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Find the value of $\delta_{ii}\delta_{jk}\delta_{jk}$.

Apparently this is $9$, but I can't see why:

$\delta_{ii} = 3$ (as in this case this is defined in $\mathbb{R}^3$)

$\delta_{jk}\delta_{jk} = \delta_{jk}$

$\delta_{ii}\delta_{jk} = 3\delta_{jk}$

but is this not just 3 x the identity matrix? Why is this a direct number?

mathim1881
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Sam Connell
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  • When you write $\delta_{jk}\delta_{jk} = \delta_{jk}$, you fail to appropriately account for the summation convention – Ben Grossmann Nov 21 '20 at 03:59
  • How do you mean? Is this not just the notation for the identity matrix multiplied by itself? – Sam Connell Nov 21 '20 at 04:01
  • Let's start with this: why is $\delta_{ii} = 3$? – Ben Grossmann Nov 21 '20 at 04:01
  • Oh I think I get it. Delta_ii is 3 because delta_ij is defined to be 1 if i = j and 0 if i =/ j. Thus it is just the Trace so delta_ii is 3. Doing the summation convention then it's 3 x 2 sums summing across k and j. 2/3 of each of these terms are 0 because j =/ k, thus it's 3 (1+1+1) = 3 x 3 = 9. Is that correct? – Sam Connell Nov 21 '20 at 04:05
  • I'm not sure I understand your first 3 sentences; it doesn't look like you explain why $\delta_{ii}$ should refer to the trace of the matrix denoted by $\delta_{ij}$. The 2 sentences after that seem correct – Ben Grossmann Nov 21 '20 at 04:10

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Note that in the summation, $i,j,k$ each appear twice. According to the Einstein summation convention, we take a sum over every variable. That is, the expression should be interpreted as $$ \sum_{i = 1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \delta_{ii} \delta_{jk}\delta_{jk} = \sum_{i = 1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \delta_{ii} \delta_{jk}. $$ We can see that this sum is indeed equal to $9$.

Ben Grossmann
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