Change of Basis Matrix under Linear Transformation

From ProofWiki
Jump to navigation Jump to search

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:


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 dimension $n$.

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

Let $u: G \to G$ be a linear operator on $G$.


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

Let:

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


Then:

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

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


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:

\(\ds \mathbf Q^{-1} \mathbf A \mathbf P\) \(=\) \(\ds \sqbrk {I_H; \sequence { {b_m}'}, \sequence {b_m} } \sqbrk {u; \sequence {b_m}, \sequence {a_n} } \sqbrk {I_G; \sequence {a_n}, \sequence { {a_n}'} }\)
\(\ds \) \(=\) \(\ds \sqbrk {I_H \circ u \circ I_G; \sequence { {b_m}'}, \sequence { {a_n}'} }\)
\(\ds \) \(=\) \(\ds \mathbf B\)

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Change_of_Basis_Matrix_under_Linear_Transformation&oldid=540959"