Invertible Matrix Corresponds with Change of Basis

Theorem

Let $R$ be a commutative ring with unity.

Let $G$ be an $n$-dimensional unitary $R$-module.

Let $\left \langle {a_n} \right \rangle$ be an ordered basis of $G$.

Let $\mathbf P = \left[{\alpha}\right]_{n}$ be a square matrix of order $n$ over $R$.

Let $\displaystyle \forall j \in \left[{1 . . n}\right]: b_j = \sum_{i=1}^n \alpha_{i j} a_i$.

Then $\left \langle {b_n} \right \rangle$ is an ordered basis of $G$ iff $\mathbf P$ is invertible.

Proof

• From Change of Basis is Invertible, if $\left \langle {b_n} \right \rangle$ is an ordered basis of $G$ then $\mathbf P$ is invertible.

• Now let $\mathbf P$ be invertible.

Then by the corollary to Linear Transformations Isomorphic to Matrix Space, there is an automorphism $u$ of $G$ which satisfies $\mathbf P = \left[{u; \left \langle {a_n} \right \rangle}\right]$.

Therefore, as $\forall j \in \left[{1 . . n}\right]: b_j = u \left({a_j}\right)$, it follows that $\left \langle {b_n} \right \rangle$ is also an ordered basis of $G$.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 29$: Theorem $29.3$