Change of Coordinate Vectors Under Linear Transformation

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Theorem

Let $R$ be a ring with unity.

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

Let $\AA, \BB$ be ordered bases of $M$ and $N$ respectively.

Let $f: M \to N$ be a linear transformation.

Let $\mathbf M_{f, \BB, \AA}$ be its matrix relative to $\AA$ and $\BB$.


Then for all $m \in M$:

$\sqbrk {\map f m}_\BB = \mathbf M_{f, \BB, \AA} \cdot \sqbrk m_\AA$

where $\sqbrk {\, \cdot \,}_-$ denotes the coordinate vector with respect to a basis.


Proof

Both sides are linear in $m$ and they coincide on the elements of $\AA$ by definition of $\mathbf M_{f, \BB, \AA}$.

So they are equal for all $m \in M$.



$\blacksquare$


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