Question

Prove Theorem $13.2\left(a\right)$. That is, prove that$\sum _{j=1}^m\sum _{k=1}^n{a}_{jk}=\sum _{k=1}^n\sum _{j=1}^m{a}_{jk}$

Short answer

This is proved using expansion of left hand side of the equation.

Step-by-step solution

Given Information

We need to prove that

$\sum _{j=1}^m\sum _{k=1}^n{a}_{jk}=\sum _{k=1}^n\sum _{j=1}^m{a}_{jk}$

Simplification

Expansion gives $\sum _{j=1}^m\sum _{k=1}^n{a}_{jk}=\sum _{j=1}^m\left(\sum _{k=1}^n{a}_{jk}\right)$

$=\sum _{j=1}^m\left(a_{j1}+a_{j2}+a_{j3}+\dots +a_{jn}\right)$

$=\left(a_{11}+a_{12}+a_{13}+\dots +a_{1n}\right)+\left(a_{21}+a_{22}+a_{23}+\dots +a_{2n}\right)+$

$+\left(a_{31}+a_{32}+a_{33}+\dots +a_{3n}\right)+\dots +\left(a_{m1}+a_{m2}+a_{m3}+\dots +a_{mn}\right)$

Grouping gives

$\sum _{j=1}^m\sum _{k=1}^n{a}_{jk}=\left(a_{11}+a_{12}+a_{13}+\dots +a_{1n}\right)+\left(a_{21}+a_{22}+a_{23}+\dots +a_{2n}\right)+$

$+\left(a_{31}+a_{32}+a_{33}+\dots +a_{3n}\right)+\dots +\left(a_{m1}+a_{m2}+a_{m3}+\dots +a_{mn}\right)$

$=\left(a_{11}+a_{21}+a_{31}+\dots +a_{m1}\right)+\left(a_{12}+a_{22}+a_{32}+\dots +a_{m2}\right)+$

$+\left(a_{13}+a_{23}+a_{33}+\dots +a_{m3}\right)+\dots +\left(a_{1n}+a_{2n}+a_{3n}+\dots +a_{mn}\right)$

$\sum _{j=1}^m\sum _{k=1}^n{a}_{jk}=\left(a_{11}+a_{21}+a_{31}+\dots +a_{m1}\right)+\left(a_{12}+a_{22}+a_{32}+\dots +a_{m2}\right)+$

$+\left(a_{13}+a_{23}+a_{33}+\dots +a_{m3}\right)+\dots +\left(a_{1n}+a_{2n}+a_{3n}+\dots +a_{mn}\right)$

$=\sum _{k=1}^n\left(a_{1k}+a_{2k}+a_{3k}+\dots +a_{mk}\right)$

$=\sum _{k=1}^n\sum _{j=1}^m{a}_{jk}$

Hence$\sum _{j=1}^m\sum _{k=1}^n{a}_{jk}=\sum _{k=1}^n\sum _{j=1}^m{a}_{jk}$