UPSC Maths 2025 Paper 1 Q1a — Solution

10 marks · Section A

Question

Can the set $\{(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, -1, 0)\}$ be extended to form a basis of the vector space $\mathbb{R}^4$ ? Justify your answer.

Technique

Test the set for linear independence (via rank), then invoke the fact that any linearly independent set in a finite-dimensional space extends to a basis.

Solution

A set can be extended to a basis of $\mathbb{R}^4$ if and only if it is linearly independent (any linearly independent subset of a finite-dimensional space extends to a basis, by the Replacement/Steinitz theorem; a linearly dependent set cannot belong to any basis). So everything hinges on independence.

Step 1 — Test independence of the three given vectors.

Form the matrix with the vectors as rows: $M=\begin{pmatrix}0 & 0 & 0 & 3\\ 1 & 1 & 0 & 0\\ 0 & 1 & -1 & 0\end{pmatrix}.$

The second row has a leading $1$ in column 1, the third row a leading $1$ in column 2, and the first row a leading $3$ in column 4. These three pivots lie in distinct columns, so the rows are linearly independent. Equivalently, suppose $a(0,0,0,3)+b(1,1,0,0)+c(0,1,-1,0)=(0,0,0,0).$ Componentwise: $b=0$ (col 1), $b+c=0$ (col 2) $\Rightarrow c=0$ , $-c=0$ (col 3, consistent), $3a=0$ (col 4) $\Rightarrow a=0$ . Hence $a=b=c=0$ and the set is linearly independent, with $\operatorname{rank} M = 3$ .

Step 2 — Conclusion on extendability.

Since the set is linearly independent and $\dim\mathbb{R}^4 = 4 > 3$ , it can be extended to a basis of $\mathbb{R}^4$ by adjoining one more vector independent of these three.

Step 3 — Exhibit an explicit extension.

Try $e_1=(1,0,0,0)$ . Form the $4\times4$ matrix and compute its determinant: $\det\begin{pmatrix}0 & 0 & 0 & 3\\ 1 & 1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 1 & 0 & 0 & 0\end{pmatrix} = 3 \neq 0.$ Expanding along the first row, only the $(1,4)$ entry $3$ survives, and its minor is nonzero, so the determinant is nonzero. The four vectors are therefore independent and form a basis.

Answer

$\boxed{\;\text{Yes. The set is linearly independent, so it extends to a basis, e.g. } \{(0,0,0,3),(1,1,0,0),(0,1,-1,0),(1,0,0,0)\}.\;}$