← 2022 Paper 1

UPSC 2022 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

Prove that any set of $n$ linearly independent vectors in a vector space $V$ of dimension $n$ constitutes a basis for $V$.

Technique

Standard “augment-and-show-dependence” proof. The argument uses the theorem that in $\dim V=n$, $n+1$ vectors are linearly dependent — itself a consequence of the dimension being well-defined.

Solution

Setup. Let $V$ be a vector space with $\dim V=n$, and let $S=\{v_1,v_2,\dots,v_n\}\subset V$ be linearly independent. To prove $S$ is a basis, we must show $S$ also spans $V$.

Step 1 — $S$ is linearly independent (given)

By hypothesis.

Step 2 — $S$ spans $V$

Let $v\in V$ be arbitrary. Consider $S'=\{v,v_1,\dots,v_n\}$, a set of $n+1$ vectors.

Claim: $S'$ is linearly dependent.

Reason: In a vector space of dimension $n$, any set of $n+1$ (or more) vectors is linearly dependent. (Standard theorem: any linearly independent set has at most $\dim V$ elements.)

Therefore, there exist scalars $\alpha,\alpha_1,\dots,\alpha_n$, not all zero, such that

Sub-claim: $\alpha\ne 0$.

Reason: If $\alpha=0$, then $(\star)$ becomes $\sum\alpha_i v_i=0$, which by linear independence of $\{v_1,\dots,v_n\}$ forces all $\alpha_i=0$. Combined with $\alpha=0$, this contradicts “not all zero”.

So $\alpha\ne 0$, and we can solve for $v$:

Hence $v$ is a linear combination of vectors in $S$, i.e., $v\in\operatorname{span}(S)$.

Since $v\in V$ was arbitrary, $\operatorname{span}(S)=V$, i.e., $S$ spans $V$.

Step 3 — Conclusion

$S$ is linearly independent (given) and spans $V$ (Step 2), so $S$ is a basis of $V$. $\blacksquare$

Answer