23.3: Change of Basis
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- 68052
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:
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.
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.
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.