In R2, fix the following points O = (1, 0), A1 = (1, 2), A2 = (0, 1), and O′ = (−1,−1), B1 = (0, 0), B2 = (2, 3). Construct the vectorized spaces Σ(O;A1,A2) and Σ(O′;B1,B2), and then use them to justify the content of (2.4.2).

Question

In R², fix the following points

O = (1, 0), [latex]A_{1}[/latex] = (1, 2), [latex]A_{2}[/latex] = (0, 1), and

[latex]O^{′}=[/latex] (−1,−1), [latex]B_{1}[/latex] = (0, 0), [latex]B_{2}[/latex] = (2, 3).

Construct the vectorized spaces Σ(O ; [latex]A_{1}[/latex] , [latex]A_{2}[/latex]) and Σ([latex]O^{′}[/latex]; [latex]B_{1}[/latex] , [latex]B_{2}[/latex]), and then use them to justify the content of (2.4.2).

Accepted Answer

Suppose

[latex] \overrightarrow{a_{1} }= \overrightarrow{OA_{1} }=\left(1,2 ight) -\left(1,0 ight)=\left(0,2 ight),[/latex]

[latex] \overrightarrow{a_{2} }= \overrightarrow{OA_{2} }=\left(0,1 ight) -\left(1,0 ight)=\left(-1,1 ight);[/latex]

[latex] \overrightarrow{b_{1} } =\overrightarrow{O^{′}B_{1} }=\left(0,0 ight)-\left(-1,-1 ight)=\left(1,1 ight),[/latex]

[latex] \overrightarrow{b_{2} } =\overrightarrow{O^{′}B_{2} }=\left(2,3 ight)-\left(-1,-1 ight)=\left(3,4 ight),[/latex]

and let

[latex] B=\left\{\overrightarrow{a_{1} }, \overrightarrow{a_{2} } ight\} , B^{′}=\left\{\overrightarrow{b_{1} }, \overrightarrow{b_{2} } ight\}[/latex]

be bases of [latex] \sum{\left(O ; A_{1} , A_{2} ight) } [/latex] and [latex] \sum{\left(O^{′} ; B_{1} , B_{2} ight) }[/latex] respectively (see Fig. 2.20).

To compute [latex]A^{B^{′} }_{B}[/latex] :

[latex] \overrightarrow{b_{1} }=\alpha _{11}\overrightarrow{a_{1} }+ \alpha _{12}\overrightarrow{a_{2} }[/latex]

[latex] \Rightarrow \left(1,1 ight)= \alpha _{11}\left(0,2 ight)+ \alpha _{12}\left(-1,-1 ight)=\left( -\alpha _{12} ,2\alpha _{11} +\alpha _{12} ight)[/latex]

[latex] \Rightarrow \left\{\begin{matrix} \alpha_{12}=-1 \ 2\alpha_{11}+\alpha_{12}=1 \end{matrix} ight.[/latex]

[latex] \Rightarrow \left\{\begin{matrix} \alpha_{12}=-1 \ \alpha_{11}=1 \end{matrix} ight.[/latex]

[latex] \Rightarrow \left[\overrightarrow{b_{1} } ight] _{B}=[/latex](1,−1).

Similarly,

[latex] \overrightarrow{b_{2} }=\alpha _{21}\overrightarrow{a_{1} }+ \alpha _{22}\overrightarrow{a_{2} }[/latex]

[latex] \Rightarrow \left(3,4 ight)=\overrightarrow{b_{2} }= \alpha _{21}\left(0,2 ight)+ \alpha _{22}\left(-1,1 ight)=\left( -\alpha _{22} ,2\alpha _{21} +\alpha _{22} ight)[/latex]

[latex] \Rightarrow \left\{\begin{matrix} \alpha_{22}=-3 \ 2\alpha_{21}+\alpha_{22}=4 \end{matrix} ight.[/latex]

[latex] \Rightarrow \left\{\begin{matrix} \alpha_{21}=\frac{7}{2} \ \alpha_{22}=-3 \end{matrix} ight.[/latex]

[latex] \Rightarrow \left[\overrightarrow{b_{2} } ight] _{B}=\left(\frac{7}{2},-3 ight).[/latex]

Putting together, then

[latex]A^{B^{′} }_{B}= \left[\begin{matrix} \left[\overrightarrow{b_{1} } ight] _{B} \ \left[\overrightarrow{b_{2} } ight] _{B} \end{matrix} ight]= \begin{bmatrix} 1 & -1 \ \frac{7}{2} & -3 \end{bmatrix}.[/latex]

To compute [latex]A^{B }_{B^{′}}[/latex] :

[latex] \overrightarrow{a_{1} }=\beta _{11}\overrightarrow{b_{1} }+\beta _{12}\overrightarrow{b_{2} }[/latex]

[latex] \Rightarrow \left(0,2 ight)= \beta _{11}\left(1,1 ight)+ \beta _{12}\left(3,4 ight)=\left(\beta _{11}+3 \beta _{12} , \beta _{11}+4 \beta _{12} ight)[/latex]

[latex] \Rightarrow \left\{\begin{matrix} \beta _{11}+3 \beta _{12}=0 \ \beta _{11}+4 \beta _{12}=2 \end{matrix} ight.[/latex]

[latex] \Rightarrow \left\{\begin{matrix} \beta _{11}=-6 \ \beta _{12}=2 \end{matrix} ight.[/latex]

[latex] \Rightarrow \left[\overrightarrow{a_{1} } ight] _{B}=[/latex](-6,2).

Also,

[latex] \overrightarrow{a_{2} }=\beta _{21}\overrightarrow{b_{1} }+\beta _{22}\overrightarrow{b_{2} }[/latex]

[latex] \Rightarrow \left(-1,1 ight)= \beta _{21}\left(1,1 ight)+ \beta _{22}\left(3,4 ight)=\left(\beta _{21}+3 \beta _{22} , \beta _{21}+4 \beta _{22} ight)[/latex]

[latex] \Rightarrow \left\{\begin{matrix} \beta _{21}+3 \beta _{22}=-1 \ \beta _{21}+4 \beta _{22}=1 \end{matrix} ight.[/latex]

[latex] \Rightarrow \left\{\begin{matrix} \beta _{21}=-7 \ \beta _{22}=2 \end{matrix} ight.[/latex]

[latex] \Rightarrow \left[\overrightarrow{a_{2} } ight] _{B}=[/latex](-7,2).

Hence,

[latex]A^{B }_{B^{′}}= \begin{bmatrix} -6 & 2 \ -7 & 2 \end{bmatrix}.[/latex]

Through actual computation, one gets

[latex]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},[/latex] and

[latex]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},[/latex]

which show that [latex]A^{B^{′} }_{B} and A^{B }_{B^{′}}[/latex] are, in fact, invertible to each other. By the way,

[latex]det A^{B^{′} }_{B} = \left|\begin{matrix} 1 & -1 \ \frac{7}{2} & -3 \end{matrix} ight| = -3+\frac{7}{2}=\frac{1}{2},[/latex]

[latex]det A^{B }_{B^{′}} = \left|\begin{matrix} -6 & 2 \ -7 & 2 \end{matrix} ight| = -12+14=2=\frac{1}{det A^{B^{′} }_{B}}[/latex]

and

[latex]\left(A^{B^{′} }_{B} ight)^{-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^{′}}[/latex]

as expected from (2.4.2).

Finally,

[latex]\overrightarrow{OO^{′} }= \left(-1,-1 ight) -\left(1,0 ight) =\left(-2,-1 ight) [/latex]

[latex]= \alpha _{1} \overrightarrow{a_{1} }+\alpha _{2} \overrightarrow{a_{2} }= \alpha _{1}\left(0,2 ight) +\alpha _{2} \left(-1,1 ight)=\left(-\alpha _{1},2\alpha _{1}+\alpha _{2} ight)[/latex]

[latex]\Rightarrow\left\{\begin{matrix} \alpha _{2}=2 \2\alpha _{1}+\alpha _{2}=-1 \end{matrix} ight.[/latex]

[latex]\Rightarrow\left\{\begin{matrix} \alpha _{1}=-\frac{3}{2} \ \alpha _{2}=2 \end{matrix} ight.[/latex]

[latex] \Rightarrow \left[O^{′} ight]_{B}=\left(-\frac{3}{2} ,2 ight) .[/latex]

While,

[latex]\overrightarrow{O^{′}O }=-\overrightarrow{OO^{′} }=\left(2,1 ight)[/latex]

[latex]= \beta _{1} \overrightarrow{b_{1} }+ \beta _{2} \overrightarrow{b_{2} }= \beta _{1}\left(1,1 ight) + \beta _{2} \left(3,4 ight)=\left( \beta _{1}+3\beta _{2},\beta _{1}+4\beta _{2} ight)[/latex]

[latex]\Rightarrow\left\{\begin{matrix} \beta _{1}+3\beta _{2}=2 \\beta _{1}+4\beta _{2}=1 \end{matrix} ight.[/latex]

[latex]\Rightarrow\left\{\begin{matrix} \beta _{1}=5 \ \beta _{2}=-1 \end{matrix} ight.[/latex]

[latex] \Rightarrow \left[O ight]_{B^{′}}=\left(5,-1 ight) .[/latex]

Now, by actual computation, we do have

[latex]- \left[O^{′} ight]_{B} A^{B }_{B^{′}}=-\left(- \frac{3}{2} 2 ight) \begin{bmatrix} -6 & 2 \ -7 & 2 \end{bmatrix}= -\left(9-14,-3+4 ight) [/latex]

[latex]=-\left(-5,1 ight) =\left(5,-1 ight) =\left[O ight]_{B^{′}}.[/latex]

The wanted formulas of changes of coordinates are

[latex]\left[P ight]_{B} = \left(- \frac{3}{2} 2 ight)+\left[P ight]_{B^{′}} \begin{bmatrix} 1 & -1 \ \frac{7}{2} & -3 \end{bmatrix} [/latex] and

[latex]\left[P ight]_{B^{′}} = \left( 5 -1 ight)+\left[P ight]_{B} \begin{bmatrix} -6 & 2 \ -7 & 2 \end{bmatrix}. [/latex]

For example, if [latex]\left[P ight]_{B^{′}} = \left( 5,2 ight) , i.e. \overrightarrow{O^{′}P } = 5 \overrightarrow{b_{1} }+2\overrightarrow{b_{2} },[/latex] then

[latex]\left[P ight]_{B} = \left(- \frac{3}{2} 2 ight)+\left( 5 2 ight) \begin{bmatrix} 1 & -1 \ \frac{7}{2} & -3 \end{bmatrix}[/latex]

[latex]= \left(- \frac{3}{2},2 ight)+\left(12,-11 ight)= \left(\frac{21}{2},-9 ight)[/latex]

which means [latex] \overrightarrow{OP } = \frac{21}{2} \overrightarrow{a_{1} }-9\overrightarrow{a_{2} }.[/latex]

Question 2.1: In R2, fix the following points O = (1, 0), A1 = (1, 2), A2 ...

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