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 \(\displaystyle{P}_{{2}}.\)
Give a full correct answer for given question
Give a full correct answer for given question
Answers (1)
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 quadratic function \(\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}\) whose graph passes through the points (1, 4), (2, 1) and (3, 4).
Assume that T is a linear transformation. Find the standard matrix of T.
\(\displaystyle{T}=\mathbb{R}^{{2}}\rightarrow\mathbb{R}^{{4}}\ {s}{u}{c}{h} \ {t}\hat \ {{T}}{\left({e}_{{1}}\right)}={\left({7},{1},{7},{1}\right)},{\quad\text{and}\quad}{T}{\left({e}_{{2}}\right)}={\left(-{8},{5},{0},{0}\right)},{w}{h}{e}{r}{e}{\ e}_{{1}}={\left({1},{0}\right)},{\quad\text{and}\quad}{e}_{{2}}={\left({0},{1}\right)}\).
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.
