Suppose that $p_0, p_1, … p_m$ are polynomials in $P_m\mathbb(F)$ such that $P_j(2) = 0$ for each $j$. Prove that $(p_0, … p_m)$ is linearly dependent.
Work so far:
Assume that it is. Then it is a basis for $P_m$. Now, I need help.
After looking around, most solutions mention that every polynomial of degree $\leq m$ can be written with list $p_0, p_1, … p_m$ (under the assumption that it is a basis. It's not, I know.) I do not understand this restriction to $\leq m$. Why do we need this restriction?
Second, most solutions use a reduction. It goes something like let $p(x) \in span(p_0(x) + \ldots + p_m(x))$. But let $x =2$. Then $p(x) = 2$. But how is this a contradiction. Yes, when $x = 2$, then $p(x) = 0$. What does this contradict with?
Here is a solution: http://www.math.ucsb.edu/~asdugas/108A_SP08/108Ahw3sol.pdf
(problem 12)
When it says "$p(2) = 0$ for any polynomial of $p(x)$ of degree less or equal to $m$. Contradiction." This doesn't make sense. What are we contradicting?
Best Answer
Consider $p(x) = x$. Then $p(2) = 2$. Can this particular polynomial be a linear combination of polynomials which all take the value $0$ at $2$? If not, then you have $m+1$ polynomials which do not span the vector space, and the vector space has dimension $m+1$. Could the polynomials be linearly independent in this case? This is basically arguing the equivalent contrapositive of the argument you listed. The argument listed says that if the $m+1$ polynomials are linearly independent, then they form a basis, so then all polynomials of degree $\leq m$ are linear combinations of the basis, and thus must satisfy $p(2) = 0$. So any polynomial of degree $\leq m$ which does not satisfy $p(2) = 0$ gives a contradiction.