tinfoQ

2021-09-29

The set $B=\left\{1+{t}^{2},t+{t}^{2},1+2t+{t}^{2}\right\}$ is a basis for ${\mathbb{P}}_{2}$. Find the coordinate vector of $p\left(t\right)=1+4t+7{t}^{2}$ relative to B.

Nathanael Webber

Given basis can be written in matrix form (first row - second power, second row - first power, third row - free coefficient):
$B=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 1& 0& 1\end{array}\right]$
Solve $Bx=p\left(t\right)$ using augmented matrix:
$\left[\begin{array}{ccc}1& 1& 1|7\\ 0& 1& 2|4\\ 1& 0& 1|1\end{array}\right]$
, multiply first row with -1 and add to third
$\sim \left[\begin{array}{ccc}1& 1& 1|7\\ 0& 1& 2|4\\ 0& -1& 0|-6\end{array}\right]$
$-{x}_{2}=-6\to {x}_{2}=6$
${x}_{2}+2{x}_{3}=4$
$6+2{x}_{3}=4$
$2{x}_{3}=4-6=-2$
${x}_{3}=-1$
${x}_{1}+{x}_{2}+{x}_{3}=7$
${x}_{1}+6-1=7$
${x}_{1}=7-5=2$
${p}_{B}=\left[\begin{array}{c}2\\ 6\\ -1\end{array}\right]$
Results:
${p}_{B}=\left[\begin{array}{c}2\\ 6\\ -1\end{array}\right]$

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