Show that {$a+bx+cx^2,a_1+b_1x+c_1x^2,a_2+b_2x+c_2x^2$} is a basis of $P_2$, if and only if {$(a,b,c),(a_1,b_1,c_1),(a_2,b_2,c_2)$} is a basis of ... Show that {$a+bx+cx^2,a_1+b_1x+c_1x^2,a_2+b_2x+c_2x^2$} is a basis of $P_2$, if and only if {$(a,b,c),(a_1,b_1,c_1),(a_2,b_2,c_2)$} is a basis of $ℝ^3$.
 A: Let's look at the coordinate transformation:
$$T:\text{R}_2[x] \rightarrow \mathbb{R}^3$$
$$T(\alpha + \beta x + \gamma x^2)=(\alpha, \beta, \gamma)$$
That's an isomorphism (why?), and as such, it transfers a basis to a basis.
A: Let $S=\{a_1+b_1x+c_1x^2,a_2+b_2x+c_2x^2,a_3+b_3x+c_3x^2\}$, and $T=\{(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b,c)\}$.
Take a vector $(p,q,r) \in \mathbb{R}^3$. Note that then $p+qx+rx^2\in P_2$.
Then
$\begin{array}{rcl}p+qx+rx^2 &=& \beta_1 (a_1+b_1x+c_1x^2)\\
&&+\beta_2 (a_2+b_2x+c_2x^2)\\
&&+\beta_3 (a_3+b_3x+c_3x^2)
\end{array}$
$\iff \left\{\begin{array}{rcl}
p=\beta_1a_1+\beta_2a_2+\beta_3a_3\\
q=\beta_1b_1+\beta_2b_2+\beta_3b_3\\
r=\beta_1c_1+\beta_2c_2+\beta_3c_3\\
\end{array}\right.$
$\begin{array}{rcl}\iff (p,q,r) &=& \beta_1 (a_1,b_1,c_1)\\
&&+\beta_2 (a_2,b_2,c_2)\\
&&+\beta_3 (a_3,b_3,c_3)
\end{array}$
So any element in $P_2$ is a linear combination of S $\iff$ any element in $\mathbb{R}^3$ is element of T.
Similar argument also can be used to show that the linear combination is unique for $S$ implies it's unique for $T$, and vice versa.
