Question

Given two bases

B={1−x,2+x,3−x+x2}

and

C={1,2+x,1+x−x2}

of P2, the vector space of polynomials of degree ≤2,

(i) find p(x)∈P2 whose coordinates with respect to B is [p(x)]B=⎡⎣⎢1 −1 3⎤⎦⎥,

(ii) find the transition (change of coordinates) matrix CMB∈R3×3 from B to C,

(iii) calculate the coordinates [p(x)]C∈R3 of p(x)∈P2 with respect to C.

Answer 1

Solution:

(i):

The coordinates of p(x)$\in$ P₂ with respect to the basis B is $[p(x)]_B=\begin{bmatrix} 1 \\-1\\3\end{bmatrix}$

$\therefore p(x)=1(1-x)-1(2+x)+3(3-x+x^2) \\=(1-2+9).1+(-1-1-3)x+3.x^2 \\=8-5x+3x^2$

(ii):

Let $1-x=a.1+b(2+x)+c(1+x-x^2)$

$\Rightarrow 1-x=(a+2b+c)+(b+c)x-cx^2$

On comparing, we get,

$a+2b+c=1 \\b+c=-1 \\-c=0 \\\Rightarrow c=0,b=-1,a=3$

So, $1-x=3(1)-1(2+x)+0.(1+x-x^2)\ ...(i)$

Also, $2+x=0(1)+1(2+x)+0.(1+x-x^2)\ ...(ii)$

Also, $3-x+x^2=4(1)+0(2+x)+(-1).(1+x-x^2)\ ...(iii)$

From (i), (ii), (iii), we get the transition matrix from B to C as:

$_CM_B=\begin{bmatrix} 3&0&4 \\-1&1&0\\0&0&-1\end{bmatrix}$

(iii):

From part (i), we have $p(x)=8-5x+3x^2$

Let $8-5x+3x^2=p(1)+q(2+x)+r(1+x-x^2)$

$\Rightarrow 8-5x+3x^2=(p+2q+r)+(q+r)x-rx^2$

On comparing, we get,

$p+2q+r=8 \\q+r=-5 \\-r=3 \\\Rightarrow r=-3,q=-2,p=15$

So, we get, $p(x)=15(1)-2(2+x)-3(1+x-x^2)$

So, the coordinates of p(x) w.r.t basis C is:

$[p(x)]_C=\begin{bmatrix} 15 \\-2\\-3\end{bmatrix}$