer 2

Consider the (surjective) linear map $f\colon P_m(\mathbf{F})\to\mathbf{F}$ defined by $p\mapsto p(2)$.

Then $p_0,p_1,\dots,p_m$ are $m+1$ elements in the kernel of $f$, which has dimension $m$ by the rank-nullity theorem and the fact that $\dim P_m(\mathbf{F})=m+1$.

Answer 3

If $p_0,p_1,\dots,p_m$ were linearly independent in $P_m(F)$, then they would form a basis of $P_m(F)$, which has dimension $m+1$. But then every polynomial $p\in P_m(F)$ would be a linear combination of $p_0,p_1,\dots,p_m$, and so $p(2)=0$. This clearly does not hold for all polynomials; it certainly does not hold for nonzero constant polynomials.