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Let ${\{a_0,...,a_n}\}$ and ${\{b_0,...,b_m}\}$ be sets of rational numbers indexed by intervals $[0;n]$ and $[0;m]$ of $\mathbb N$.

How to prove that:

$\displaystyle \sum_{i=0}^n \Big ( a_i \displaystyle \sum_{j=0}^m b_j \Big)=\displaystyle \sum_{k=0}^{n+m} \Big(\displaystyle \sum a_ib_j:i+j=k \Big)$

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    In the right-hand sum, you need to restrict the values which i and j can take on or else they can take values for where there are no elements (e.g., $a_{n+1}$). – 2017-01-17
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    @marty cohen I defined $i$ to be a variable for indices from $[0;n]$ set and $j$ to be similar variable for $[0;m]$. – 2017-01-17
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    Where does it matter that $a_i,b_j$ are rational vs. real, complex, or matrices for that matter? Edited to remove the `rational-numbers` tag. – 2017-01-19

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Without loss of generality we can assume $m=n$, since we can set $a_i=0$ for $i>m$ and $b_j=0$ for $j>n$. We want to show:

$$\sum_{i,j=0}^n a_ib_j = \sum_{k=0}^{2n} \sum_{i,j = 0\\i+j=k}^n a_ib_j$$

For each $k$ consider the partial sum:

$$S_k := \sum_{i,j = 0\\i+j=k}^n a_ib_j$$

We can divide $\sum_{i,j=0}^na_ib_j$ into partial sums like $S_k$. To sum over all the values $a_ib_j$, we can sum over all $S_k$ (associativity of $+$). If $k$ is not between $0$ and $2n$, the partial sum is empty, since $i$ and $j$ can only be between $0$ and $n$. Hence $\sum_{i,j=0}^na_ib_j = \sum_{k=0}^{2n} S_k$.