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.
(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)\n\\\\=(1-2+9).1+(-1-1-3)x+3.x^2\n\\\\=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\n\\\\b+c=-1\n\\\\-c=0\n\\\\\\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\n\\\\q+r=-5\n\\\\-r=3\n\\\\\\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}"
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