Relative Matrix of Composition of Linear Transformations/Proof 1

Theorem

Let $R$ be a ring with unity.

Let $M, N, P$ be free $R$-modules of finite dimension $m, n, p > 0$ respectively.

Let $\AA, \BB, \CC$ be ordered bases of $M, N, P$.

Let $f: M \to N$ and $g : N \to P$ be linear transformations, and $g \circ f$ be their composition.

Let $\mathbf M_{f, \BB, \AA}$ and $\mathbf M_{g, \CC, \BB}$ be their matrices relative to $\AA, \BB$ and $\BB, \CC$ respectively.

Then the matrix of $g \circ f$ relative to $\AA$ and $\CC$ is:

$\mathbf M_{g \mathop \circ f, \CC, \AA} = \mathbf M_{g, \CC, \BB} \cdot \mathbf M_{f, \BB, \AA}$

Proof

Let $m \in M$, and $\sqbrk m_\AA$ be its coordinate vector with respect to $\AA$.

On the one hand:

\(\ds \sqbrk {\map g {\map f m} }_\CC\)

\(=\)

\(\ds \mathbf M_{g \mathop \circ f, \CC, \AA} \cdot \sqbrk m_\AA\)

Change of Coordinate Vectors Under Linear Mapping applied to $g \circ f$

On the other hand:

\(\ds \sqbrk {\map g {\map f m} }_\CC\)

\(=\)

\(\ds \mathbf M_{g, \CC, \BB} \cdot \sqbrk {\map f m}_\BB\)

Change of Coordinate Vectors Under Linear Mapping applied to $g$

\(\ds \)

\(=\)

\(\ds \mathbf M_{g, \CC, \BB} \cdot \mathbf M_{f, \BB, \AA} \cdot \sqbrk m_\AA\)

Change of Coordinate Vectors Under Linear Mapping applied to $f$

Thus:

$\forall m \in M: \paren {\mathbf M_{g \mathop \circ f, \CC, \AA} - \mathbf M_{g, \CC, \BB} \cdot \mathbf M_{f, \BB, \AA} } \cdot \sqbrk m_\AA = 0$

The result follows.

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$\blacksquare$