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

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 $$G$$.