5.5 Coordinates and the Transition Matrix
Definition 5.5.1. Let \(B=\{x_1,x_2,\dots ,x_n\}\) be a basis for a vector space \(V\) and let \(X\in V\). If \(\alpha _1, \alpha _2,\dots , \alpha _n\) are scalars such that \(X=\alpha _1x_1+\alpha _2x_2+\cdots +\alpha _nx_n\), then \(\alpha _1, \alpha _2,\dots , \alpha _n\) are called the coordinates of \(X\) related to the basis \(B\).
Example 5.5.2. Find the coordinates of the vector \(V=(-1,3,5)\) relative to the basis \(B=\{(1,1,0), (0,1,1), (1,1,1)\}\) for \(\mathbb {R}^3\).
Solution. Let \(a,b,c\in \mathbb {R}\) such that \((-1,3,5)=a(1,1,0)+b(0,1,1)+c(1,1,1)\). Then we obtain the system \begin {align*} a+c &=-1\\ a+b+c &=3\\ b+c &=5 \end {align*}
from this we obtain the coordinates \(a=-2, b=4\) and \(c=1\).
Definition 5.5.3. Suppose that \(B=\{x_1,x_2,\dots ,x_n\}\) and \(B'=\{y_1,y_2,\dots ,y_n\}\) are bases for a vector space \(V\).
Let \(v\in V\) be a vector such that
\[V=\sum ^n_{i=1}\alpha _ix_i\qquad \text {and}\qquad v=\sum ^n_{i=1}\beta _iy_i\]
where \(\alpha =(\alpha _1,\alpha _2,\dots ,\alpha _n)\) and \(\beta =(\beta _1,\beta _2,\dots ,\beta _n)\) are the coordinate vectors for \(V\) with respect to the basis \(B\) and \(B'\) respectively. Then \(n\)
by \(n\) matrix \(A\) such that \(A\alpha ^t=\beta ^t\) is called the transition matrix from \(B\) to \(B'\).
Example 5.5.4. Let \(B=\{(1,1,0), (0,1,1), (1,1,1)\}\) and \(B'=\{(2,0,3), (1,1,2), (0,2,7)\}\) be two bases for \(\mathbb {R}^3\).
Find the transition matrix from \(B\) to \(B'\).
Solution. The transition matrix from \(B\) to \(B'\) is the matrix whose columns are the coordinate
vectors of the vectors of \(B\) relative to the vectors in \(B'\). Thus if \((1,1,0)=a(2,0,3)+b(1,1,2)+c(0,2,7)\) then
\[a=-\frac {1}{8},\hspace {0.5cm}b=\frac {5}{4},\hspace {0.5cm} c=-\frac {1}{8}\]
are coordinates of \((1,1,0)\) relative to \(B'\).
Similarly, the coordinates for \((0,1,1)\) relative to \(B'\) are
\[a=\frac {-5}{12},\hspace {0.5cm} b=\frac {5}{6},\hspace {0,5cm}c=\frac {1}{12}\]
and the coordinates of \((1,1,1)\) relative to \(B'\) are
\[a=\frac {-1}{6},\hspace {0.5cm} b=\frac {4}{3},\hspace {0.5cm} c=\frac {-1}{6}.\]
Therefore, the transition matrix from \(B\) to \(B'\) is
\[ A= \begin {pmatrix} \frac {-1}{8}&\frac {-5}{12}&\frac {-1}{6}\\\\ \frac {5}{4}&\frac {5}{6}&\frac {4}{3}\\\\ \frac {-1}{8}&\frac {1}{12}&\frac {-1}{6}\\ \end {pmatrix} \]
Note that \(A^{-1}= \begin {pmatrix} 12&4&20\\ -2&0&2\\ -10&-3&-20\\ \end {pmatrix} \) is the transition matrix from \(B'\) to \(B\).
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