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 $\AA$, $\BB$ 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