Change of Basis Matrix under Linear Transformation/Converse/Corollary
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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