Question 2.1: In R2, fix the following points O = (1, 0), A1 = (1, 2), A2 ...
In R², fix the following points
O = (1, 0), A_{1} = (1, 2), A_{2} = (0, 1), and
O^{′}= (−1,−1), B_{1} = (0, 0), B_{2} = (2, 3).
Construct the vectorized spaces Σ(O ; A_{1} , A_{2}) and Σ(O^{′}; B_{1} , B_{2}), and then use them to justify the content of (2.4.2).
Suppose
\overrightarrow{a_{1} }= \overrightarrow{OA_{1} }=\left(1,2\right) -\left(1,0\right)=\left(0,2\right),
\overrightarrow{a_{2} }= \overrightarrow{OA_{2} }=\left(0,1\right) -\left(1,0\right)=\left(-1,1\right);
\overrightarrow{b_{1} } =\overrightarrow{O^{′}B_{1} }=\left(0,0\right)-\left(-1,-1\right)=\left(1,1\right),
\overrightarrow{b_{2} } =\overrightarrow{O^{′}B_{2} }=\left(2,3\right)-\left(-1,-1\right)=\left(3,4\right),
and let
B=\left\{\overrightarrow{a_{1} }, \overrightarrow{a_{2} }\right\} , B^{′}=\left\{\overrightarrow{b_{1} }, \overrightarrow{b_{2} }\right\}
be bases of \sum{\left(O ; A_{1} , A_{2}\right) } and \sum{\left(O^{′} ; B_{1} , B_{2}\right) } respectively (see Fig. 2.20).
To compute A^{B^{′} }_{B} :
\overrightarrow{b_{1} }=\alpha _{11}\overrightarrow{a_{1} }+ \alpha _{12}\overrightarrow{a_{2} }
\Rightarrow \left(1,1\right)= \alpha _{11}\left(0,2\right)+ \alpha _{12}\left(-1,-1\right)=\left( -\alpha _{12} ,2\alpha _{11} +\alpha _{12}\right)
\Rightarrow \left\{\begin{matrix} \alpha_{12}=-1 \\ 2\alpha_{11}+\alpha_{12}=1 \end{matrix} \right.
\Rightarrow \left\{\begin{matrix} \alpha_{12}=-1 \\ \alpha_{11}=1 \end{matrix} \right.
\Rightarrow \left[\overrightarrow{b_{1} }\right] _{B}=(1,−1).
Similarly,
\overrightarrow{b_{2} }=\alpha _{21}\overrightarrow{a_{1} }+ \alpha _{22}\overrightarrow{a_{2} }
\Rightarrow \left(3,4\right)=\overrightarrow{b_{2} }= \alpha _{21}\left(0,2\right)+ \alpha _{22}\left(-1,1\right)=\left( -\alpha _{22} ,2\alpha _{21} +\alpha _{22}\right)
\Rightarrow \left\{\begin{matrix} \alpha_{22}=-3 \\ 2\alpha_{21}+\alpha_{22}=4 \end{matrix} \right.
\Rightarrow \left\{\begin{matrix} \alpha_{21}=\frac{7}{2} \\ \alpha_{22}=-3 \end{matrix} \right.
\Rightarrow \left[\overrightarrow{b_{2} }\right] _{B}=\left(\frac{7}{2},-3\right).
Putting together, then
A^{B^{′} }_{B}= \left[\begin{matrix} \left[\overrightarrow{b_{1} }\right] _{B} \\ \left[\overrightarrow{b_{2} }\right] _{B} \end{matrix} \right]= \begin{bmatrix} 1 & -1 \\ \frac{7}{2} & -3 \end{bmatrix}.
To compute A^{B }_{B^{′}} :
\overrightarrow{a_{1} }=\beta _{11}\overrightarrow{b_{1} }+\beta _{12}\overrightarrow{b_{2} }
\Rightarrow \left(0,2\right)= \beta _{11}\left(1,1\right)+ \beta _{12}\left(3,4\right)=\left(\beta _{11}+3 \beta _{12} , \beta _{11}+4 \beta _{12}\right)
\Rightarrow \left\{\begin{matrix} \beta _{11}+3 \beta _{12}=0 \\ \beta _{11}+4 \beta _{12}=2 \end{matrix} \right.
\Rightarrow \left\{\begin{matrix} \beta _{11}=-6 \\ \beta _{12}=2 \end{matrix} \right.
\Rightarrow \left[\overrightarrow{a_{1} }\right] _{B}=(-6,2).
Also,
\overrightarrow{a_{2} }=\beta _{21}\overrightarrow{b_{1} }+\beta _{22}\overrightarrow{b_{2} }
\Rightarrow \left(-1,1\right)= \beta _{21}\left(1,1\right)+ \beta _{22}\left(3,4\right)=\left(\beta _{21}+3 \beta _{22} , \beta _{21}+4 \beta _{22}\right)
\Rightarrow \left\{\begin{matrix} \beta _{21}+3 \beta _{22}=-1 \\ \beta _{21}+4 \beta _{22}=1 \end{matrix} \right.
\Rightarrow \left\{\begin{matrix} \beta _{21}=-7 \\ \beta _{22}=2 \end{matrix} \right.
\Rightarrow \left[\overrightarrow{a_{2} }\right] _{B}=(-7,2).
Hence,
A^{B }_{B^{′}}= \begin{bmatrix} -6 & 2 \\ -7 & 2 \end{bmatrix}.
Through actual computation, one gets
A^{B^{′} }_{B} A^{B }_{B^{′}}= \begin{bmatrix} 1 & -1 \\ \frac{7}{2} & -3 \end{bmatrix} \begin{bmatrix} -6 & 2 \\ -7 & 2 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=I_{2}, and
A^{B }_{B^{′}} A^{B^{′} }_{B} = \begin{bmatrix} -6 & 2 \\ -7 & 2 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ \frac{7}{2} & -3 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}=I_{2},
which show that A^{B^{′} }_{B} and A^{B }_{B^{′}} are, in fact, invertible to each other. By the way,
det A^{B^{′} }_{B} = \left|\begin{matrix} 1 & -1 \\ \frac{7}{2} & -3 \end{matrix} \right| = -3+\frac{7}{2}=\frac{1}{2},
det A^{B }_{B^{′}} = \left|\begin{matrix} -6 & 2 \\ -7 & 2 \end{matrix} \right| = -12+14=2=\frac{1}{det A^{B^{′} }_{B}}
and
\left(A^{B^{′} }_{B}\right)^{-1} = \frac{1}{det A^{B^{′} }_{B}} \begin{bmatrix} -3 & 1 \\ -\frac{7}{2} & 1 \end{bmatrix} = 2 \begin{bmatrix} -3 & 1 \\ -\frac{7}{2} & 1 \end{bmatrix}= \begin{bmatrix} -6 & 2 \\ -7 & 2 \end{bmatrix}=A^{B }_{B^{′}}
as expected from (2.4.2).
Finally,
\overrightarrow{OO^{′} }= \left(-1,-1\right) -\left(1,0\right) =\left(-2,-1\right)
= \alpha _{1} \overrightarrow{a_{1} }+\alpha _{2} \overrightarrow{a_{2} }= \alpha _{1}\left(0,2\right) +\alpha _{2} \left(-1,1\right)=\left(-\alpha _{1},2\alpha _{1}+\alpha _{2}\right)
\Rightarrow\left\{\begin{matrix} \alpha _{2}=2 \\2\alpha _{1}+\alpha _{2}=-1 \end{matrix} \right.
\Rightarrow\left\{\begin{matrix} \alpha _{1}=-\frac{3}{2} \\ \alpha _{2}=2 \end{matrix} \right.
\Rightarrow \left[O^{′}\right]_{B}=\left(-\frac{3}{2} ,2\right) .
While,
\overrightarrow{O^{′}O }=-\overrightarrow{OO^{′} }=\left(2,1\right)
= \beta _{1} \overrightarrow{b_{1} }+ \beta _{2} \overrightarrow{b_{2} }= \beta _{1}\left(1,1\right) + \beta _{2} \left(3,4\right)=\left( \beta _{1}+3\beta _{2},\beta _{1}+4\beta _{2}\right)
\Rightarrow\left\{\begin{matrix} \beta _{1}+3\beta _{2}=2 \\\beta _{1}+4\beta _{2}=1 \end{matrix} \right.
\Rightarrow\left\{\begin{matrix} \beta _{1}=5 \\ \beta _{2}=-1 \end{matrix} \right.
\Rightarrow \left[O\right]_{B^{′}}=\left(5,-1\right) .
Now, by actual computation, we do have
– \left[O^{′}\right]_{B} A^{B }_{B^{′}}=-\left(- \frac{3}{2} 2\right) \begin{bmatrix} -6 & 2 \\ -7 & 2 \end{bmatrix}= -\left(9-14,-3+4\right)
=-\left(-5,1\right) =\left(5,-1\right) =\left[O\right]_{B^{′}}.
The wanted formulas of changes of coordinates are
\left[P\right]_{B} = \left(- \frac{3}{2} 2\right)+\left[P\right]_{B^{′}} \begin{bmatrix} 1 & -1 \\ \frac{7}{2} & -3 \end{bmatrix} and
\left[P\right]_{B^{′}} = \left( 5 -1\right)+\left[P\right]_{B} \begin{bmatrix} -6 & 2 \\ -7 & 2 \end{bmatrix}.
For example, if \left[P\right]_{B^{′}} = \left( 5,2\right) , i.e. \overrightarrow{O^{′}P } = 5 \overrightarrow{b_{1} }+2\overrightarrow{b_{2} }, then
\left[P\right]_{B} = \left(- \frac{3}{2} 2\right)+\left( 5 2\right) \begin{bmatrix} 1 & -1 \\ \frac{7}{2} & -3 \end{bmatrix}
= \left(- \frac{3}{2},2\right)+\left(12,-11\right)= \left(\frac{21}{2},-9\right)
which means \overrightarrow{OP } = \frac{21}{2} \overrightarrow{a_{1} }-9\overrightarrow{a_{2} }.
