# Simple Tensorial Equation

I am reading a paper and the author says he solves Equation below:

$k_{ij} = \frac{1}{2}(K_{ij}+K_{ji}+K_{kk}\delta_{ij})$

To do so, he then says

"solving for $K_{kk}$ we get"

$K_{kk}=\frac{1}{2}k_{kk}$

Why it happens this way, if we put $i=j=k$ then it comes $K_{kk}=\frac{2}{3}k_{kk}$ not the one that author mentions.

Where I am making mistake?

• Does $K_{kk}$ mean the trace of $K$ or does it mean $K$ evaluated at the repeated index $k$? (in other words, is the author [ab]using Einstein summation convention?) – Willie Wong Nov 3 '16 at 19:31
• @WillieWong My apology if did not explain it, he means Einstein summation convention not trace of matrix. – Soyol Nov 3 '16 at 21:16

Since the index $k$ on the right side of the first equation does not appear on the left side, the author is obviously using the summation convention, that is, $$K_{kk} \equiv \sum_{\alpha = 1}^N K_{\alpha \alpha}$$ where $N$ is the dimension of the indices and $\alpha$ of course is summed over explicitly. Similarly, $k_{kk}$ has an implied sum over values of the index $k$.
Under that interpretation, the author gets from the first equation to an expression like the second by multiplying both sides by $\delta_{ij}$ and summing over $i$ and $j$. When you do this you get $$k_{kk} = \frac12\left( K_{kk} + K_{kk} + K_{kk} \sum_{i=1}^N\sum_{j=1}^N\delta_{ij}\right) = \left( 1+\frac{N}2\right)K_{kk}$$ or $$K_{kk} = \frac2{2+N} k_{kk}$$
The author would be right if he is working in two dimensions. You very likely forgot to sum over $i$ and $j$ in $\sum_{i=1}^N\sum_{j=1}^N\delta_{ij}$.
• Thank you this makes sense, and my mistake was as you mentioned. And to confirm your guess, yes author works in 2d, so $N=2$ – Soyol Nov 3 '16 at 21:19