MHB How to Expand and Simplify the Expression of Kronecker Delta?

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The discussion centers on expanding and simplifying the expression δ_ijδ_ij. The user, Jason, believes his solution is correct, showing that δ_ijδ_ij simplifies to δ_ii, which equals 3 in three dimensions under the Einstein summation convention. Participants confirm that Jason's working is accurate and clarify that "expand" typically means writing out all terms, which leads to a simplified result. The conversation highlights the importance of understanding the context of linear algebra and the implications of summation conventions. Overall, Jason's approach effectively addresses the problem as posed.
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Hi, I'm working on a problem stated as:
Expand the following expression and simplify where possible
$$
\delta_{ij}\delta_{ij}
$$

I'm pretty sure this is correct, but not sure that I am satisfying the expand question. I'm not up to speed in linear algebra (taking a continuum mechanics course) - the question could be asking for $\hat{e}$ or matrix type expansion...

my solution:
\begin{alignat}{3}
\delta_{ij}\delta_{ij} & = & \delta_{ij}\delta_{ji}\\
& = & \delta_{ii}\\
& = & 3
\end{alignat}

Any suggestions for how to expand - or does this answer the question?
Thanks, Jason
 
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Your working is correct if you're in 3 dimensions and you're using the Einstein summation convention.
 
Thanks. What do you think they mean by "Expand", then?
 
When you expand a sum like that, you're writing out every term, essentially. And then because they combine as you've shown, it collapses down to a nice compact number.
 
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