Proving Basis of Dual Space: V* in P_n(F)

[jclawson709]

Homework Statement

Let V = P_n(F), and let c_0, c_1,..., c_n be distinct scalars in F. For 0 <= i <= n, define f_i(p(x)) = p(c_i). Prove that {f_0, f_1,..., f_n} is a basis for V*. Hint: Apply any linear combination of this set that equals the zero transformation to p(x) = (x-c_1)*(x-c_2)*...*(x-c_n), and deduce that the first coefficient is zero.

Homework Equations

The Attempt at a Solution

I really have no clue where to start on this one. First off, I don't even see how doing what the hint says would prove it to be a basis. Is the hint implying that these scalars are the roots of the polynomial? Also I'm not sure how I would pick a linear combination of the set that equals 0. I was thinking of using c_i - a*c_j, where a is just another scalar which causes a*c_j = c_i, so that it equals zero and the resulting transformation equals zero. Not too sure about that reasoning though, and how it would play into the hint the book gives.

 

[micromass]

OK, let's do this thing. Don't think about the hint just yet. First, we'll start of with some definitions. What exactly is a basis of a vector space? This will be what we'll try to prove...

 

[jclawson709]

ok so yeah, a basis is a set of vectors that are linearly independent. so if {f_0,...,f_n} is a basis then a_0f_0 + a_1f_1 + ... + a_nf_n = 0 if and only if a_0, a_1, ..., a_n are all zero, right?

 

[micromass]

ok so yeah, a basis is a set of vectors that are linearly independent.

Hmm, that's a start but it's not completely true. A basis is certainly a set of linearly independent vectors, BUT there's another very important requirement: you will want the vectors to span the space! But let's not deal with spanning the space just now, let's focus on the linearly independent part:

so if {f_0,...,f_n} is a basis then a_0f_0 + a_1f_1 + ... + a_nf_n = 0 if and only if a_0, a_1, ..., a_n are all zero, right?

Right, so you take $$a_0,...,a_n$$ such that $$a_0f_0+...a_nf_n=0$$. Our aim is to show that $$a_0=...=a_n=0$$.

This is where the hint comes in, in order to show that $$a_0=...=a_n=0$$, you'll want to define the polynomial $$p(x)=(x-c_1)...(x-c_n)$$. Now, calculate what $$(a_0f_0+...+a_nf_n)(p(x))$$ equals.