# Does the set p_1=1,p_2=1−3x,p_3=x,p_4=x^2 form a basis for P2?

Apr 4, 2017

See below.

#### Explanation:

Any ${P}_{2} \left(x\right) = a {x}^{2} + b x + c$ can be represented as

$\alpha {p}_{1} + \beta {p}_{2} + \gamma {p}_{3} + \delta {p}_{4}$

or

$a {x}^{2} + b x + c = \alpha + \beta - \left(3 \beta - \gamma\right) x + \delta {x}^{2}$

with

$\left\{\begin{matrix}\alpha + \beta = a \\ - 3 \beta + \gamma = b \\ \delta = c\end{matrix}\right.$

or

$\left(\begin{matrix}0 & 0 & 0 & 1 \\ 0 & - 3 & 1 & 0 \\ 1 & 1 & 0 & 0\end{matrix}\right) \left(\begin{matrix}\alpha \\ \beta \\ \gamma \\ \delta\end{matrix}\right) = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)$

or

$\left(\begin{matrix}0 & 0 & 1 \\ 0 & - 3 & 0 \\ 1 & 1 & 0\end{matrix}\right) \left(\begin{matrix}\alpha \\ \beta \\ \delta\end{matrix}\right) = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right) - \gamma \left(\begin{matrix}0 \\ 1 \\ 0\end{matrix}\right)$

Here $\gamma$ is a free parameter and can be set to $0$. Also

$\left(\begin{matrix}0 & 0 & 1 \\ 0 & - 3 & 0 \\ 1 & 1 & 0\end{matrix}\right)$ is invertible so a minimal basis is formed by

${p}_{1} , {p}_{2} , {p}_{4}$ or ${p}_{1} , {p}_{3} , {p}_{4}$