I want to understand the how to interpret the matrices which are represented by index notation. Here is my matrix

$𝜎_{𝑖𝑗}+𝜎_{𝑖𝑘}𝑤_{𝑘𝑗}−𝑤_{𝑖𝑘} 𝜎_{𝑘𝑗}$

All the matrices in the equation are 3x3 matrices. Is $\sigma_{ik}$ and $\sigma_{kj}$ the same or is one the transpose of the other, and how do you identify that? The same question goes for $𝑤_{𝑘𝑗}$ and $𝑤_{𝑖𝑘}$. This is an equation from hypoelasticity model. I also understand that it is not a good practice to repeat the indices in an equation. Is that the reason they used different indices? I have got to know the basic sense of index notation, and I am in the process of learning more. It would be great if anyone can shed some light on this. Thank you.


I don't know about the specific equation in hand but a common way to represent matrix elements is to using $\sigma_{ij}$ to mean the element of a matrix, say $\Sigma$ at the $i$th row and the $j$th column.

Then $\sigma_{ik}\omega_{kj}$ is the product of the element of $\Sigma$ matrix at the $i$th row and the $k$th column and that of $\Omega$ matrix at the $k$th row and the $j$th column. This is useful because of matrix multiplication:

$(\Sigma \times \Omega)_{ij} = \sum_{k}{\sigma_{ik}\omega_{kj}}$

There's nothing fancy here: this is just the definition of the product of two matrices.

To give you a concrete example, suppose $\Sigma = \begin{bmatrix} 1 & -2 \\ 13 & 9 \end{bmatrix}$, then $\sigma_{11}=1, \sigma_{12}=-2, \sigma_{21}=13$ and $\sigma_{22}=9$.

  • $\begingroup$ Thanks for the reply. I definitely agree with your reply which address the issue of matrix multiplication. However I would like to understand that if you have two matrix - $w_{kj}$ and $w_{ik}$, is it correct to say that they are transpose of each other? (I say this by seeing that 'k' represents row in one matrix and column in another). Am I seeing it with the correct perspective? $\endgroup$ – Willis Aug 26 '15 at 23:19
  • $\begingroup$ So the transpose is just the rows and columns flipped. If $\Omega^T$ is the transpose of $\Omega$, then we can say $(\Omega^T)_{ij}=(\Omega)_{ji}=\omega_{ji}$. Does that make sense? $\endgroup$ – MahoganyConvergence Aug 26 '15 at 23:24
  • $\begingroup$ Yes thats true, it does make sense, this is the general definition of a transpose of a matrix. Let me put it this way, in the material I am reading I have one equation which says $C_{ij}=ω_{ij}ω_{ik}$ and they have written down the same thing to be equivalent to $C=ω^Tω$. How do they get that? $\endgroup$ – Willis Aug 27 '15 at 0:34
  • $\begingroup$ What they mean by $C = \Omega^T \times \Omega$ is that $c_{ij}=\sum_{k}{(\Omega^T)_{ik}(\Omega)_{kj}}=\sum_{k}{\omega_{ki}\omega_{kj}}$. I'm careful in using capital letters to denote full matrices and small letters for individual elements to avoid confusion. $\endgroup$ – MahoganyConvergence Aug 27 '15 at 0:39
  • $\begingroup$ Alright. Appreciate for that distinction you made between the full matrices and the elements. I got the picture. Thank you. $\endgroup$ – Willis Aug 27 '15 at 0:45

It seems that it is Einstein's sum notation. So you have actually $A = (a_{ij})_{1\le i, j \le 3}$ with $$ a_{ij} = \sigma_{ij} + \sum_{k=1}^3 \left( \sigma_{ik} w_{kj} - w_{ik} \sigma_{kj} \right), $$ or $$ A = \Sigma + \Sigma W - W \Sigma, $$ with $\Sigma = (\sigma_{ij})_{1\le i, j \le 3}$ and $W = (w_{ij})_{1\le i, j \le 3}$.

  • $\begingroup$ I was about to question whether the repetition of the extra $ij$ was necessary but I suppose it is, since you've to label the numbers and also to Karl they're position in the matrix. $\endgroup$ – snulty Aug 27 '15 at 9:16
  • $\begingroup$ @snulty where exactly? like $A=(a_{ij})_{ij}$? $\endgroup$ – user251257 Aug 27 '15 at 11:07
  • $\begingroup$ Yes that's it! Would it be better to write however that $(A)_{ij}=a_{ij}$? I just realised that while writing on my phone, my first comment has some typos, like I believe Karl should be label and they're should be their. $\endgroup$ – snulty Aug 27 '15 at 12:33
  • $\begingroup$ @snulty they are the same. Both notations are common. I was too lazy to write $(a_{ij})_{1\le i, j \le 3}$ $\endgroup$ – user251257 Aug 27 '15 at 12:36
  • $\begingroup$ Ah that's fine, I understood the meaning anyway! And yes, there doesn't seem to be much need anyway to write the $(a_{ij})_{1\le i, j \le 3}$ anyway :) $\endgroup$ – snulty Aug 27 '15 at 12:46

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