Question

# Let gamma = {t^2 - t + 1, t + 1, t^2 + 1} and beta = {t^2 + t + 4, 4t^2 - 3t + 2, 2t^2 + 3} be ordered bases for P_2(R).Find the change of coordinate matrix Q

Alternate coordinate systems

Let $$\displaystyle\gamma={\left\lbrace{t}^{{2}}-{t}+{1},{t}+{1},{t}^{{2}}+{1}\right\rbrace}{\quad\text{and}\quad}\beta={\left\lbrace{t}^{{2}}+{t}+{4},{4}{t}^{{2}}-{3}{t}+{2},{2}{t}^{{2}}+{3}\right\rbrace}{b}{e}{\quad\text{or}\quad}{d}{e}{r}{e}{d}{b}{a}{s}{e}{s}{f}{\quad\text{or}\quad}{P}_{{2}}{\left({R}\right)}.$$ Find the change of coordinate matrix Q that changes $$\beta \text{ coordinates into } \gamma-\text{ coordinates}$$

2020-11-02

Let $$\displaystyle{t}^{{2}}+{t}+{4}=\gamma_{{1}}{\left({t}^{{2}}-{t}+{1}\right)}+\gamma_{{2}}{\left({t}={1}\right)}+\gamma_{{3}}{\left({t}^{{2}}+{1}\right)}$$
$$\displaystyle\Rightarrow\gamma_{{1}}+\gamma_{{3}}={1},-\gamma_{{1}}+\gamma_{{2}}={1},\gamma_{{1}}+\gamma_{{2}}+\gamma_{{3}}={4}$$
$$\displaystyle\Rightarrow{1}+\gamma_{{2}}={4}$$
$$\displaystyle\Rightarrow\gamma_{{2}}={3}$$
$$\displaystyle-\gamma_{{1}}+\gamma_{{2}}={1}\Rightarrow-\gamma_{{1}}+{3}={1}\Rightarrow\gamma_{{1}}={2}$$
$$\displaystyle\gamma_{{1}}+\gamma_{{3}}={1}\Rightarrow{2}+\gamma_{{3}}={1}\Rightarrow\gamma_{{3}}=-{1}$$ Let $$\displaystyle{4}{t}^{{2}}-{3}{r}+{2}=\gamma_{{1}}{\left({t}^{{2}}-{t}+{1}\right)}+\gamma_{{2}}{\left({t}+{1}\right)}+\gamma_{{3}}{\left({t}^{{2}}+{1}\right)}$$
$$\displaystyle\Rightarrow\gamma_{{1}}+\gamma_{{3}}={4},-\gamma_{{1}}+\gamma_{{2}}=-{3},\gamma_{{1}}+\gamma_{{2}}+\gamma_{{3}}={2}$$
$$\displaystyle\Rightarrow{4}+\gamma_{{2}}={2}$$
$$\displaystyle\Rightarrow\gamma_{{2}}=-{2}$$
$$\displaystyle-\gamma_{{1}}+\gamma_{{2}}=-{3}\Rightarrow-\gamma_{{1}}-{2}=-{3}\Rightarrow\gamma_{{1}}={1}$$
$$\displaystyle\gamma_{{1}}+\gamma_{{3}}={4}\Rightarrow{1}+\gamma_{{3}}={4}\Rightarrow\gamma_{{3}}={3}$$ Let $$\displaystyle{2}{t}^{{2}}+{3}=\gamma_{{1}}{\left({t}^{{2}}-{t}+{1}\right)}+\gamma_{{2}}{\left({t}+{1}\right)}+\gamma_{{3}}{\left({t}^{{2}}+{1}\right)}$$
$$\displaystyle\Rightarrow\gamma_{{1}}+\gamma_{{3}}={2},-\gamma_{{1}}+\gamma_{{2}}={0},\gamma_{{1}}+\gamma_{{2}}+\gamma_{{3}}={3}$$
$$\displaystyle\Rightarrow{2}+\gamma_{{2}}={3}$$
$$\displaystyle\Rightarrow\gamma_{{2}}={1}$$
$$\displaystyle-\gamma_{{1}}+\gamma_{{2}}={0}\Rightarrow-\gamma_{{1}}+{1}={0}{R}{i}{>}\leftrightarrow{o}{w}\gamma_{{1}}={1}$$
$$\displaystyle\gamma_{{1}}+\gamma_{{3}}={2}\Rightarrow{1}+\gamma_{{3}}={2}\Rightarrow\gamma_{{3}}={1}$$
$$\therefore Q = \left(\begin{array}{c}2 & 1 & 1 \\ 3 & -2 & 1 & \\-1 & 3 & 1\end{array}\right)$$