UPSC Maths 2025 Paper 1 Q1b — Solution

10 marks · Section A

Question

Find the range, rank, kernel and nullity of the linear transformation $T : \mathbb{R}^4 \to \mathbb{R}^3$ given by $T(x, y, z, w) = (x - w, y + z, z - w)$ .

Technique

Write the matrix of $T$ ; then range = column space, rank = rank of the matrix, kernel = null space, with the Rank–Nullity theorem as a check.

Solution

Step 1 — Matrix of $T$ (standard bases).

$A=\begin{pmatrix}1 & 0 & 0 & -1\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & -1\end{pmatrix},\qquad T(\mathbf v)=A\mathbf v.$ The columns are the images of $e_1,e_2,e_3,e_4$ : $T(e_1)=(1,0,0)$ , $T(e_2)=(0,1,0)$ , $T(e_3)=(0,1,1)$ , $T(e_4)=(-1,0,-1)$ .

Step 2 — Rank. $A$ is already in echelon form, with pivots in columns 1, 2, 3 (the rows $(1,0,0,-1)$ , $(0,1,1,0)$ , $(0,0,1,-1)$ are independent). Hence $\operatorname{rank}(T)=3.$

Step 3 — Range. Since $\operatorname{rank}=3=\dim\mathbb{R}^3$ , the range is all of $\mathbb{R}^3$ : $\operatorname{Range}(T)=\mathbb{R}^3,\qquad T \text{ is onto.}$ A basis is $\{(1,0,0),(0,1,0),(0,0,1)\}$ .

Step 4 — Kernel. Solve $A\mathbf v=\mathbf 0$ : $x-w=0,\qquad y+z=0,\qquad z-w=0.$ Let $w=t$ (free). Then $x=t$ , $z=t$ , $y=-z=-t$ . So $\ker(T)=\{\,t(1,-1,1,1):t\in\mathbb{R}\,\}=\operatorname{span}\{(1,-1,1,1)\}.$

Step 5 — Nullity. $\dim\ker(T)=1$ , so $\operatorname{nullity}(T)=1$ .

Check (Rank–Nullity): $\operatorname{rank}+\operatorname{nullity}=3+1=4=\dim\mathbb{R}^4$ . ✓

Answer

$\boxed{\;\operatorname{Range}(T)=\mathbb{R}^3,\quad \operatorname{rank}=3,\quad \ker(T)=\operatorname{span}\{(1,-1,1,1)\},\quad \operatorname{nullity}=1.\;}$