Change of Basis Matrix under Linear Transformation/Converse/Corollary

Corollary to Change of Basis Matrix under Linear Transformation: Converse
Let $R$ be a commutative ring with unity. Let $G$ be a free unitary $R$-module of finite dimensions $n$.

Let $\sequence {a_n}$ be an ordered basis of $G$.

Let $\mathbf A$ and $\mathbf B$ be square matrices of order $n$ over $R$.

Let there exist an invertible matrix $\mathbf P$ of order $n$ such that:
 * $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$

Then there exist:
 * a linear operator $u$ on $G$
 * an ordered basis $\sequence { {a_n}'}$ of $G$

such that:


 * $\mathbf A = \sqbrk {u; \sequence {a_n} }$
 * $\mathbf B = \sqbrk {u; \sequence { {a_n}'} }$

where $\sqbrk {u; \sequence {a_n} }$ denotes the matrix of $u$ relative to $\sequence {a_n}$.

Proof
Follows directly from Change of Basis Matrix under Linear Transformation: Converse.