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Let { 1 - x, x, 1 + x^2, x - x^3, x^4 } be the basis for the vector space of po…

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Wize University Linear Algebra Textbook > Vector Spaces and Subspaces

Coordinates

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Let {1−x,x,1+x2,x−x3,x4}\{ 1 - x, x, 1 + x^2, x - x^3, x^4 \} be the basis for the vector space of polynomials with degree at most 4. What is the coordinate vector for vv in this basis?
v=4−x2+x3−2x4v = 4 - x^2 + x^3 - 2x^4

More Coordinates Questions:
Coordinates
Practice: Coordinates
Find the coordinates of the polynomial p(x)=5+x−3x2−3x3p(x)=5+x-3x^2-3x^3 relative to the following basis of P3P_3:
B={(1+x),  (x+x2),  (x2+x3),  (1−x3)}B = \{(1+x),\ \ (x+x^2),\ \ (x^2+x^3),\ \ (1-x^3)\}
Practice: Coordinates
Find the coordinates of the polynomial p(x)=5+x−3x2−3x3p(x)=5+x-3x^2-3x^3 relative to the following basis of P3P_3:
{(1+x),(x+x2),(x2+x3),(1−x3)}\{(1+x),(x+x^2),(x^2+x^3),(1-x^3)\}
Verify if the following vectors form a basis for R3\mathbb{R}^3:
{[112],[1−21],[2−11]}\left\{ \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix},\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} \right\}
19.4F_Mid_Builder_$\tkcth{8.5.}\tkco{9}$__$\tkco{Mock 3}$
Taken together,
v⃗1=[11]\vv_1 = \colvec{1}{1} and v⃗2=[1−1]\vv_2 = \colvec{1}{-1}
constitute a basis forR2\mathbb{R}^2. Determine the coordinates [x⃗]v=[xv1xv2][ \vx ]_{v} = \colvec{\bcth{x_{v_1}}}{\bct{x_{v_2}}}of the vector x⃗=[21]\vx = \colvec{2}{1} relative to this basis.
Concept Clarifier: Finding Coordinate Vectors
The following is a basis for R2\mathbb{R}^2 .
{(11),(−20)}\left \{ \left ( \begin{array}{c} 1 \\ 1 \end{array} \right ), \left ( \begin{array}{c} -2 \\0 \end{array} \right ) \right \}
If v⃗=(2,3)\vec{v} = (2, 3) is a coordinate vector in this basis, what is v⃗\vec{v} in coordinates with respect to the basis
19.4F_Mid_Builder_$\tkcth{8.4.16}$_$\tkcth{mock 1 ✓}$
Find the coordinates (c1, c2)(c_1,\, c_2) for the point P=(8,11)P = (8,11) in terms of the nonstandard basis vectors e⃗1=<1, 1>\ve{1} = \rowvec{1}{1} and e⃗2=<6, 9>\ve{2} = \rowvec{6}{9}.
Coordinates
Find the coordinates of the polynomial p(x)=2−x+x2p(x)=2-x+x^2 relative to the basis {1+x,1+x2,x+x2}\{1+x,1+x^2,x+x^2\} of P2P_2
19.4F_133_8.2_Mock_F1_$\tkco{eg6}$_$\key{Final}$_Builder_$\tkcth{8.2.}\tkcf{6}$_
The set {[21]}\left\{ \colvec{2}{1} \right\} is a basis for the subspace TT, where T⊆R2T \subseteq \mathbb{R}^2.
Give the position vector r⃗P\vr_{_{P}} (in the standard basis) corresponding to the point in TT whose coordinate is given by PT=(3)P_{T} = (3).
If r⃗=[xy]\vr = \colvec{\bco{x}}{\bct{y}}, then
Coordinates
Practice: Coordinates
Find the coordinates of the polynomial p(x)=5+x−3x2−3x3p(x)=5+x-3x^2-3x^3 relative to the following basis of P3P_3:
B={(1+x),  (x+x2),  (x2+x3),  (1−x3)}B = \{(1+x),\ \ (x+x^2),\ \ (x^2+x^3),\ \ (1-x^3)\}
Coordinates
Practice: Coordinates
Let B1={[−120],[3−52],[4−73]}B_1= \left\{ \begin{bmatrix} -1\\ 2 \\0\\ \end{bmatrix} , \begin{bmatrix} 3\\ -5\\ 2\\ \end{bmatrix} , \begin{bmatrix} 4\\ -7\\ 3\\ \end{bmatrix} \right\} and B2={[2−31],[11−1],[−1−30]}B_2= \left\{ \begin{bmatrix} 2\\ -3 \\1\\ \end{bmatrix} , \begin{bmatrix} 1\\ 1\\ -1\\ \end{bmatrix} , \begin{bmatrix} -1\\ -3\\ 0\\ \end{bmatrix} \right\} be two bases of R3\mathbb{R}^3.
Suppose [x⃗]B1=[−48−7][\vec x]_{\small B_1} = \begin{bmatrix} -4\\ 8\\ -7\\ \end{bmatrix}.
Example: Change of Basis
Example: Change of Basis
Suppose we have 2 ordered bases: A=((1,−1),(−2,1)),B=((1,1),(1,2))A=((1,-1),(-2,1)),B=((1,1),(1,2)) .
Compute the change of basis matrix CA→BC_{A\rightarrow B}, and if[v⃗]A=(−3,−4)[\vec v]_A=(-3,-4), then find [v⃗]B[\vec v]_B .
Practice Question: Change of Basis
Suppose we have an ordered basis B=((1,1,1),(2,0,−1),(−1,1,0))B=((1,1,1),(2,0,-1),(-1,1,0)) . Let S=((1,0,0),(0,1,0),(0,0,1))S=((1,0,0),(0,1,0),(0,0,1)) be the standard ordered basis. Compute the change of basis matrix CS→BC_{S\rightarrow B} , and use it to find (2,−1,2)B(2,-1,2)_B
(i.e. find [v⃗]B[\vec v]_B , if [v⃗]S=(2,−1,2)[\vec v]_S=(2,-1,2) ).
Consider the vector v=(1,2,3)v = (1, 2, 3) with respect to the standard unit vector basis. What is the coordinate vector for vv in the following basis:
{(110),(010),(−10−2)}\left \{ \left ( \begin{array}{c} 1 \\ 1\\ 0 \end{array} \right ), \left ( \begin{array}{c} 0 \\1 \\ 0 \end{array} \right ) , \left ( \begin{array}{c} -1 \\ 0 \\ -2 \end{array} \right ) \right \}