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$\begingroup$

I am sure this is trivial to most, but wanted to confirm that this is how you write the sum for this sequence:

$$a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31}+a_{14}b_{41}+\cdots+a_{1j}b_{i1}=\sum_{n=1}^j\sum_{m=1}^ia_{1n}b_{m1}$$

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    Are $a_{11}b_{21}$ and $a_{12}b_{11}$ terms of the sum? I would usually expect one of these to be listed _between_ $a_{11}b_{11}$ and $a_{12}b_{21}$ in the sum on the left side. My guess is you meant to include these terms but got hasty when writing the sum. But that is only a guess, and I could be wrong. – 2017-01-10
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    @HugHes Can you please clarify what the LHS means? There is no obvious pattern implied by the $\cdots$: comparing the first two terms suggests that the second and third indices advance together, but the last term doesn't fit the pattern! – 2017-01-10
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    I added some more terms I hope they clear it up – 2017-01-10
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    @HugHes : The problem is that your last term is neither $a_{1j} b_{j1}$ nor $a_{1i} b_{i1}$, but $a_{1j} b_{i1},$ so that the second index on $a$ doesn't match the first index on $b,$ although they match in the first four terms you've written. Consequently it is unclear whether you intended them always to match, or if not, then what you intended instead. $\qquad$ – 2017-01-10
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    Yeah I can see the confusion, however the index for a is not supposed to match the index for b, otherwise the sum could be simply written as $$\sum_{n=1}^ia_{1n}b_{n1}$$ or $$\sum_{n=1}^ja_{1n}b_{n1}$$ – 2017-01-10

2 Answers 2

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This looks OK to me. I have two suggestions.

First, use $m$ and $n$ for the limits and $i$ and $j$ for the running indices. That's standard usage and will help your readers.

Second, write out each side independently for some small values of $m$ and $n$ and check that you get the same $mn$ terms in each case.

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If what you mean is $\sum a_{1n}b_{m1}$ for $1 \le n \le i$ and $1 \le m \le j$, then you are correct.

I think that you would have to show more terms in the sum before I would be confident of this result.

Also, this look like a partial matrix pruduct.

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    Thanks for the help, the way you have written it is more clear but is more difficult to work with. And you are correct about the partial matrix product. – 2017-01-10