# 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.

• $P_2$ means polynomials of degree 2? – user348749 Feb 21 '17 at 8:31

$p = c_0 p_0 + c_1 p_1 + c_2 p_2 \neq 0$ 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.