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# Find the change-of-basis matrix$P_{C \leftarrow B}$from the given ordered basis B to the given ordered basis C of the vector space V.$\left. \begin{array} { l } { V = P _ { 2 } ( \mathbb { R } ) ; } \\ { B = \left \{ 2 + x ^ { 2 } , - 1 - 6 x + 8 x ^ { 2 } , - 7 - 3 x - 9 x ^ { 2 } \right\} }; \\ { C = \left \{1 + x , - x + x ^ { 2 } , 1 + 2 x ^ { 2 } \right\} }. \end{array} \right.$

Solution

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Consider the ordered basis $B=\left\{2+x^2,-1-6x+8x^2,-7-3x-9x^2\right\}$ and $C=\left\{1+x,-x+x^2,1+2x^2\right\}$ for $P_3$.

In order to find $\left[2+x^2\right]_C$, $\left[-1-6x+8x^2\right]_C$ and $\left[-7-3x-9x^2\right]_C$ we must solve the equations

$a_1(1+x)+a_2(-x+x^2)+a_3(1+2x^2)=2+x^2$, $b_1(1+x)+b_2(-x+x^2)+b_3(1+2x^2)=-1-6x+8x^2$ and $c_1(1+x)+c_2(-x+x^2)+c_3(1+2x^2)=-7-3x-9x^2$.

Since $a_1(1+x)+a_2(-x+x^2)+a_3(1+2x^2)=a_1+a_1x-a_2x+a_2x^2+a_3+2a_3x^2=(a_1+a_3)+(a_1-a_2)x+(a_2+2a_3)x^2$ the first equation requires that $a_1+a_3=2$, $a_1-a_2=0$ and $a_2+2a_3=1$.

Then we have that $(-2)(a_1+a_3)+a_1-a_2+a_2+2a_3=(-2)\cdot 2+0+1$ which gives us that $-a_1=-3$ and then $a_1=3$.

Since $a_1=3$, $a_1+a_3=2$ and $a_1-a_2=0$ we have that $3+a_3=2$ and $3-a_2=0$ and then $a_2=3$ and $a_3=-1$.

Since $b_1(1+x)+b_2(-x+x^2)+b_3(1+2x^2)=b_1+b_1x-b_2x+b_2x^2+b_3+2b_3x^2=(b_1+b_3)+(b_1-b_2)x+(b_2+2b_3)x^2$ the second equation requires that $b_1+b_3=-1$, $b_1-b_2=-6$ and $b_2+2b_3=8$.

Then we have that $(-2)(b_1+b_3)+b_1-b_2+b_2+2b_3=(-2)\cdot(-1)-6+8$ which gives us that $-b_1=4$ and then $b_1=-4$.

Since $b_1=-4$, $b_1+b_3=-1$ and $b_1-b_2=-6$ we have that $-4+b_3=-1$ and $-4-b_2=-6$ and then $b_2=2$ and $b_3=3$.

Since $c_1(1+x)+c_2(-x+x^2)+c_3(1+2x^2)=c_1+c_1x-c_2x+c_2x^2+c_3+2c_3x^2=(c_1+c_3)+(c_1-c_2)x+(c_2+2c_3)x^2$ the third equation requires that $c_1+c_3=-7$, $c_1-c_2=-3$ and $c_2+2c_3=-9$.

Then we have that $(-2)(c_1+c_3)+c_1-c_2+c_2+2c_3=(-2)\cdot(-7)-3-9$ which gives us that $-c_1=2$ and then $c_1=-2$.

Since $c_1=-2$, $c_1+c_3=-7$ and $c_1-c_2=-3$ we have that $-2+c_3=-7$ and $-2-c_2=-3$ and then $c_2=1$ and $c_3=-5$.

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