UPSC 2024 Maths Optional Paper 1 Q1b — Step-by-Step Solution

10 marks · Section A

Rank and nullity; rank-nullity theorem · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

Let $T:\mathbb{R}^3\to\mathbb{R}^3$ be a linear operator and $B=\{v_1,v_2,v_3\}$ be a basis of $\mathbb{R}^3$ over $\mathbb{R}$ . Suppose that $Tv_1=(1,1,0)$ , $Tv_2=(1,0,-1)$ , $Tv_3=(2,1,-1)$ . Find a basis for the range space and null space of $T$ .

Technique

Range = span of the images of basis vectors; rank–nullity gives the null-space dimension; coordinate equations in basis $B$ find the kernel vector.

Solution

Step 1 — Range space.

Since $\{v_1,v_2,v_3\}$ is a basis of the domain, $\operatorname{Range}(T)=\operatorname{span}\{Tv_1,Tv_2,Tv_3\}$ .

Observe $Tv_3=(2,1,-1)=(1,1,0)+(1,0,-1)=Tv_1+Tv_2$ . So $Tv_3$ is dependent on $Tv_1,Tv_2$ . Meanwhile $Tv_1=(1,1,0)$ and $Tv_2=(1,0,-1)$ are linearly independent (neither is a scalar multiple of the other).

Therefore $\operatorname{Range}(T)=\operatorname{span}\{Tv_1,Tv_2\}$ , with $\dim\operatorname{Range}(T)=2$ .

Step 2 — Null space.

By rank–nullity, $\dim\ker T=3-2=1$ .

A vector $x=\alpha v_1+\beta v_2+\gamma v_3\in\ker T$ iff

\alpha(1,1,0)+\beta(1,0,-1)+\gamma(2,1,-1)=(0,0,0).

Componentwise:

\alpha+\beta+2\gamma=0,\quad \alpha+\gamma=0,\quad -\beta-\gamma=0.

From equation 2: $\alpha=-\gamma$ . From equation 3: $\beta=-\gamma$ . These satisfy equation 1 automatically. Setting $\gamma=1$ : $(\alpha,\beta,\gamma)=(-1,-1,1)$ , so the null vector in the domain is $-v_1-v_2+v_3$ .

Verification. $T(-v_1-v_2+v_3)=-Tv_1-Tv_2+Tv_3=-(1,1,0)-(1,0,-1)+(2,1,-1)=(0,0,0)$ ✓.

Answer

\boxed{\;\text{Range basis: }\{(1,1,0),\,(1,0,-1)\},\quad \dim\operatorname{Range}(T)=2.\;}

\boxed{\;\text{Null space basis: }\{v_3-v_1-v_2\},\quad \dim\ker T=1.\;}