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UPSC 2016 Maths Optional Paper 1 Q2a-ii — Step-by-Step Solution

6 marks · Section A

Matrix of a linear transformation · Linear Algebra · asked 10× in 14 yrs · Read the full method →

Question

If T:P2(x)→P3(x)T:P_2(x)\to P_3(x) is such that T(f(x))=f(x)+5∫0xf(t) dtT(f(x))=f(x)+5\displaystyle\int_0^x f(t)\,dt, then choosing {1,1+x,1−x2}\{1,1+x,1-x^2\} and {1,x,x2,x3}\{1,x,x^2,x^3\} as bases of P2(x)P_2(x) and P3(x)P_3(x) respectively, find the matrix of TT.

Technique

Apply the operator to each domain basis polynomial, then write its coordinates in the codomain basis; here the codomain basis is standard, so coordinates are coefficients.

Solution

Step 1 — Apply TT to each domain basis polynomial

T(f)=f+5∫0xf dtT(f)=f+5\int_0^x f\,dt.

f=1f=1: ∫0x1 dt=x\int_0^x 1\,dt=x, so T(1)=1+5xT(1)=1+5x.

f=1+xf=1+x: ∫0x(1+t) dt=x+x22\int_0^x(1+t)\,dt=x+\tfrac{x^2}{2}, so T(1+x)=(1+x)+5(x+x22)=1+6x+52x2T(1+x)=(1+x)+5\big(x+\tfrac{x^2}{2}\big)=1+6x+\tfrac52 x^2.

f=1−x2f=1-x^2: ∫0x(1−t2) dt=x−x33\int_0^x(1-t^2)\,dt=x-\tfrac{x^3}{3}, so T(1−x2)=(1−x2)+5(x−x33)=1+5x−x2−53x3T(1-x^2)=(1-x^2)+5\big(x-\tfrac{x^3}{3}\big)=1+5x-x^2-\tfrac53 x^3.

Step 2 — Coordinates in the codomain basis {1,x,x2,x3}\{1,x,x^2,x^3\}

Since the codomain basis is the standard one, coordinates are just the coefficients:

T(1)→(1, 5, 0, 0)T(1+x)→(1, 6, 52, 0)T(1−x2)→(1, 5, −1, −53).\begin{aligned}T(1)&\to(1,\,5,\,0,\,0)\\ T(1+x)&\to(1,\,6,\,\tfrac52,\,0)\\ T(1-x^2)&\to(1,\,5,\,-1,\,-\tfrac53).\end{aligned}

Step 3 — Assemble the matrix (columns = images)

Answer

  [T]=[111565052−100−53](4×3).  \boxed{\;[T]=\begin{bmatrix}1 & 1 & 1\\ 5 & 6 & 5\\ 0 & \tfrac52 & -1\\ 0 & 0 & -\tfrac53\end{bmatrix}\quad(4\times3).\;}
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