Q:

Prove that the following vectors form a basis of $\Bbb R^3$

Question states:
"If $p_0, p_1, p_2$ are independent polynomials in $P_2$ (set of all polynomials of degree $2$ or less) and $x_0, x_1,$ and $x_2$ are distinct real numbers, show that the three column vectors
$$
        \begin{matrix}
        p_0(x_0) \\
        p_1(x_0) \\
        p_2(x_0) \\
        \end{matrix}, \qquad 
        \begin{matrix}
        p_0(x_1) \\
        p_1(x_1) \\
        p_2(x_1) \\
        \end{matrix}, \qquad
        \begin{matrix}
        p_0(x_2) \\
        p_1(x_2) \\
        p_2(x_2) \\
        \end{matrix}
$$
are linearly independent.
My attempt: Denote the three vectors as $v_1, v_2, v_3$, respectively. I tried showing that $c_1v_1 + c_2v_2 + c_3v_3 = 0$ has only the trivial solution, but this was futile. I tried showing that I tried showing that the transpose of matrix $A$, where the columns of $A$ are the column vectors above, is invertible. This was a futile attempt also.

A:

$p = c_0 p_0 + c_1 p_1 + c_2 p_2 \neq 0$ for any constants $c_0, c_1, c_2$ and is a polynomial of degree at most 2. Thus $p$ can not have more than 2 roots. If the rows are dependent, we can find constants such that $c_0p_0(x_i) + c_1p_1(x_i) + c_2p_2(x_i) = 0$ for $i  =0,1,2$, a contradiction.

