Product of Change of Basis Matrices

Theorem
Let $R$ be a ring with unity.

Let $M$ be a free $R$-module of finite dimension $n>0$.

Let $\mathcal A$, $\mathcal B$ and $\mathcal C$ be ordered bases of $M$.

Let $\mathbf M_{\mathcal A,\mathcal B}$, $\mathbf M_{\mathcal B,\mathcal C}$ and $\mathbf M_{\mathcal A,\mathcal C}$ be the change of basis matrices from $\mathcal A$ to $\mathcal B$, $\mathcal B$ to $\mathcal C$ and $\mathcal A$ to $\mathcal C$ respectively.

Then $\mathbf M_{\mathcal A,\mathcal C} = \mathbf M_{\mathcal A,\mathcal B} \cdot \mathbf M_{\mathcal B,\mathcal C}$

Proof
Let $m\in M$.

Let $[m]_{\mathcal A}$ be its coordinate vector relative to $\mathcal A$, and similary for $\mathcal B$ and $\mathcal C$.

By:
 * Transformation of Coordinate Vector in Terms of Matrix of Change of Basis
 * Matrix Multiplication is Associative:


 * $[m]_{\mathcal A} = \mathbf M_{\mathcal A,\mathcal C} \cdot [m]_{\mathcal C}$
 * $[m]_{\mathcal A} = \mathbf M_{\mathcal A,\mathcal B} \cdot [m]_{\mathcal B} = \mathbf M_{\mathcal A,\mathcal B} \cdot \mathbf M_{\mathcal B,\mathcal C}

\cdot [m]_{\mathcal C}$

Thus $(\mathbf M_{\mathcal A,\mathcal C} - \mathbf M_{\mathcal A,\mathcal B} \cdot \mathbf M_{\mathcal B,\mathcal C}) \cdot [m]_{\mathcal C} = 0$.

Because $m$ is arbitrary, $\mathbf M_{\mathcal A,\mathcal C} - \mathbf M_{\mathcal A,\mathcal B} \cdot \mathbf M_{\mathcal B,\mathcal C} = 0$.