Prove that linear functionals on the space of polynomial functions is a basis for $V^{*}.$ Here is one of those examples from Linear Algebra by Kenneth Hoffman, Ray Kunze(p. 100):

Let $V$ be the vector space of all polynomial functions from $R$ into $R$ which have degree less than or equal to 2. Let $t_{1}, t_{2},$ and $t_{3}$ be any three distinct real numbers, and let $$L_{i}(p) = p(t_{i}).$$ Then $L_{1}, L_{2},$ and $L_{3}$ are linear functionals on $V$. These functionals are linearly independent; for, suppose $$L=c_{1}L_{1}+c_{2}L_{2}+c_{3}L_{3}.$$ if $L=0$, i.e. if $L(p)=0$ for each $p$ in $V$, then applying $L$ to the particular polynomial 'functions' $1, x, x^{2},$ we obtain $$\begin{align} c_{1}+c_{2}+c_{3}&=0\\t_{1}c_{1}+t_{2}c_{2}+t_{3}c_{3}&=0\\t^{2}_{1}c_{1}+t^{2}_{2}c_{2}+t^{2}_{3}c_{3}&=0 \end{align}$$ From this it follows that $c_{1}=c_{2}=c_{3}=0$, because (as a short computation shows) the matrix $$\begin{bmatrix}1&1&1\\ t_{1}&t_{2}&t_{3}\\ t^{2}_{1}&t^{2}_{2}&t^{2}_{3}\end{bmatrix}$$ is invertible when $t_{1},t_{2},$ and $t_{3}$ are distinct...

Although I agree the scalars $c_{i}$ must be $0$, I don't see how it leads to the conclusion that the $L_{i}$ are linearly independent. In other words, is it safe to say that $L_{i}(1), L_{i}(x), L_{i}(x^2)$ sufficiently represent $L_{i}(p)$ for every $p$? It seems to me that the choice of polynomial functions above is somewhat related to the conclusion, but I'm not able to confidently say that's true. Any comments or hints would be greatly appreciated.
 A: This is a classical interpolation fact: a polynomial of degree $\leq n$ is completely determined by its values at $n+1$ distinct points.
Which, in itself, is a generalization of the fact that there's only one line through two distinct points.
A: $L_1, L_2, L_3$ are vectors (read: linear functionals) in the vector space $V^*$. By the definition of linear independence in a vector space, $L_1, L_2, L_3$ are independent iff
$$
\underbrace{0 = c_1 L_1 + c_2 L_2 + c_3 L_3}_{(1)} \implies \underbrace{c_1 = c_2 = c_3 = 0}_{(2)}.
$$
So to prove independence, we start with equation $(1)$ and must prove that this implies equation $(2)$.
Since $0, L_1, L_2, L_3$ are linear functionals, $(1)$ implies that 
$$
0(p) = c_1 L_1(p) + c_2 L_2(p) + c_3 L_3(p) \qquad \forall p \in V.
$$
In particular, the above is true for $p = 1$, $p = x$, and $p = x^2$. Writing these three equations down gives the matrix equation 
$$
\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} L_1(1) & L_2(1) & L_3(1) \\ L_1(x) & L_2(x) & L_3(x) \\ L_1(x^2) & L_2(x^2) & L_3(x^2)\end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ t_1 & t_2 & t_3 \\ t_1^2 & t_2^2 & t_3^2 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}
$$
which you agree implies $c_i = 0$. Recall that $c_i = 0$ is equation $(2)$. So we have used equation $(1)$ to prove equation $(2)$, which establishes linear independence of $L_1, L_2, L_3$.
(It is worth noting that, as you say, the choice of polynomials $p$ is important; had we not chosen the $p$ (and $t_i$) wisely, the matrix equation we'd get would not imply $c_i = 0$, and we would not have proved linear independence of the $L_i$. This is not the essence of the linear independence argument, though; it is the specific machinery that gets us there.)
