There is this exercise that goes like: Let $x_1,x_2,x_3,x_4$ ∈ R be such that $x_1 < x_2 < x_3 < x_4$ and consider the polynomials $p_1(x), p_2(x), p_3(x), p_4(x) ∈ P_3\ $ defined by :
$p_1(x) = \frac{(x−x_2)(x−x_3)(x−x_4)}{(x_1 − x_2)(x_1 − x_3)(x_1 − x_4)}$
$p_2(x) = \frac{(x−x_1)(x−x_3)(x−x_4)}{(x_2 − x_1)(x_2 − x_3)(x_2 − x_4)}$
$p_3(x) = \frac{(x−x_1)(x−x_2)(x−x_4)}{(x_3 − x_1)(x_3 − x_2)(x_3 − x_4)}$
$p_4(x) =\frac{(x−x_1)(x−x_2)(x−x_3)}{(x_4 − x_1)(x_4 − x_2)(x_4 − x_3)}$
Show that $T: P_3 \to R^4$ defined by
$$T(p) =(p(x_1) , p(x_2) , p(x_3), p(x_4) )$$
for $p \in P_3$ is an isomorphism.
How do I do that?
We'll prove that $\{p_1,p_2,p_3,p_4\}$ is a linearly independent set of $P^3$. Notice first that $p_i(x_j) = \delta_{ij}$. Let
$$ \alpha_1p_1(x)+\alpha_1p_2(x)+\alpha_3p_3(x)+\alpha_4p_4(x) = 0 $$
At $x = x_1$, only $p_1\neq 0$, so $\alpha_1$ = 0. We do that for $x_2$, $x_3$ and $x_4$ to complete the proof. Since $\dim P^3 = 4$, this is a basis, implying that we can write some arbitrary $p\in P^3$ as
$$ p(x) = c_1p_1(x)+c_1p_2(x)+c_3p_3(x)+c_4p_4(x) $$
Again, notice that $p(x_i) = c_i p_i(x_i) = c_i$ for each $i$. This makes the $T$ linear and well defined. It is easy to see that its kernel is zero dimensional, because
\begin{align} T(p) = 0 \iff&(p(x_1),p(x_2),p(x_3),p(x_4) = 0 \\ \iff& p(x_i) =c_i = 0 \text{ for each } i \end{align}
And since $\dim\mathbb{R}^4 = \dim P^3 = 4$, the rank-nullity theorem says that $\dim(\text{im}(T))= 4$. This completes the proof.