coordinate x^2 to basis {x^2 +x, x+1, x^2 +1} at P


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Question Number 54939 by gunawan last updated on 15/Feb/19
$coordinate x^2 to basis {x^2 +x, x+1, x^2 +1} at P_2 is...$
$coordinate x^{2} to basis {x^{2} + x, x + 1, x^{2} + 1}$ $at P_{2} is . . .$
Commented by Abdo msup. last updated on 15/Feb/19
$let find α ,β and γ with verify x^2 =α(x^2 +x)+β(x+1)+γ(x^2 +1) ∀x ⇒ x^2 =(α+γ)x^(2 ) +(α+β)x +β+γ ⇒ { ((α+γ =1)),((α+β=0 and β+γ =0 ⇒α−γ =0 ⇒γ=α ⇒α=(1/2))) :} =γ and β=−γ =−(1/2) so x^2 =(1/2)(x^2 +x)−(1/2)(x+1)+(1/2)(x^2 +1) so x^2 ((1/2),−(1/2),(1/2)) in that basis .$
$l e t f i n d α, β a n d γ w i t h v e r i f y$ $x^{2} = α (x^{2} + x) + β (x + 1) + γ (x^{2} + 1) \forall x \Rightarrow$ $x^{2} = (α + γ) x^{2} + (α + β) x + β + γ \Rightarrow$ ${\begin{cases} α + γ = 1 \\ α + β = 0 a n d β + γ = 0 \Rightarrow α - γ = 0 \Rightarrow γ = α \Rightarrow α = \frac{1}{2} \end{cases}$ $= γ a n d β = - γ = - \frac{1}{2} s o$ $x^{2} = \frac{1}{2} (x^{2} + x) - \frac{1}{2} (x + 1) + \frac{1}{2} (x^{2} + 1)$ $s o x^{2} (\frac{1}{2}, - \frac{1}{2}, \frac{1}{2}) i n t h a t b a s i s .$
Commented by gunawan last updated on 15/Feb/19

$thank y o u very much Sir$
Commented by Abdo msup. last updated on 15/Feb/19

$y o u a r e w e l c o m e .$
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