Saltar al contenido principal
LibreTexts Español

23.3: Cambio de Bases

  1. Última actualización
  2. Guardar como PDF
  • Page ID
    115245

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
    %matplotlib inline
import matplotlib.pylab as plt
import numpy as np
import sympy as sym
from urllib.request import urlretrieve
sym.init_printing(use_unicode=True)
urlretrieve('https://raw.githubusercontent.com/colbrydi/jupytercheck/master/answercheck.py', 
            'answercheck.py');

    Consideremos ahora las siguientes dos bases en\(R^2\):

    \[B_1 = \{(1,2), (3,-1)\} \nonumber \]

    \[B_2 = \{(3,1), (5,2)\} \nonumber \]

    La transformación de la “base estándar” a\(B_1\) y se\(B_2\) puede definir como los vectores de columna\(P_1\) y de la\(P_2\) siguiente manera:

    B1 = np.matrix([[1,2],[3,-1]]).T
P1 = np.linalg.inv(B1)

sym.Matrix(P1)
    B2 = np.matrix([[3,1],[5,2]]).T
P2 = np.linalg.inv(B2)

sym.Matrix(P2)
    Hacer esto

    Encuentra la matriz de transición\(T\) que tomará puntos en la representación de\(B_1\) coordenadas y los pondrá en\(B_2\) coordenadas.

    NOTA esto es análogo al problema de la cinemática del robot. Queremos representar puntos en un sistema de coordenadas diferente.

    # Put your answer to the above question here.
    from answercheck import checkanswer

checkanswer.matrix(T,'dcc03ddff982e29eea6dd52ec9088986')
    Pregunta

    Dado\(u_{B_1} = \left[ \begin{matrix} 2 \\ 1 \end{matrix} \right]\) (un punto nombrado\(u\) en el sistema de\(B_1\) coordenadas) y su matriz de transición calculada\(T\), ¿cuál es el mismo punto expresado en la\(B_2\) base (es decir, qué es\(u_{B2}\))? Almacene su respuesta en una variable llamada ub2 para su comprobación.

    ub1 = np.matrix([[2],[1]])
sym.Matrix(ub1)
    ##Put your code here
    from answercheck import checkanswer

checkanswer.vector(ub2,'9a5fe29254c07cf59ebdffcaba679917')

    Hay tres bases\(B_1\),\(B_2\), y\(B_3\). Tenemos la matriz\(P_{12}\) de transición de\(B_1\) a\(B_2\) y la matriz\(P_{23}\) de transición de\(B_2\) a\(B_3\). En\(R^n\), podemos calcular la matriz de transición como\(P_{12}=B_2^{-1}B_1,\quad P_{23}=B_3^{-1}B_2\).

    Entonces podemos encontrar todas las demás matrices de transición.

    \(P_{13} = B_3^{-1}B_1=B_3^{-1}B_2*B_2^{-1}B_1= P_{23}P_{12}\)

    \(P_{21} = B_1^{-1}B_2 = (B_2^{-1}B_1)^{-1}=P_{12}^{-1}\)

    \(P_{32} = B_2^{-1}B_3 = (B_3^{-1}B_2)^{-1}=P_{23}^{-1}\)

    \(P_{31} = B_1^{-1}B_3 = (B_3^{-1}B_1)^{-1}=P_{13}^{-1}=(P_{23}P_{12})^{-1}=P_{12}^{-1}P_{23}^{-1}\)

    El resultado es cierto para espacios vectoriales generales y puede extenderse a muchas bases.

    This page titled 23.3: Cambio de Bases is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Dirk Colbry via source content that was edited to the style and standards of the LibreTexts platform.