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$$\left(\sum_{i=1}^3 \sum_{j=1}^3 \delta_{ij}\right)\left(\sum_{m=1}^3 \sum_{n=1}^3 \delta_{mn}\right)$$ eq. 1. Should be:

I am just confused because I am getting $3 \times 3$.

I was reading a book. It it stated that

$$\left(\sum_{i=1}^3 \sum_{j=1}^3 \delta_{ij}\right)\left(\sum_{m=1}^3 \sum_{n=1}^3 \delta_{mn}\right)=81$$. Any advice? I am getting $3 \times 3$ Maybe the book is wrong. Just want to confirm.

Thanks in advance!

Rafael Acosta
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  • attached in a picture is the question – Rafael Acosta Apr 20 '21 at 17:15
  • Hi, welcome to Mathematics Stack Exchange! There are blind and visually impaired users of this site who use screen readers to interact with it. These screen readers can't extract information from images. Please update the question by typing the LaTeX of the equation from your screenshot. – Arya McCarthy Apr 20 '21 at 17:17
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    I would agree: the answer is $3 \times 3$. Is there an answer you think it should be instead? – Theo Bendit Apr 20 '21 at 17:21
  • @Theo Bendit I edited the question. If you can confirm it. It will be very helpful!! – Rafael Acosta Apr 20 '21 at 17:37
  • Ah, then you are very much correct, and the book is very much incorrect (barring any transcription errors). Each double sum being multiplied is summing the entries in the $3 \times 3$ identity matrix, which is indeed $3$. – Theo Bendit Apr 20 '21 at 17:39

1 Answers1

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We just need to show that $$\sum_{i=1}^3 \sum_{j=1}^3 \delta_{ij} = 3.$$ Expanding the summation terms, we get \begin{align*} \sum_{i=1}^3 (\delta_{i1} + \delta_{i2} + \delta_{i3}) &= (\delta_{11} + \delta_{12} + \delta_{13}) + (\delta_{21} + \delta_{22} + \delta_{23}) + (\delta_{31} + \delta_{32} + \delta_{33}) \\ &= (1 + 0 + 0) + (0 + 1 + 0) + (0 + 0 + 1) = 3. \end{align*}

Theo Bendit
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