# Determine if left{ begin{bmatrix}2 -4 1 end{bmatrix},begin{bmatrix}-3 5 -1 end{bmatrix} right} is a basis for Col begin{bmatrix}1 -3 1 end{bmatrix},begin{bmatrix}-3 7 -2 end{bmatrix}

Question
Alternate coordinate systems
Determine if $$\displaystyle\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}\backslash-{4}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{3}\backslash{5}\backslash-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}$$ is a basis for
$$\displaystyle{C}{o}{l}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash-{3}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{3}\backslash{7}\backslash-{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$

2021-02-26
$$\displaystyle\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}\backslash-{4}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{3}\backslash{5}\backslash-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}$$ Since PSK\not{\exists} any C \ in R such that \begin{bmatrix} +2 \\ -4 \\ 1 \end{bmatrix}, \begin{bmatrix} -3 \\ 5 \\ -1 \end{bmatrix} . So \begin{bmatrix} 2 \\ -4 \\ 1 \end{bmatrix}, \begin{bmatrix} -3 \\ 5 \\ -1 \end{bmatrix}ZSK are linearly independent. Now for $$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash-{3}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},={C}_{{1}}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}\backslash-{4}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}+{C}_{{2}}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{3}\backslash{5}\backslash-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\Rightarrow{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{2}{C}_{{1}}-{3}{C}_{{2}}={1}-{\left({1}\right)}\backslash-{4}{C}_{{1}}+{5}{C}_{{2}}=-{3}-{\left({2}\right)}\backslash{C}_{{1}}-{C}_{{2}}={1}-{\left({3}\right)}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ From (3) $$\displaystyle{C}_{{1}}={C}_{{2}}+{1},$$ From (1) $$\displaystyle{2}{C}_{{2}}+{2}-{3}{C}_{{2}}={1}\Rightarrow-{C}_{{2}}=-{1}\Rightarrow{C}_{{2}}={1}$$ Thus $$\displaystyle{C}_{{1}}={2}$$ So $$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash-{3}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={2}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}\backslash-{4}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}+{\left({1}\right)}{x}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{3}\backslash{5}\backslash-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$ Against for $$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{3}\backslash{7}\backslash-{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={x}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}\backslash-{1}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}+{y}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{3}\backslash{5}\backslash-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{R}{i}{>}\leftrightarrow{o}{w}{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{2}{x}-{3}{y}=-{3}-{\left({4}\right)}\backslash-{4}{x}+{5}{y}=+{7}-{\left({5}\right)}\backslash{x}-{y}={2}-{\left({6}\right)}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ From (6) $$\displaystyle{x}={y}-{2}$$ From (4) $$\displaystyle{2}{y}-{4}-{3}{y}=-{3}\Rightarrow-{y}={1}\Rightarrow{y}=-{1}$$
$$\displaystyle\Rightarrow{x}=-{1}-{2}=-{3}$$ From (5) $$\displaystyle-{4}{x}{\left(-{3}\right)}+{5}{\left(-{1}\right)}={12}-{5}={7}$$ So $$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{3}\backslash{7}\backslash{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={\left(-{3}\right)}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}\backslash-{1}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}+{\left(-{1}\right)}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{3}\backslash{5}\backslash-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$ So \left\{ \begin{bmatrix}2 \\ -4 \\ 1 \end{bmatrix}, \begin{bmatrix}-3 \\ 5 \\ -1 \end{bmatrix} \right\} generate col \begin{bmatrix}1 & -3 \\ -3 & 7 \\ 1 & 2 \end{bmatrix} . So \left\{ \begin{bmatrix}2 \\ -4 \\ 1 \end{bmatrix}, \begin{bmatrix}-3 \\ 5 \\ -1 \end{bmatrix} \right\} forms a basis of col \begin{bmatrix}1 & -3 \\ -3 & 7 \\ 1 & 2 \end{bmatrix}.

### Relevant Questions

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 $$\displaystyle{M}_{{22}},{S}=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&-{1}\backslash{0}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{1}\backslash{1}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}\backslash{0}&-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}\backslash-{1}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},{v}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{3}\backslash-{2}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}.$$
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}},$$ its representation in the coordinate system defined by B
All bases considered in these are assumed to be ordered bases. In Exercise, compute the coordinate vector of v with respect to the giving basis S for V. V is $$\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}.$$
Since we will be using various bases and the coordinate systems they define, let's review how we translate between coordinate systems. a. Suppose that we have a basis$$\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}}$$
Consider the following two bases for $$\displaystyle{R}^{{3}}$$ :
$$\displaystyle\alpha\:=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}\backslash{1}\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}-{1}\backslash{0}\backslash{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}{3}\backslash{1}\backslash-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}{\quad\text{and}\quad}\beta\:=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{1}\backslash{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}-{2}\backslash{3}\backslash{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}{2}\backslash{3}\backslash-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}$$ If $$\displaystyle{\left[{x}\right]}_{{\alpha}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{2}\backslash-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}_{{\alpha}}{t}{h}{e}{n}{f}\in{d}{\left[{x}\right]}_{{\beta}}$$
(that is, express x in the $$\displaystyle\beta$$ coordinates).
Give a full correct answer for given question 1- Let W be the set of all polynomials $$\displaystyle{a}+{b}{t}+{c}{t}^{{2}}\in{P}_{{{2}}}$$ such that $$\displaystyle{a}+{b}+{c}={0}$$ Show that W is a subspace of $$\displaystyle{P}_{{{2}}},$$ find a basis for W, and then find dim(W) 2 - Find two different bases of $$\displaystyle{R}^{{{2}}}$$ so that the coordinates of $$\displaystyle{b}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{5}\backslash{3}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$ are both (2,1) in the coordinate system defined by these two bases
Consider the bases $$\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}}}$$