Linear mapping/Matrix/Composition/Fact/Proof

Proof

We consider the commutative diagram

${\begin{matrix}K^{p}&{\stackrel {M_{\mathfrak {v}}^{\mathfrak {u}}(\psi )}{\longrightarrow }}&K^{n}&{\stackrel {M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )}{\longrightarrow }}&K^{m}&\\\!\!\!\!\!\Psi _{\mathfrak {u}}\downarrow &&\!\!\!\!\!\Psi _{\mathfrak {v}}\downarrow &&\downarrow \Psi _{\mathfrak {w}}\!\!\!\!\!&&\\U&{\stackrel {\psi }{\longrightarrow }}&V&{\stackrel {\varphi }{\longrightarrow }}&W&\!\!\!\!\!,\\\end{matrix}}$

where the commutativity rests on the identities

${}\varphi \circ \Psi _{\mathfrak {v}}=\Psi _{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\,$

from fact. The (inverse) coordinate mappings ${}\psi _{\mathfrak {u}},\psi _{\mathfrak {v}},\psi _{\mathfrak {w}}$ are bijective. Therefore, we have

${}M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )=\Psi _{\mathfrak {w}}^{-1}\circ \varphi \circ \Psi _{\mathfrak {v}}\,.$

Hence, we get altogether

${}{\begin{aligned}M_{\mathfrak {w}}^{\mathfrak {u}}(\varphi \circ \psi )&=\Psi _{\mathfrak {w}}^{-1}\circ (\varphi \circ \psi )\circ \Psi _{\mathfrak {u}}\\&={\left(\Psi _{\mathfrak {w}}^{-1}\circ \varphi \right)}\circ {\left(\Psi _{\mathfrak {v}}\circ \Psi _{\mathfrak {v}}^{-1}\right)}\circ {\left(\psi \circ \Psi _{\mathfrak {u}}\right)}\\&={\left(\Psi _{\mathfrak {w}}^{-1}\circ \varphi \circ \Psi _{\mathfrak {v}}\right)}\circ {\left(\Psi _{\mathfrak {v}}^{-1}\circ \psi \circ \Psi _{\mathfrak {u}}\right)}\\&=M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ M_{\mathfrak {v}}^{\mathfrak {u}}(\psi ),\end{aligned}}$

where we have everywhere compositions of mappings. Due to exercise, the composition of linear mappings on standard spaces corresponds to the matrix multiplication.

To fact