How to Expand and Simplify the Expression of Kronecker Delta?

[jasonmcc]

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

 

[Ackbach]

Your working is correct if you're in 3 dimensions and you're using the Einstein summation convention.

 

[jasonmcc]

Thanks. What do you think they mean by "Expand", then?

 

[Ackbach]

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.

 

[blue_raver22]

Hello Jason,

Your solution is correct. The expansion of $\delta_{ij}\delta_{ij}$ is indeed equal to $\delta_{ii}$, which simplifies to 3. This is because the Kronecker delta symbol $\delta_{ij}$ is equal to 1 when $i=j$ and 0 otherwise. Therefore, when we multiply two Kronecker delta symbols together, we are essentially counting the number of times $i$ and $j$ are equal, which in this case is equal to the number of dimensions, which is 3.

There are different ways to expand this expression, depending on what you are trying to achieve. For example, if you are trying to express the Kronecker delta symbol in terms of the identity matrix, you can write it as $\delta_{ij} = \textbf{I}_{ij}$, where $\textbf{I}$ is the identity matrix. In this case, the expansion would be:

\begin{align}
\delta_{ij}\delta_{ij} & = \textbf{I}_{ij}\textbf{I}_{ji}\\
& = \textbf{I}_{ii}\\
& = \textbf{I}
\end{align}

Alternatively, you could also expand the expression using Einstein summation notation, which is commonly used in continuum mechanics. In this notation, the Kronecker delta symbol is represented as $\delta_{ij} = \delta_{ii} = g_{ij}$, where $g_{ij}$ is the metric tensor. In this case, the expansion would be:

\begin{align}
\delta_{ij}\delta_{ij} & = \delta_{ii}\delta_{ij}\\
& = g_{ii}g_{ij}\\
& = g_{ij}
\end{align}

I hope this helps! Let me know if you have any other questions. Keep up the good work in your continuum mechanics course.

Best,