Change of Basis is Invertible

Theorem
Let $R$ be a ring with unity.

Let $M$ be a free $R$-module.

Let $\left \langle {a_n} \right \rangle$ and $\left \langle {b_n} \right \rangle$ be bases of $M$.

Let $\mathbf P$ be the matrix corresponding to the change of basis from $\left \langle {a_n} \right \rangle$ to $\left \langle {b_n} \right \rangle$.

Then $\mathbf P$ is invertible, and its inverse $\mathbf P^{-1}$ is the matrix corresponding to the change of basis from $\left \langle {b_n} \right \rangle$ to $\left \langle {a_n} \right \rangle$.

Proof
Let $\mathbf P = \left[{\alpha}\right]_{n}$ be the matrix $\left[{I_G; \left \langle {a_n} \right \rangle, \left \langle {b_n} \right \rangle}\right]$ corresponding to the change of basis from $\left \langle {b_n} \right \rangle$ to $\left \langle {a_n} \right \rangle$.

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

Thus the matrix corresponding to the change of basis from $\left \langle {a_n} \right \rangle$ to $\left \langle {b_n} \right \rangle$ is also the matrix:
 * $\left[{v; \left \langle {a_n} \right \rangle}\right]$

where $v$ is the endomorphism of $G$ which satisfies $\forall k \in \left[{1 \,.\,.\, n}\right]: v \left({a_k}\right) = b_k$.

The result follows from Product of Change of Basis Matrices and Change of Basis Matrix Between Equal Bases:


 * $\left[{I_G; \left \langle {a_n} \right \rangle, \left \langle {b_n} \right \rangle}\right] \left[{I_G; \left \langle {b_n} \right \rangle, \left \langle {a_n} \right \rangle}\right] = \left[{I_G; \left \langle {a_n} \right \rangle, \left \langle {a_n} \right \rangle}\right] = I_n$


 * $\left[{I_G; \left \langle {b_n} \right \rangle, \left \langle {a_n} \right \rangle}\right] \left[{I_G; \left \langle {a_n} \right \rangle, \left \langle {b_n} \right \rangle}\right] = \left[{I_G; \left \langle {b_n} \right \rangle, \left \langle {b_n} \right \rangle}\right] = I_n$