Change of Basis Matrix under Linear Transformation

Theorem
Let $R$ be a commutative ring with unity.

Let $G$ and $H$ be free unitary $R$-modules of finite dimensions $n, m > 0$ respectively.

Let $\sequence {a_n}$ and $\sequence { {a_n}'}$ be ordered bases of $G$.

Let $\sequence {b_m}$ and $\sequence { {b_m}'}$ be ordered bases of $H$.

Let $u: G \to H$ be a linear transformation.

Let $\sqbrk {u; \sequence {b_m}, \sequence {a_n} }$ denote the matrix of $u$ relative to $\sequence {a_n}$ and $\sequence {b_m}$.

Let:
 * $\mathbf A = \sqbrk {u; \sequence {b_m}, \sequence {a_n} }$
 * $\mathbf B = \sqbrk {u; \sequence { {b_m}'}, \sequence { {a_n}'} }$

Then:
 * $\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$

where:
 * $\mathbf P$ is the matrix corresponding to the change of basis from $\sequence {a_n}$ to $\sequence { {a_n}'}$
 * $\mathbf Q$ is the matrix corresponding to the change of basis from $\sequence {b_m}$ to $\sequence { {b_m}'}$.

Converse
The converse is also true:

Proof
We have $u = I_H \circ u \circ I_G$

and $\mathbf Q^{-1} = \sqbrk {I_H; \sequence { {b_m}'}, \sequence {b_m} }$.

Thus by Set of Linear Transformations is Isomorphic to Matrix Space: