Product of Change of Basis Matrices
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Theorem
Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n>0$.
Let $\AA$, $\BB$ and $\CC$ be ordered bases of $M$.
Let $\mathbf M_{\AA, \BB}$, $\mathbf M_{\BB, \CC}$ and $\mathbf M_{\AA, \CC}$ be the change of basis matrices from $\AA$ to $\BB$, $\BB$ to $\CC$ and $\AA$ to $\CC$ respectively.
Then:
- $\mathbf M_{\AA, \CC} = \mathbf M_{\AA, \BB} \cdot \mathbf M_{\BB, \CC}$
Proof
Let $m \in M$.
Let $\sqbrk m_\AA$ be its coordinate vector relative to $\AA$, and similarly for $\BB$ and $\CC$.
On the one hand:
\(\ds \sqbrk m_\AA\) | \(=\) | \(\ds \mathbf M_{\AA, \CC} \cdot \sqbrk m_\CC\) | Change of Coordinate Vector Under Change of Basis |
On the other hand:
\(\ds \sqbrk m_AA\) | \(=\) | \(\ds \mathbf M_{\AA, \BB} \cdot \sqbrk m_\BB\) | Change of Coordinate Vector Under Change of Basis | |||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf M_{\AA, \BB} \cdot \mathbf M_{\BB, \CC} \cdot \sqbrk m_\CC\) | Change of Coordinate Vector Under Change of Basis |
Thus:
- $\forall m \in M: \paren {\mathbf M_{\AA, \CC} - \mathbf M_{\AA, \BB} \cdot \mathbf M_{\BB, \CC} } \cdot \sqbrk m_\CC = 0$
Because $m$ is arbitrary, the result follows.
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$\blacksquare$
Also see
- Relative Matrix of Composition of Linear Transformations, an analogous result for linear transformations, of which this is a special case