Moving factors out of sums (Lagrange identity proof)

Question

Was reading through Lagrange Identity Proof.

However, one thing the proof assumes is

$$\sum_{i=1}^p\sum_{j=1}^q a_i b_j=\sum_{i=1}^pa_i\sum_{j=1}^qb_j$$

which seems intuitive - but I wonder if anyone could demonstrate to me via a quick proof why this holds.

Answer 1

You're expected to prove

$$\sum_{i=1}^p\sum_{j=1}^qa_ib_j=\sum_{i=1}^pa_i\sum_{j=1}^qb_j\iff$$

$$a_1b_1+a_1b_2+\ldots +a_1b_q+a_2b_1+\ldots+a_2b_q+\ldots a_pb_1+\ldots a_pb_q=$$

$$=(a_1+\ldots +a_p)(b_1+\ldots + b_q)$$

Perhaps opening up things it looks clearer. I'd go by mathematical induction with $\;p+q\;$ : the first non-completely trivial case is $\;p+q=2\,,\,\,p,q\ge 1\iff p=q=1\;$ , and here we have:

$$a_1b_1=(a_1)(b_1)$$

The inductive step, assuming for $\;p+q\ge 2\;$ , goes either

$$\sum_{i=1}^p\sum_{j=1}^{q+1}a_ib_j=\sum_{i=1}^p\sum_{j=1}^qa_ib_j+a_1b_{q+1}+\ldots+a_pb_{q+1}=\;\ldots etc.$$

and the second other case with $\;p+1\;$ ...

Answer 2

Basically you overread the parenthesis I guess:

$$ \sum_{i=1}^p\sum_{j=1}^p a_ib_j = \sum_{i=1}^p a_i\left(b_1+b_2+\cdots+b_q\right) = \sum_{i=1}^p a_i \left(\sum_{j=1}^q b_j\right) = \sum_{i=1}^pa_i\sum_{j=1}^qb_j $$

Answer 3

\begin{align*} (a_1+\dots+a_p)(b_1+\dots+b_q) & = (a_1b_1+a_1b_2+\dots+a_1b_q) + \dots + (a_pb_1+\dots+a_pb_q) \\ & = \sum^q_{j=1} a_1b_j + \dots + \sum^q_{j=1}a_pb_j = \sum^p_{i=1} \sum^q_{j=1} a_ib_j \end{align*}