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.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.5$: Corollary