23.3: Change of Basis

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%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');

Now consider the following two bases in \(R^2\):

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

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

The transformation from the “standard basis” to \(B_1\) and \(B_2\) can be defined as the column vectors \(P_1\) and \(P_2\) as follows:

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)

Do This

Find the transition matrix \(T\) that will take points in the \(B_1\) coordinate representation and put them into \(B_2\) coordinates.

NOTE this is analogous to the robot kinematics problem. We want to represent points in a different coordinate system.

# Put your answer to the above question here.

from answercheck import checkanswer

checkanswer.matrix(T,'dcc03ddff982e29eea6dd52ec9088986')

Question

Given \(u_{B_1} = \left[ \begin{matrix} 2 \\ 1 \end{matrix} \right]\) (a point named \(u\) in the \(B_1\) coordinate system) and your calculated transition matrix \(T\), what is the same point expressed in the \(B_2\) basis (i.e. what is \(u_{B2}\))? Store your answer in a variable named ub2 for checking.

ub1 = np.matrix([[2],[1]])
sym.Matrix(ub1)

##Put your code here

from answercheck import checkanswer

checkanswer.vector(ub2,'9a5fe29254c07cf59ebdffcaba679917')

There are three bases \(B_1\), \(B_2\), and \(B_3\). We have the transition matrix \(P_{12}\) from \(B_1\) to \(B_2\) and the transition matrix \(P_{23}\) from \(B_2\) to \(B_3\). In \(R^n\), we can compute the transition matrix as \(P_{12}=B_2^{-1}B_1,\quad P_{23}=B_3^{-1}B_2\).

Then we can find all other transition matrices.

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

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

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

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

The result is true for general vector spaces and can be extended to many bases.

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