Linear mapping/Matrix/Composition/Fact/Proof2

Proof

We consider the chain of mappings

$U{\stackrel {\psi }{\longrightarrow }}V{\stackrel {\varphi }{\longrightarrow }}W.$

Suppose that ${}\psi$ is described by the ${}n\times p$-matrix ${}B=(b_{jk})_{jk}$, and that ${}\varphi$ is described by the ${}m\times n$-matrix ${}A={\left(a_{ij}\right)}_{ij}$ (with respect to the bases). The composition ${}\varphi \circ \psi$ has the following effect on the base vector ${}u_{k}$.

${}{\begin{aligned}{\left(\varphi \circ \psi \right)}{\left(u_{k}\right)}&=\varphi {\left(\psi {\left(u_{k}\right)}\right)}\\&=\varphi {\left(\sum _{j=1}^{n}b_{jk}v_{j}\right)}\\&=\sum _{j=1}^{n}b_{jk}\varphi (v_{j})\\&=\sum _{j=1}^{n}b_{jk}{\left(\sum _{i=1}^{m}a_{ij}w_{i}\right)}\\&=\sum _{i=1}^{m}{\left(\sum _{j=1}^{n}a_{ij}b_{jk}\right)}w_{i}\\&=\sum _{i=1}^{m}c_{ik}w_{i}.\end{aligned}}$

Here, these coefficients ${}c_{ik}=\sum _{j=1}^{n}a_{ij}b_{jk}$ are just the entries of the product matrix ${}A\circ B$.

To fact