Question

let B={1-t,t-t²,2-2t+t²} be an ordered basis for p₂

let p(t)=3+t-6t²

1. find [p(t)]_B

Answer 1

Denote: $p_1=1-t$, $p_2=t-t^2$, $p_3=2-2t+t^2$. The aim is to find real numbers $\alpha_1,\alpha_2,\alpha_3$ satisfying: $p(t)=\alpha_1p_1+\alpha_2p_2+\alpha_3p_3$. We get the following equations for $\alpha_1$, $\alpha_2$ and $\alpha_3$ by considering coefficients near $t^2$, $t$, $1$: $-6=\alpha_3-\alpha_2$, $1=-\alpha_1+\alpha_2-2\alpha_3$ and $3=\alpha_1+2\alpha_3$. From the first and third equation we get: $\alpha_2=\alpha_3+6$, $\alpha_1=3-2\alpha_3$. By substituting it in the second equation we get: $1=-(3-2\alpha_3)+(\alpha_3+6)-2\alpha_3$. Thus, $\alpha_3=-2$, $\alpha_2=4$, $\alpha_1=7$.

Answer: $p(t)=7(1-t)+4(t-t^2)-2(2-2t+t^2)$.