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UPSC 2024 Maths Optional Paper 1 Q1a — Step-by-Step Solution

10 marks · Section A

Bases and dimension; coordinates in a basis · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

Let $H$ be a subspace of $\mathbb{R}^4$ spanned by the vectors $v_1=(1,-2,5,-3)$, $v_2=(2,3,1,-4)$, $v_3=(3,8,-3,-5)$. Find a basis and dimension of $H$, and extend the basis of $H$ to a basis of $\mathbb{R}^4$.

Technique

Row-reduce the matrix formed by $v_1,v_2,v_3$ to find the rank; then adjoin standard basis vectors to extend.

Solution

Step 1 — Row reduce to find rank.

Form the matrix with $v_1,v_2,v_3$ as rows:

$R_2\to R_2-2R_1$: $(0,7,-9,2)$. $R_3\to R_3-3R_1$: $(0,14,-18,4)$. $R_3\to R_3-2R_2$: $(0,0,0,0)$.

The echelon form has two non-zero rows, so $\operatorname{rank}M=2$, hence $\dim H=2$. The two pivot rows correspond to $v_1$ and $v_2$, which are linearly independent, so $\{v_1,v_2\}$ is a basis of $H$.

Step 2 — Extend to a basis of $\mathbb{R}^4$.

Adjoin the standard vectors $e_3=(0,0,1,0)$ and $e_4=(0,0,0,1)$. Verify $\{v_1,v_2,e_3,e_4\}$ spans $\mathbb{R}^4$ by computing the $4\times4$ determinant:

Expand along rows 3 and 4: only the $(3,3)$ entry is non-zero (value $1$), and its cofactor reduces to

The full determinant is $7\ne 0$, so the four vectors are linearly independent and span $\mathbb{R}^4$.

Answer