← 2013 Paper 1

UPSC Maths 2013 Paper 1 Q2a-i — Solution

10 marks · Section A

Question

Let $P_n$ denote the vector space of all real polynomials of degree atmost $n$ and $T:P_2\to P_3$ be a linear transformation given by

Find the matrix of $T$ with respect to the bases $\{1,x,x^2\}$ and $\{1,x,1+x^2,1+x^3\}$ of $P_2$ and $P_3$ respectively. Also, find the null space of $T$.

Technique

Compute $T$ on each domain-basis element; expand each image in the codomain basis (matching $x^{k}$ coefficients); assemble columns. Nullspace via the integral-vanishing argument.

Solution

Step 1 — Apply $T$ to each domain-basis element

Step 2 — Express each image in the codomain basis $\{1,\,x,\,1+x^{2},\,1+x^{3}\}$

Let $e_1=1,\;e_2=x,\;e_3=1+x^{2},\;e_4=1+x^{3}$. For each image, find scalars $(a,b,c,d)$ with $a\,e_1+b\,e_2+c\,e_3+d\,e_4=\text{image}$:

$T(1)=x$: $a+bx+c(1+x^{2})+d(1+x^{3})=x$ forces $b=1$, $c=0$, $d=0$, and then $a+c+d=0\Rightarrow a=0$. Coordinates: $(0,\,1,\,0,\,0)$.

$T(x)=\tfrac{x^{2}}{2}$: matching coefficients of $x^{0},x,x^{2},x^{3}$: $a+c+d=0,\;b=0,\;c=\tfrac{1}{2},\;d=0$, so $a=-\tfrac{1}{2}$. Coordinates: $\bigl(-\tfrac{1}{2},\,0,\,\tfrac{1}{2},\,0\bigr)$.

$T(x^{2})=\tfrac{x^{3}}{3}$: $a+c+d=0,\;b=0,\;c=0,\;d=\tfrac{1}{3}$, so $a=-\tfrac{1}{3}$. Coordinates: $\bigl(-\tfrac{1}{3},\,0,\,0,\,\tfrac{1}{3}\bigr)$.

Step 3 — Assemble the matrix (images as columns)

Step 4 — Null space of $T$

$p(x)\in\ker T\iff\int_0^{x}p(t)\,dt=0\;\forall x$. Differentiating both sides (using FTC), $p(x)=0$ identically. So

Answer