Matrix Corresponding to Change of Basis under Linear Transformation

Theorem

Let $R$ be a ring with unity.

Let $G$ and $H$ be free $R$-modules of finite dimensions $n,m>0$ respectively.

Let $\left \langle {a_n} \right \rangle$ and $\left \langle {{a_n}'} \right \rangle$ be ordered bases of $G$.

Let $\left \langle {b_m} \right \rangle$ and $\left \langle {{b_m}'} \right \rangle$ be ordered bases of $H$.

Let $u: G \to H$ be a linear transformation, and let $\left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right]$ be the matrix of $u$ relative to $\left \langle {a_n} \right \rangle$ and $\left \langle {b_m} \right \rangle$.

Let:

$\mathbf A = \left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right]$

$\mathbf B = \left[{u; \left \langle {{b_m}'} \right \rangle, \left \langle {{a_n}'} \right \rangle}\right]$

Then:

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

where:

$\mathbf P$ is the matrix corresponding to the change of basis from $\left \langle {a_n} \right \rangle$ to $\left \langle {{a_n}'} \right \rangle$

$\mathbf Q$ is the matrix corresponding to the change of basis from $\left \langle {b_m} \right \rangle$ to $\left \langle {{b_m}'} \right \rangle$.

Converse

The converse is also true:

Let $R$ be a commutative ring with unity.

Let $G$ be an $n$-dimensional unitary $R$-module, and let $\left \langle {a_n} \right \rangle$ be an ordered basis of $G$.

Let $H$ be an $m$-dimensional unitary $R$-module, and let $\left \langle {b_m} \right \rangle$ 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 exists a linear transformation $u: G \to H$, and ordered bases $\left \langle {{a_n}'} \right \rangle$ and $\left \langle {{b_m}'} \right \rangle$ of $G$ and $H$ respectively such that:

$\mathbf A = \left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right]$

$\mathbf B = \left[{u; \left \langle {{b_m}'} \right \rangle, \left \langle {{a_n}'} \right \rangle}\right]$

Corollary

Let $R$ be a commutative ring with unity.

Let $G$ be an $n$-dimensional unitary $R$-module, and let $\left \langle {a_n} \right \rangle$ and $\left \langle {{a_n}'} \right \rangle$ be ordered bases of $G$.

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

Let:

$\mathbf A = \left[{u; \left \langle {a_n} \right \rangle}\right]$

$\mathbf B = \left[{u; \left \langle {{a_n}'} \right \rangle}\right]$

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

where $\mathbf P$ is the matrix corresponding to the change of basis from $\left \langle {a_n} \right \rangle$ to $\left \langle {{a_n}'} \right \rangle$.

Corollary to Converse

Let $R$ be a commutative ring with unity.

Let $G$ be an $n$-dimensional unitary $R$-module, and let $\left \langle {a_n} \right \rangle$ and $\left \langle {{a_n}'} \right \rangle$ 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 exists a linear operator $u$ on $G$, and an ordered basis $\left \langle {{a_n}'} \right \rangle$ of $G$ such that:

$\mathbf A = \left[{u; \left \langle {a_n} \right \rangle}\right]$

$\mathbf B = \left[{u; \left \langle {{a_n}'} \right \rangle}\right]$.

Proof

We have $u = I_H \circ u \circ I_G$

and $\mathbf Q^{-1} = \left[{I_H; \left \langle {{b_m}'} \right \rangle, \left \langle {b_m} \right \rangle}\right]$.

Thus by Linear Transformations Isomorphic to Matrix Space:

\(\ds \mathbf Q^{-1} \mathbf A \mathbf P\)

\(=\)

\(\ds \left[{I_H; \left \langle { {b_m}'} \right \rangle, \left \langle {b_m} \right \rangle}\right] \left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right] \left[{I_G; \left \langle {a_n} \right \rangle, \left \langle { {a_n}'} \right \rangle}\right]\)

\(\ds \)

\(=\)

\(\ds \left[{I_H \circ u \circ I_G; \left \langle { {b_m}'} \right \rangle, \left \langle { {a_n}'} \right \rangle}\right]\)

\(\ds \)

\(=\)

\(\ds \mathbf B\)

$\blacksquare$

Proof of Converse

Let $\mathbf P = \left[{\alpha}\right]_n, \mathbf Q = \left[{\beta}\right]_m$.

Let:

$\forall j \in \left[{1 \,.\,.\, n}\right]: {a_j}' = \displaystyle \sum_{i \mathop = 1}^n \alpha_{i j} a_i$

$\forall j \in \left[{1 \,.\,.\, m}\right]: {b_j}' = \displaystyle \sum_{i \mathop = 1}^m \beta_{i j} b_i$

Then $\left \langle {{a_n}'} \right \rangle$ and $\left \langle {{b_m}'} \right \rangle$ are ordered bases of $G$ and $H$ respectively by Invertible Matrix Corresponds with Change of Basis.

Also we have:

$\mathbf P$ is the matrix corresponding to the change in basis from $\left \langle {a_n} \right \rangle$ to $\left \langle {{a_n}'} \right \rangle$

$\mathbf Q$ is the matrix corresponding to the change in basis from $\left \langle {b_m} \right \rangle$ to $\left \langle {{b_m}'} \right \rangle$

so $\mathbf Q^{-1}$ is the matrix corresponding to the change in basis from $\left \langle {{b_m}'} \right \rangle$ to $\left \langle {b_m} \right \rangle$

Let $\mathcal L_R \left({G, H}\right)$ be the set of all linear transformations from $G$ to $H$.

By Linear Transformations Isomorphic to Matrix Space, there exists $u \in \mathcal L_R \left({G, H}\right)$ such that $\mathbf A = \left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right]$.

But then, by the main result, $\left[{u; \left \langle {{b_m}'} \right \rangle, \left \langle {{a_n}'} \right \rangle}\right] = \mathbf Q^{-1} \mathbf A \mathbf P = \mathbf B$.

$\blacksquare$

Proof of Corollary

Follows directly.

$\blacksquare$

Proof of Corollary to Converse

Follows directly.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 29$: Theorem $29.4, \ 29.5$ and corollaries