# Give a full correct answer for given question

Question
Alternate coordinate systems

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 $$b= \begin{bmatrix} 5\\ 3 \end{bmatrix}$$ are both (2,1) in the coordinate system defined by these two bases

2020-11-02

Consider $$W = \left \{ a + bt + ct^{2}; a + b + c = 0 \right \}$$ Yes.W is a subspace of $$\displaystyle{P}_{{{2}}}.$$ Since $$\displaystyle{0}\epsilon{W}.$$ If $$\displaystyle{x}+{y}{t}+{z}{t}^{{{2}}}{\quad\text{and}\quad}{l}+{m}{t}+{n}{t}^{{{2}}}$$ are elements of $$\displaystyle{P}_{{{2}}},$$ then $$\displaystyle{\left({x}+{l}\right)}+{\left({y}+{m}\right)}{t}+{\left({z}+{n}\right)}{t}^{{2}}\epsilon{W}.$$ So w is closed under vector addtion. Let $$\displaystyle{r}\epsilon{\mathbb{{{R}}}}$$ and $$\displaystyle{a}+{b}{t}+{c}{t}^{{2}}\epsilon{W}$$ then $$\displaystyle{r}{\left({a}+{b}{t}+{c}{t}{2}\right)}={r}{a}+{r}{b}{t}+{r}{c}{t}^{{2}}\epsilon{W}.$$ Therefore w is closed under scalar multiplication . Therefore W is a subspace of $$\displaystyle{P}_{{2}}.$$

### Relevant Questions

Given the full and correct answer the two bases of $$B1 = \left\{ \left(\begin{array}{c}2\\ 1\end{array}\right),\left(\begin{array}{c}3\\ 2\end{array}\right) \right\}$$
$$B_2 = \left\{ \left(\begin{array}{c}3\\ 1\end{array}\right),\left(\begin{array}{c}7\\ 2\end{array}\right) \right\}$$
find the change of basis matrix from $$\displaystyle{B}_{{1}}\to{B}_{{2}}$$ and next use this matrix to covert the coordinate vector
$$\overrightarrow{v}_{B_1} = \left(\begin{array}{c}2\\ -1\end{array}\right)$$ of v to its coodirnate vector
$$\overrightarrow{v}_{B_2}$$

To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}$$

(a) Find the bases and dimension for the subspace $$H = \left\{ \begin{bmatrix} 3a + 6b -c\\ 6a - 2b - 2c \\ -9a + 5b + 3c \\ -3a + b + c \end{bmatrix} ; a, b, c \in R \right\}$$ (b) Let be bases for a vector space V,and suppose (i) Find the change of coordinate matrix from B toD. (ii) Find $$\displaystyle{\left[{x}\right]}_{{D}}{f}{\quad\text{or}\quad}{x}={3}{b}_{{1}}-{2}{b}_{{2}}+{b}_{{3}}$$

The given system of inequality:
$$\displaystyle{\left\lbrace\begin{matrix}{y}<{9}-{x}^{2}\\{y}\ge{x}+{3}\end{matrix}\right.}$$
Also find the coordinates of all vertices, and check whether the solution set is bounded.

Consider the following two bases for $$\displaystyle{R}^{{3}}$$ :
$$\alpha := \left\{ \begin{bmatrix} 2 \\ 1\\ 3 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix}3 \\ 1 \\ -1 \end{bmatrix} \right\} and\ \beta := \left\{ \begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix}-2 \\ 3\\ 1 \end{bmatrix}, \begin{bmatrix}2 \\ 3\\ -1 \end{bmatrix} \right\}$$ If $$[x]_{\alpha} = \begin{bmatrix}1 \\ 2 \\-1 \end{bmatrix}_{\alpha} then\ find\ [x]_{\beta}$$
(that is, express x in the $$\displaystyle\beta$$ coordinates).

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 $$R^2, S = \left\{ \begin{bmatrix}1 \\ 0 \end{bmatrix}\begin{bmatrix} 0 \\1 \end{bmatrix} \right\}, v = \begin{bmatrix} 3 \\-2 \end{bmatrix}$$

The system of equation \begin{cases}2x + y = 1\\4x +2y = 3\end{cases} by graphing method and if the system has no solution then the solution is inconsistent. Given: The linear equations is \begin{cases}2x + y = 1\\4x +2y = 3\end{cases}
The quadratic function $$\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}$$ whose graph passes through the points (1, 4), (2, 1) and (3, 4).
$$\displaystyle{T}=\mathbb{R}^{{2}}\rightarrow\mathbb{R}^{{4}}\ {s}{u}{c}{h} \ {t}\hat \ {{T}}{\left({e}_{{1}}\right)}={\left({7},{1},{7},{1}\right)},{\quad\text{and}\quad}{T}{\left({e}_{{2}}\right)}={\left(-{8},{5},{0},{0}\right)},{w}{h}{e}{r}{e}{\ e}_{{1}}={\left({1},{0}\right)},{\quad\text{and}\quad}{e}_{{2}}={\left({0},{1}\right)}$$.
A line L through the origin in $$\displaystyle\mathbb{R}^{{3}}$$ can be represented by parametric equations of the form x = at, y = bt, and z = ct. Use these equations to show that L is a subspase of $$RR^3$$  by showing that if $$v_1=(x_1,y_1,z_1)\ and\ v_2=(x_2,y_2,z_2)$$  are points on L and k is any real number, then $$kv_1\ and\ v_1+v_2$$  are also points on L.