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UPSC Maths 2013 Paper 1 Q2a-ii — Solution

8 marks · Section A

Question

Let $V$ be an $n$-dimensional vector space and $T:V\to V$ be an invertible linear operator. If $\beta=\{X_1,X_2,\ldots,X_n\}$ is a basis of $V$, show that $\beta'=\{TX_1,TX_2,\ldots,TX_n\}$ is also a basis of $V$.

Technique

Standard invertible-linear-operator argument; injectivity drives independence, surjectivity drives spanning. Both follow from $T$ being a bijection.

Solution

Strategy. Show $\beta'$ is linearly independent and spans $V$. Both facts follow directly from $T$ being injective (for independence) and surjective (for spanning) — i.e., invertible.

Step 1 — $\beta'$ is linearly independent

Suppose $c_1,\ldots,c_n\in\mathbb F$ satisfy

By linearity of $T$,

Since $T$ is invertible, it is in particular injective, so $\sum c_i X_i\in\ker T=\{0\}$:

Since $\beta$ is a basis (hence linearly independent), $c_1=\cdots=c_n=0$.

So $\beta'$ is linearly independent.

Step 2 — $\beta'$ spans $V$

Let $v\in V$. Since $T$ is invertible (hence surjective), there exists $u\in V$ with $T(u)=v$. Since $\beta$ is a basis, write $u=\sum c_i X_i$. Then

So $V\subseteq\operatorname{span}(\beta')$.

Step 3 — Conclude

$\beta'$ has $n=\dim V$ elements and is linearly independent (Step 1), so it automatically spans $V$ (a linearly independent set of size $\dim V$ in $V$ is a basis). Alternatively, by Step 2 it spans; combining with Step 1 gives the basis property.

Answer