# Change of Basis Matrix under Linear Transformation/Converse

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

Let $\sequence {b_m}$ be an ordered basis of $H$.

Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over $R$.

Let there exist:

an invertible matrix $\mathbf P$ of order $n$
an invertible matrix $\mathbf Q$ of order $m$

such that:

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

Then there exist:

a linear transformation $u: G \to H$
ordered bases $\sequence { {a_n}'}$ and $\sequence { {b_m}'}$ of $G$ and $H$ respectively

such that:

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

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

### Corollary

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

Let:

$\mathbf P = \sqbrk \alpha_n$
$\mathbf Q = \sqbrk \beta_m$

Let:

$\forall j \in \closedint 1 n: {a_j}' = \ds \sum_{i \mathop = 1}^n \alpha_{i j} a_i$
$\forall j \in \closedint 1 m: {b_j}' = \ds \sum_{i \mathop = 1}^m \beta_{i j} b_i$
$\sequence { {a_n}'}$ and $\sequence { {b_m}'}$ are ordered bases of $G$ and $H$ respectively.

Also we have:

$\mathbf P$ is the matrix corresponding to the change in basis from $\sequence {a_n}$ to $\sequence { {a_n}'}$
$\mathbf Q$ is the matrix corresponding to the change in basis from $\sequence {b_m}$ to $\sequence { {b_m}'}$
so $\mathbf Q^{-1}$ is the matrix corresponding to the change in basis from $\sequence { {b_m}'}$ to $\sequence {b_m}$

Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.

By Set of Linear Transformations is Isomorphic to Matrix Space, there exists $u \in \map {\LL_R} {G, H}$ such that:

$\mathbf A = \sqbrk {u; \sequence {b_m}, \sequence {a_n} }$

But then, by Change of Basis Matrix under Linear Transformation:

$\sqbrk {u; \sequence { {b_m}'}, \sequence { {a_n}'} } = \mathbf Q^{-1} \mathbf A \mathbf P = \mathbf B$

$\blacksquare$