# The coordinate vector of \displaystyle{\left[{\mathbf{{{p}}}}\ {\left({t}\right)}\ =\ {6}\ +\ {3}{t}\ -{t}^{{{2}}}\right]} relative to the basis \displaystyle{m}{a}{t}{h}{s}{c}{r}{\left\lbrace{B}\right\rbrace}\ ={\left\lbrace{1}\ +\ {t},\ {1}\ +\ {t}^{{{2}}},\ {t}\ +\ {t}^{{{2}}}\right\rbrace}

Question
Alternate coordinate systems
The coordinate vector of $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left[{\left\lbrace{\mathbf{{{\left\lbrace{\left\lbrace{p}\right\rbrace}\right\rbrace}}}}\right\rbrace}\ {\left\lbrace{\left({\left\lbrace{t}\right\rbrace}\right)}\right\rbrace}\ =\ {\left\lbrace{6}\right\rbrace}\ +\ {\left\lbrace{3}\right\rbrace}{\left\lbrace{t}\right\rbrace}\ -{\left\lbrace{t}\right\rbrace}^{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}\right]}\right\rbrace}$$ relative to the basis $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{m}\right\rbrace}{\left\lbrace{a}\right\rbrace}{\left\lbrace{t}\right\rbrace}{\left\lbrace{h}\right\rbrace}{\left\lbrace{s}\right\rbrace}{\left\lbrace{c}\right\rbrace}{\left\lbrace{r}\right\rbrace}{\left\lbrace\le{f}{t}{\left\lbrace{\left\lbrace{B}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}\right\rbrace}\ ={\left\lbrace\le{f}{t}{\left\lbrace{\left\lbrace{1}\right\rbrace}\ +\ {\left\lbrace{t}\right\rbrace},\ {\left\lbrace{1}\right\rbrace}\ +\ {\left\lbrace{t}\right\rbrace}^{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}},\ {\left\lbrace{t}\right\rbrace}\ +\ {\left\lbrace{t}\right\rbrace}^{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{r}{i}{g}{h}{t}\right\rbrace}\right\rbrace}$$

2020-12-18
Any arbitrary vector in $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\mathbb{{{\left\lbrace{\left\lbrace{P}\right\rbrace}_{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}\right\rbrace}}}}\right\rbrace}$$ can be written as,
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left[{\left\lbrace{\mathbf{{{\left\lbrace{\left\lbrace{p}\right\rbrace}\right\rbrace}}}}\right\rbrace}\ {\left\lbrace{\left({\left\lbrace{t}\right\rbrace}\right)}\right\rbrace}\ =\ {\left\lbrace{a}\right\rbrace}\ {\left\lbrace{\left(\ {\left\lbrace{1}\right\rbrace}\ +\ {\left\lbrace{t}\right\rbrace}^{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}\ \right)}\right\rbrace}\ +\ {\left\lbrace{b}\right\rbrace}\ {\left\lbrace{\left(\ {\left\lbrace{t}\right\rbrace}\ +\ {\left\lbrace{t}\right\rbrace}^{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}\ \right)}\right\rbrace}\ +\ {\left\lbrace{c}\right\rbrace}\ {\left\lbrace{\left(\ {\left\lbrace{1}\right\rbrace}\ +\ {\left\lbrace{2}\right\rbrace}\ {\left\lbrace{t}\right\rbrace}\ +\ {\left\lbrace{t}\right\rbrace}^{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}\ \right)}\right\rbrace}\right]}\right\rbrace}.$$
Thus, for the vector $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left[{\left\lbrace{\mathbf{{{\left\lbrace{\left\lbrace{p}\right\rbrace}\right\rbrace}}}}\right\rbrace}\ {\left\lbrace{\left(\ {\left\lbrace{t}\right\rbrace}\ \right)}\right\rbrace}={\left\lbrace{6}\right\rbrace}\ +\ {\left\lbrace{3}\right\rbrace}\ {\left\lbrace{t}\right\rbrace}\ -\ {\left\lbrace{t}\right\rbrace}^{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}\right]}\right\rbrace}$$ can be written as,
1) $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left[{\left\lbrace{a}\right\rbrace}\ {\left\lbrace{\left(\ {\left\lbrace{1}\right\rbrace}\ +\ {\left\lbrace{t}\right\rbrace}\ \right)}\right\rbrace}\ +\ {\left\lbrace{b}\right\rbrace}\ {\left\lbrace{\left(\ {\left\lbrace{1}\right\rbrace}\ +\ {\left\lbrace{t}\right\rbrace}^{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}\ \right)}\right\rbrace}\ +\ {\left\lbrace{c}\right\rbrace}\ {\left\lbrace{\left(\ {\left\lbrace{t}\right\rbrace}\ +\ {\left\lbrace{t}\right\rbrace}^{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}\ \right)}\right\rbrace}={\left\lbrace{6}\right\rbrace}\ +\ {\left\lbrace{3}\right\rbrace}{\left\lbrace{t}\right\rbrace}\ -\ {\left\lbrace{t}\right\rbrace}^{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}\right]}\right\rbrace}$$
On comparing the terms of both the sides in the above equation (1) gives the following equations.
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{a}\right\rbrace}\ +\ {\left\lbrace{b}\right\rbrace}={\left\lbrace{6}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{a}\right\rbrace}\ +\ {\left\lbrace{c}\right\rbrace}={\left\lbrace{3}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{b}\right\rbrace}\ +\ {\left\lbrace{c}\right\rbrace}=-{\left\lbrace{1}\right\rbrace}$$
The representation of the linear system in a matrix is,
$$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{1}&{0}\backslash{1}&{0}&{1}\backslash{0}&{1}&{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{a}\backslash{c}\backslash{d}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{6}\backslash{3}\backslash-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$
The solution of the above system gives as by solving the augmented matrix
$$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{1}&{0}&{6}\backslash{1}&{0}&{1}&{3}\backslash{0}&{1}&{1}&-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$
Use Elementary row transformations to reduce the augmented matrix to row reduced Echelon form.
Step 1:
$$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{1}&{0}&\ \ {6}\backslash{1}&{0}&{1}&\ \ {3}\backslash{0}&{1}&{1}&-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{u}{n}{d}{e}{r}\rightarrow{\left\lbrace{R}_{{{2}}}\ \rightarrow\ {R}_{{{2}}}\ -\ {R}_{{{1}}}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{1}&{0}&\ \ \ {6}\backslash{0}&-{1}\ \ \ &{1}&-{3}\backslash{0}&{1}&{1}&-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$
Step 2
$$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{1}&{0}&\ \ \ {6}\backslash{0}&-{1}\ \ \ &{1}&-{3}\backslash{0}&{1}&{1}&-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{u}{n}{d}{e}{r}\rightarrow{\left\lbrace{R}_{{{3}}}\ \rightarrow\ {R}_{{{3}}}\ +\ {R}_{{{2}}}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{1}&{0}&\ \ \ {6}\backslash{0}&-{1}\ \ \ &{1}&-{3}\backslash{0}&{0}&{2}&-{4}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$
Step 3:
$$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{1}&{0}&\ \ \ {6}\backslash{0}&-{1}\ \ \ &{1}&-{3}\backslash{0}&{0}&{2}&-{4}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{u}{n}{d}{e}{r}\rightarrow{\left\lbrace{R}_{{{1}}}\ \rightarrow\ {R}_{{{1}}}\ +\ {R}_{{{2}}}\ -{\frac{{{1}}}{{{2}}}}\ {R}_{{{3}}}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}&{0}&\ \ \ {5}\backslash{0}&-{1}\ \ \ &{1}&-{3}\backslash{0}&{0}&{2}&-{4}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$
Step 4:
$$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}&{0}&\ \ \ {5}\backslash{0}&-{1}\ \ \ &{1}&-{3}\backslash{0}&{0}&{2}&-{4}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{u}{n}{d}{e}{r}\rightarrow{\left\lbrace{R}_{{{2}}}\ \rightarrow\ {R}_{{{2}}}\ -{\frac{{{1}}}{{{2}}}}\ {R}_{{{3}}}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}&{0}&\ \ \ {5}\backslash{0}&{1}&{0}&\ \ \ {1}\backslash{0}&{0}&{1}&-{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$
Step 5:
$$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}&{0}&\ \ \ {5}\backslash{0}&-{1}\ \ \ &{1}&-{1}\backslash{0}&{0}&{2}&-{4}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{u}{n}{d}{e}{r}\rightarrow{\left\lbrace{R}_{{{2}}}\ \rightarrow\ -{1}\ \times\ {R}_{{{2}}}\right\rbrace}_{{{R}_{{{3}}}\rightarrow{\frac{{{1}}}{{{2}}}}\ {R}_{{{3}}}}}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}&{0}&\ \ \ {5}\backslash{0}&{1}&{0}&\ \ \ {1}\backslash{0}&{0}&{1}&-{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$
Thus, the solution is $$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{a}\backslash{c}\backslash{d}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\ \ \ \ {5}\backslash\ \ \ \ {1}\backslash-{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$
Therefore, the coordinate vector $$\displaystyle{\mathbf{{{\left[{p}\right]}_{{{m}{a}{t}{h}{s}{c}{r}{\left\lbrace{B}\right\rbrace}}}}}}\ {i}{s}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\ \ \ \ {5}\backslash\ \ \ \ {1}\backslash-{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$

### Relevant Questions

The system of equation $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{b}\right\rbrace}{\left\lbrace{e}\right\rbrace}{\left\lbrace{g}\right\rbrace}\in{\left\lbrace\le{f}{t}{\left\lbrace{\left\lbrace{c}\right\rbrace}{\left\lbrace{a}\right\rbrace}{\left\lbrace{s}\right\rbrace}{\left\lbrace{e}\right\rbrace}{\left\lbrace{s}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}\right\rbrace}{\frac{{{n}}}{{{\left\lbrace{2}\right\rbrace}}}}\backslash-{\frac{{{\left\lbrace{\left\lbrace{n}\right\rbrace}+{\left\lbrace{1}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}{\left\lbrace{e}\right\rbrace}{\left\lbrace{n}\right\rbrace}{\left\lbrace{d}\right\rbrace}{\left\lbrace\le{f}{t}{\left\lbrace{\left\lbrace{c}\right\rbrace}{\left\lbrace{a}\right\rbrace}{\left\lbrace{s}\right\rbrace}{\left\lbrace{e}\right\rbrace}{\left\lbrace{s}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}\right\rbrace}$$ by graphing method and if the system has no solution then the solution is inconsistent.
Given: The linear equations is $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{y}\right\rbrace}={\left\lbrace{1}\right\rbrace}{\left\lbrace\quad\text{and}\quad\right\rbrace}{\left\lbrace{4}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{2}\right\rbrace}{\left\lbrace{y}\right\rbrace}={\left\lbrace{3}\right\rbrace}.$$
All bases considered in these are assumed to be ordered bases. In Exercise, compute the coordinate vector of v with respect to the giving basis S for V. V is $$\displaystyle{R}^{{2}},{S}=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},{v}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{3}\backslash-{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\$$
Consider the following vectors in $$\displaystyle{R}^{{4}}:$$ $$\displaystyle{v}_{{1}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{1}\backslash{1}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{v}_{{2}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}\backslash{1}\backslash{1}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{v}_{{3}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}\backslash{0}\backslash{1}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{v}_{{4}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}\backslash{0}\backslash{0}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$ a. Explain why $$\displaystyle{B}=\le{f}{t}{\left\lbrace{v}_{{1}},{v}_{{2}},{v}_{{3}},{v}_{{4}}{r}{i}{g}{h}{t}\right\rbrace}$$
forms a basis for $$\displaystyle{R}^{{4}}.$$ b. Explain how to convert $$\displaystyle\le{f}{t}{\left\lbrace{x}{r}{i}{g}{h}{t}\right\rbrace}_{{B}},$$ the representation of a vector x in the coordinates defined by B, into x, its representation in the standard coordinate system. c. Explain how to convert the vector x into,$$\displaystyle{\left\lbrace{x}\right\rbrace}_{{B}},$$ its representation in the coordinate system defined by B
Consider the bases $$\displaystyle{B}={\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}\backslash{3}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{3}\backslash{5}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\right)}{o}{f}{R}^{{2}}{\quad\text{and}\quad}{C}={\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{1}\backslash{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{0}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}\backslash{1}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\right)}{o}{f}{R}^{{3}}$$.
and the linear maps PSKS \in L (R^2, R^3) and T \in L(R^3, R^2) given given (with respect to the standard bases) by $$\displaystyle{\left[{S}\right]}_{{{E},{E}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}&-{1}\backslash{5}&-{3}\backslash-{3}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{\quad\text{and}\quad}{\left[{T}\right]}_{{{E},{E}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&-{1}&{1}\backslash{1}&{1}&-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$ Find each of the following coordinate representations. $$\displaystyle{\left({a}\right)}{\left[{S}\right]}_{{{B},{E}}}$$
Consider the bases $$\displaystyle{B}={\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}\backslash{3}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{3}\backslash{5}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\right)}{o}{f}{R}^{{2}}{\quad\text{and}\quad}{C}={\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{1}\backslash{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{0}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}\backslash{1}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\right)}{o}{f}{R}^{{3}}$$.
and the linear maps PSKS \in L (R^2, R^3) and T \in L(R^3, R^2) given given (with respect to the standard bases) by $$\displaystyle{\left[{S}\right]}_{{{E},{E}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}&-{1}\backslash{5}&-{3}\backslash-{3}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{\quad\text{and}\quad}{\left[{T}\right]}_{{{E},{E}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&-{1}&{1}\backslash{1}&{1}&-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$ Find each of the following coordinate representations. $$\displaystyle{\left({b}\right)}{\left[{S}\right]}_{{{E},{C}}}$$
$$\displaystyle{\left({c}\right)}{\left[{S}\right]}_{{{B},{C}}}$$
Given the elow bases for R^2 and the point at the specified coordinate in the standard basis as below, (40 points) $$\displaystyle{B}{1}=\le{f}{t}{\left\lbrace{\left({1},{0}\right)},{\left({0},{1}\right)}{r}{i}{g}{h}{t}\right\rbrace}&{M}{S}{K}{B}{2}={\left({1},{2}\right)},{\left({2},-{1}\right)}{r}{i}{g}{h}{t}\rbrace{\left({1},{7}\right)}={3}^{\cdot}{\left({1},{2}\right)}-{\left({2},{1}\right)}{N}{S}{K}{B}{2}={\left({1},{1}\right)},{\left(-{1},{1}\right)}{\left({3},{7}={5}^{\cdot}{\left({1},{1}\right)}+{2}^{\cdot}{\left(-{1},{1}\right)}{N}{S}{K}{B}{2}={\left({1},{2}\right)},{\left({2},{1}\right)}{\left({0},{3}\right)}={2}^{\cdot}{\left({1},{2}\right)}-{2}^{\cdot}{\left({2},{1}\right)}{N}{S}{K}{\left({8},{10}\right)}={4}^{\cdot}{\left({1},{2}\right)}+{2}^{\cdot}{\left({2},{1}\right)}{N}{S}{K}{B}{2}={\left({1},{2}\right)},{\left(-{2},{1}\right)}{\left({0},{5}\right)}={N}{S}{K}{\left({1},{7}\right)}=\right.}$$ a. Use graph technique to find the coordinate in the second basis. (10 points) b. Show that each basis is orthogonal. (5 points) c. Determine if each basis is normal. (5 points) d. Find the transition matrix from the standard basis to the alternate basis. (15 points)
All bases considered in these are assumed to be ordered bases. In Exercise, compute coordinate vector v with respect to the giving basis S for V. V is $$\displaystyle{M}_{{22}},{S}=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&-{1}\backslash{0}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{1}\backslash{1}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}\backslash{0}&-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}\backslash-{1}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},{v}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{3}\backslash-{2}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}.$$
Since we will be using various bases and the coordinate systems they define, let's review how we translate between coordinate systems. a. Suppose that we have a basis$$\displaystyle{B}={\left\lbrace{v}_{{1}},{v}_{{2}},\ldots,{v}_{{m}}\right\rbrace}{f}{\quad\text{or}\quad}{R}^{{m}}$$. Explain what we mean by the representation {x}g of a vector x in the coordinate system defined by B. b. If we are given the representation $$\displaystyle{\left\lbrace{x}\right\rbrace}_{{B}},$$ how can we recover the vector x? c. If we are given the vector x, how can we find $$\displaystyle{\left\lbrace{x}\right\rbrace}_{{B}}$$? d. Suppose that BE is a basis for R^2. If {x}_B = \begin{bmatrix}1 \\ -2 \end{bmatrix}ZSK find the vector x. e. If $$\displaystyle{x}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}\backslash-{4}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{f}\in{d}{\left\lbrace{x}\right\rbrace}_{{B}}$$
All bases considered in these are assumed to be ordered bases. In Exercise, compute the coordinate vector of v with respect to the giving basis S for V. V is $$\displaystyle{M}_{{22}},{S}=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}\backslash{0}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{0}\backslash{1}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{1}\backslash{0}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{0}\backslash{0}&{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},{v}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}\backslash-{1}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}.$$
(a) Find the bases and dimension for the subspace $$\displaystyle{H}=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{3}{a}+{6}{b}-{c}\backslash{6}{a}-{2}{b}-{2}{c}\backslash-{9}{a}+{5}{b}+{3}{c}\backslash-{3}{a}+{b}+{c}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{a},{b},{c}\in{R}{r}{i}{g}{h}{t}\right\rbrace}$$ (b) Let be bases for a vector space V,and suppose (i) Find the change of coordinate matrix from B toD. (ii) Find $$\displaystyle{\left[{x}\right]}_{{D}}{f}{\quad\text{or}\quad}{x}={3}{b}_{{1}}-{2}{b}_{{2}}+{b}_{{3}}$$