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 $B(p_1(x),p_2(x),p_3(x),p_4(x))$ is a basis of $P_3$ and that the column vector $p(x)=(p_1(x),p_2(x),p_3(x),p_4(x))$ is an isomorphism.
Can someone help me step by step please?
You need to show these polynomials are linearly independent.
Let $$ c_1 p_1(x)+c_2p_2(x)+c_3 p_3(x)+c_4p_4(x)=0$$
Evaluate the above at $x=x_1$ and you get $c_1=0$
Evaluate the above at $x=x_2$ and you get $c_2=0$
Similarly we get $c_3=0$ and $c_4=0$
Now define the function $$F:P_3 \to P(x)$$ by $$ F(a_1+a_2x+a_3x^3+a^4x^3)= (a_1p_1(x),a_2p_2(x),a_3p_3(x),a_4p_4(x))$$
It is not hard to see that F is an isomorphism.