Question Number 54939 by gunawan last updated on 15/Feb/19
$$\mathrm{coordinate}\:{x}^{\mathrm{2}} \:\mathrm{to}\:\mathrm{basis}\:\left\{{x}^{\mathrm{2}} +{x},\:{x}+\mathrm{1},\:{x}^{\mathrm{2}} +\mathrm{1}\right\} \\ $$$$\mathrm{at}\:\mathrm{P}_{\mathrm{2}} \:\mathrm{is}… \\ $$
Commented by Abdo msup. last updated on 15/Feb/19
$${let}\:{find}\:\:\alpha\:,\beta\:{and}\:\:\gamma\:{with}\:{verify}\: \\ $$$${x}^{\mathrm{2}} =\alpha\left({x}^{\mathrm{2}} \:+{x}\right)+\beta\left({x}+\mathrm{1}\right)+\gamma\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\:\:\forall{x}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} =\left(\alpha+\gamma\right){x}^{\mathrm{2}\:} \:+\left(\alpha+\beta\right){x}\:+\beta+\gamma\:\Rightarrow \\ $$$$\begin{cases}{\alpha+\gamma\:=\mathrm{1}}\\{\alpha+\beta=\mathrm{0}\:\:\:{and}\:\beta+\gamma\:=\mathrm{0}\:\Rightarrow\alpha−\gamma\:=\mathrm{0}\:\Rightarrow\gamma=\alpha\:\Rightarrow\alpha=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$=\gamma\:\:{and}\:\beta=−\gamma\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\:{so}\: \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} \:+{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right) \\ $$$${so}\:{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}},−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\:{in}\:{that}\:{basis}\:. \\ $$
Commented by gunawan last updated on 15/Feb/19
$$\mathrm{thank}\:{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir} \\ $$
Commented by Abdo msup. last updated on 15/Feb/19
$${you}\:{are}\:{welcome}. \\ $$