← 2016 Paper 1

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 13 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)+50xf(t)dtT(f(x))=f(x)+5\displaystyle\int_0^x f(t)\,dt, then choosing {1,1+x,1x2}\{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+50xfdtT(f)=f+5\int_0^x f\,dt.

f=1f=1: 0x1dt=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=1x2f=1-x^2: 0x(1t2)dt=xx33\int_0^x(1-t^2)\,dt=x-\tfrac{x^3}{3}, so T(1x2)=(1x2)+5(xx33)=1+5xx253x3T(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(1x2)(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]=[11156505210053](4×3).  \boxed{\;[T]=\begin{bmatrix}1 & 1 & 1\\ 5 & 6 & 5\\ 0 & \tfrac52 & -1\\ 0 & 0 & -\tfrac53\end{bmatrix}\quad(4\times3).\;}
We post more of this — worked solutions, CSAT trap breakdowns, guide chapters — a few times a week on Telegram. Free, no sign-in. Join

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.