# Interraption: To show that the system \displaystyle\dot{{r}}={r}{\left({1}-{r}^{2}\right)},\dot{\theta}={1} is equivalent to \displaystyle\dot{{x}}={x}-{y}-{x}{\left({x}^{2}+{y}^{2}\right)},\dot{{y}}={x}+{y}-{y}{\left({x}^{2}+{y}^{2}\right)} for polar to Cartesian coordinates. A limit cycle is a closed trajectory. Isolated means that neighboring trajectories are not closed. A limit cycle is said to be unstable or half stable, if all neighboring trajectories approach the lemin cycle. These systems oscillate even in the absence of external periodic force.

Question
Alternate coordinate systems
Interraption: To show that the system $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{r}\right\rbrace}}}={\left\lbrace{r}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{r}\right\rbrace}^{{{2}}}\right)}\right\rbrace},\dot{{\theta}}={\left\lbrace{1}\right\rbrace}$$ is equivalent to $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{x}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}-{\left\lbrace{y}\right\rbrace}-{\left\lbrace{x}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace},\dot{{{\left\lbrace{y}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}+{\left\lbrace{y}\right\rbrace}-{\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace}$$ for polar to Cartesian coordinates.
A limit cycle is a closed trajectory. Isolated means that neighboring trajectories are not closed.
A limit cycle is said to be unstable or half stable, if all neighboring trajectories approach the lemin cycle.
These systems oscillate even in the absence of external periodic force.

2020-11-04
The coordinate system can be either the Cartesian system or the polar system. In case of the polar system, a point is determined by a distance and angle from a reference point while as in Cartesian coordinate system each point is determined by a pair of numerical coordinates.
It is given that $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}={\left\lbrace{r}\right\rbrace}{\cos{{\left\lbrace\theta\right\rbrace}}}{\left\lbrace\quad\text{and}\quad\right\rbrace}{\left\lbrace{y}\right\rbrace}={\left\lbrace{r}\right\rbrace}{\sin{{\left\lbrace\theta\right\rbrace}}}$$
Square and add the above equations.
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}={\left\lbrace{r}\right\rbrace}^{{{2}}}{\left\lbrace{\left\lbrace{\cos}\right\rbrace}^{{{2}}}\theta\right\rbrace}+{\left\lbrace{r}\right\rbrace}^{{{2}}}{\left\lbrace{\left\lbrace{\sin}\right\rbrace}^{{{2}}}\theta\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}={\left\lbrace{r}\right\rbrace}^{{{2}}}{\left\lbrace{\left({\left\lbrace{\left\lbrace{\cos}\right\rbrace}^{{{2}}}\theta\right\rbrace}+{\left\lbrace{\left\lbrace{\sin}\right\rbrace}^{{{2}}}\theta\right\rbrace}\right)}\right\rbrace}$$
Substitute $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{\left\lbrace{\cos}\right\rbrace}^{{{2}}}\theta\right\rbrace}+{\left\lbrace{\left\lbrace{\sin}\right\rbrace}^{{{2}}}\theta\right\rbrace}\right)}\right\rbrace}={\left\lbrace{1}\right\rbrace},$$ so that
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}={\left\lbrace{r}\right\rbrace}^{{{2}}}$$
Now,
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}={\left\lbrace{r}\right\rbrace}{\cos{{\left\lbrace\theta\right\rbrace}}}$$
Differentiate.
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{x}\right\rbrace}}}=\dot{{{\left\lbrace{r}\right\rbrace}}}\theta-{\left\lbrace{r}\right\rbrace}\theta{\sin{{\left\lbrace\theta\right\rbrace}}}$$
Substitute $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{r}\right\rbrace}}}={\left\lbrace{r}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{r}\right\rbrace}^{{{2}}}\right)}\right\rbrace},{\cos{{\left\lbrace\theta\right\rbrace}}}={\frac{{{x}}}{{{\left\lbrace{r}\right\rbrace}}}}{\left\lbrace\quad\text{and}\quad\right\rbrace}{\left\lbrace{y}\right\rbrace}\ \text{for}\ {\left\lbrace{r}\right\rbrace}{\sin{{\left\lbrace\theta\right\rbrace}}}$$ into the above equation.
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{x}\right\rbrace}}}={\left\lbrace{\left({\frac{{{\left\lbrace{\left\lbrace{x}\right\rbrace}{\left\lbrace{\left({\left\lbrace{r}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{r}\right\rbrace}^{{{2}}}\right)}\right\rbrace}\right)}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{r}\right\rbrace}}}}\right)}\right\rbrace}-{\left\lbrace{y}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{x}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{r}\right\rbrace}^{{{2}}}\right)}\right\rbrace}-{\left\lbrace{y}\right\rbrace}$$
Substitute $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}^{{{2}}}={\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{x}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{x}\right\rbrace}^{{{2}}}-{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace}-{\left\lbrace{y}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{x}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}-{\left\lbrace{y}\right\rbrace}-{\left\lbrace{x}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace}$$
Thus, it is shown that $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{x}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}-{\left\lbrace{y}\right\rbrace}-{\left\lbrace{x}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace}$$
Now,
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\sin{{\left\lbrace\theta\right\rbrace}}}$$
Differentiate.
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{y}\right\rbrace}}}=\dot{{{\left\lbrace{r}\right\rbrace}}}{\sin{{\left\lbrace\theta\right\rbrace}}}+{\left\lbrace{r}\right\rbrace}\theta{\cos{{\left\lbrace\theta\right\rbrace}}}$$
Substitute $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{r}\right\rbrace}}}={\left\lbrace{r}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{r}\right\rbrace}^{{{2}}}\right)}\right\rbrace},{\sin{{\left\lbrace\theta\right\rbrace}}}={\frac{{{y}}}{{{\left\lbrace{r}\right\rbrace}}}},\dot{{\theta}}={\left\lbrace{1}\right\rbrace}{\left\lbrace\quad\text{and}\quad\right\rbrace}{\left\lbrace{x}\right\rbrace}={\left\lbrace{r}\right\rbrace}{\cos{{\left\lbrace\theta\right\rbrace}}}$$ into the above equation.
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{y}\right\rbrace}}}={\left\lbrace{\left({\frac{{{\left\lbrace{\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{r}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{r}\right\rbrace}^{{{2}}}\right)}\right\rbrace}\right)}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{r}\right\rbrace}}}}\right)}\right\rbrace}+{\left\lbrace{x}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{y}\right\rbrace}}}={\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{r}\right\rbrace}^{{{2}}}\right)}\right\rbrace}+{\left\lbrace{x}\right\rbrace}$$
But $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}^{{{2}}}={\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{y}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}+{\left\lbrace{y}\right\rbrace}-{\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace}$$
Therefore, it is shown that $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{y}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}+{\left\lbrace{y}\right\rbrace}-{\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace}$$
For system $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{r}\right\rbrace}}}={\left\lbrace{r}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{r}\right\rbrace}^{{{2}}}\right)}\right\rbrace},\dot{{\theta}}={\left\lbrace{1}\right\rbrace}\ \text{is equivalent}\ \to\dot{{{\left\lbrace{x}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}-{\left\lbrace{y}\right\rbrace}-{\left\lbrace{x}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace},$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{y}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}+{\left\lbrace{y}\right\rbrace}-{\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace}$$ is shown below.

### Relevant Questions

To determine:
a) The origin $$\displaystyle{\left[\begin{array}{cc} {0}&\ {0}\end{array}\right]}$$ is a critical point of the systems.
$$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}{\quad\text{and}\quad}{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$. Futhermore, it is a center of the corresponding linear system.
b) The systems $$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}{\quad\text{and}\quad}{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$ are almost linear.
c) To prove: $$\displaystyle{\left[{\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}\ {<}\ {0}\right]}{\quad\text{and}\quad}{\left[{r}\rightarrow\ {0}\ {a}{s}\ {t}\rightarrow\ \infty\right]},$$ hence the critical point for the system $$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$ is asymptotically stable and the solution of the initial value problem for $$\displaystyle{\left[{r}\ {w}{i}{t}{h}\ {r}={r}_{{{0}}}\ {a}{t}\ {t}={0}\right]}$$ becomes unbounded as $$\displaystyle{\left[{t}\rightarrow{\frac{{{1}}}{{{2}}}}\ {r}{\frac{{{2}}}{{{0}}}}\right]}$$, hence the critical point for the system $$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$ is unstable.
To determine:
a) Whether the statement, " The point with Cartesian coordinates $$\displaystyle{\left[\begin{array}{cc} -{2}&\ {2}\end{array}\right]}$$ has polar coordinates $$\displaystyle{\left[{b}{f}{\left({2}\sqrt{{{2}}},\ {\frac{{{3}\pi}}{{{4}}}}\right)}\ {\left({2}\sqrt{{{2}}},{\frac{{{11}\pi}}{{{4}}}}\right)}\ {\left({2}\sqrt{{{2}}},\ -{\frac{{{5}\pi}}{{{4}}}}\right)}\ {\quad\text{and}\quad}\ {\left(-{2}\sqrt{{2}},\ -{\frac{{\pi}}{{{4}}}}\right)}\right]}$$ " is true or false.
b) Whether the statement, " the graphs of $$\displaystyle{\left[{r}{\cos{\theta}}={4}\ {\quad\text{and}\quad}\ {r}{\sin{\theta}}=\ -{2}\right]}$$ intersect exactly once " is true or false.
c) Whether the statement, " the graphs of $$\displaystyle{\left[{r}={4}\ {\quad\text{and}\quad}\ \theta={\frac{{\pi}}{{{4}}}}\right]}$$ intersect exactly once ", is true or false.
d) Whether the statement, " the point $$\displaystyle{\left[\begin{array}{cc} {3}&{\frac{{\pi}}{{{2}}}}\end{array}\right]}{l}{i}{e}{s}{o}{n}{t}{h}{e}{g}{r}{a}{p}{h}{o}{f}{\left[{r}={3}{\cos{\ }}{2}\ \theta\right]}$$ " is true or false.
e) Whether the statement, " the graphs of $$\displaystyle{\left[{r}={2}{\sec{\theta}}\ {\quad\text{and}\quad}\ {r}={3}{\csc{\theta}}\right]}$$ are lines " is true or false.
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
The reason ehy the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}$$ lies on the polar graph $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}$$ even though it does not satisfy the equation.
To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle\ {x}=\ {r}{\cos{\theta}}$$
$$\displaystyle\ {y}=\ {r}{\sin{\theta}}$$
b. From rectangular to polar:
$$\displaystyle{r}=\pm\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}$$
$$\displaystyle{\cos{\theta}}={\frac{{{x}}}{{{r}}}},{\sin{\theta}}={\frac{{{y}}}{{{r}}}},{\tan{\theta}}={\frac{{{x}}}{{{y}}}}$$
Calculation:
Given: equation in rectangular-coordinate is $$\displaystyle{y}={x}$$.
Converting into equivalent polar equation -
$$\displaystyle{y}={x}$$
Put $$\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},$$
$$\displaystyle\Rightarrow\ {r}{\sin{\theta}}={r}{\cos{\theta}}$$
$$\displaystyle\Rightarrow\ {\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}={1}$$
$$\displaystyle\Rightarrow\ {\tan{\theta}}={1}$$
Thus, desired equivalent polar equation would be $$\displaystyle\theta={1}$$
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 $$\displaystyle{R}^{{2}},{S}=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},{v}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{3}\backslash-{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\$$
Since we will be using various bases and the coordinate systems they define, let's review how we translate between coordinate systems. a. Suppose that we have a basis$$\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}.$$
Consider the following two bases for $$\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).