# Linear mapping/Matrix/Composition/Fact/Proof2

${\displaystyle U{\stackrel {\psi }{\longrightarrow }}V{\stackrel {\varphi }{\longrightarrow }}W.}$
Suppose that ${\displaystyle {}\psi }$ is described by the ${\displaystyle {}n\times p}$-matrix ${\displaystyle {}B=(b_{jk})_{jk}}$, and that ${\displaystyle {}\varphi }$ is described by the ${\displaystyle {}m\times n}$-matrix ${\displaystyle {}A={\left(a_{ij}\right)}_{ij}}$ (with respect to the bases). The composition ${\displaystyle {}\varphi \circ \psi }$ has the following effect on the base vector ${\displaystyle {}u_{k}}$.
{\displaystyle {}{\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 ${\displaystyle {}c_{ik}=\sum _{j=1}^{n}a_{ij}b_{jk}}$ are just the entries of the product matrix ${\displaystyle {}A\circ B}$.