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 ...

Question

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 $ℝ^3$.

Answer 1

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.

Answer 2

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.