Change of Basis is Invertible

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

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

The result follows from:


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


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