# Summation Convention – Substitution Rule

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1. Feb 16, 2016

### FluidStu

Hello. I'm new to this forum. I'm starting a PhD – it's going to be a big long journey through the jungle that is CFD. I would like to arm myself with some tools before entering. The machete is Cartesian Tensors.

I know the rules regarding free suffix's and dummy suffixes, but I'm having trouble proving the substitution rule:
δi j δj k = δi k

Assuming a range of 3 for each component, I will choose that the free suffix's have the following values: i=3, k=1. Therefore, using the implicit summation rule for the dummy index:
δi j δj k = δ3 1 δ1 1 + δ3 2 δ2 1 + δ3 3 δ3 1 = ?

Is my expansion correct?

Many thanks.

Last edited by a moderator: Feb 17, 2016
2. Feb 16, 2016

### blue_leaf77

You can visualize this equation by remembering that kronecker delta symbol actually represents the matrix elements of an identity matrix.

3. Feb 17, 2016

### HallsofIvy

Staff Emeritus
No, It is not because, since "i" and "k" are NOT repeated your result should have "i" and "k" in it, only "j" is replaced by numbers:
δijδjk= δi1δ1k+ δi2δ2k+ δi3δ3k

4. Feb 24, 2016

### FluidStu

Ok, that makes sense (since the index which appears once simultaneously represents all the possible values of that index).

However, from your expansion, I do not understand how one could end up with δik. This was really the point in my question – I would like to see step-wise how δijδjk = δik.

5. Feb 24, 2016

### blue_leaf77

What do you get when multiplying an identity matrix with another identity matrix?

6. Feb 24, 2016

### FluidStu

I * I = I

7. Feb 24, 2016

### blue_leaf77

Now, imagine you have a 3x3 matrices $A$, $B$, and $C$ such that $AB=C$. How do you write the element $c_{ik}$ of $C$ in terms of the elements of $A$ and $B$?

8. Feb 24, 2016

### FluidStu

cik = aiibik + aijbjk + aikbkk

9. Feb 24, 2016

### blue_leaf77

Almost, just that the index to be summed must be written as numbers, not as indices as you did there.

10. Feb 24, 2016

### FluidStu

Do you mean:

cik = a11b13 + a12b23 + a13b33 ?

11. Feb 24, 2016

### blue_leaf77

No. The left most and right most indices in each term in right hand side should be equal to the indices in the left hand side, which is $ik$. Only the two two middle indices changes due to the summation.
Having fixed this, you should be able to see how that equation becomes of when $A=B=C=I$.

12. Feb 24, 2016

### HallsofIvy

Staff Emeritus
I presume this is the "Kronecker delta"- $\delta_{ii}= 1$, $\delta_{ij}= 0$ if $i\ne j$.
Then $\delta_{ij}\delta_{jk}= \delta_{i1}\delta_{1k}+ \delta_{i2}\delta_{2j}+ \delta_{i3}\delta_{3j}$
If i= 1, k= 1, then only the first term is non-zero and it is 1: [itex]\delta_{11}= 1+ 0+ 0= 1[/quote].
If i= 1, k= 2, then at least one of the factors in every term is 0: [itex]\delta_{12}= 0+ 0+ 0= 0[/quote].
If i= 1, k= 3, then at least one of the factors in every term is 0: [itex]\delta_{13}= 0+ 0+ 0= 0.

If i= 2, k= 2, then only the second term is non-zero and it is 1: [itex]\delta_{22}= 0+ 1+ 0= 1.
etc.

13. Feb 24, 2016

### FluidStu

Ok, so then cik = ai1b1k + ai2b2k + ai3b3k?

In this way I can see that A=B=C=I. However, I still don't see how this leads to the proof that δijδjk= δik. If you want to expand on this for others visiting the thread, feel free. However, I can see it clearly from this:

In all the cases where i = k, we end up with 1, while in all the cases where i ≠ k, we end up with 0. In other words, we end up with δik. Very good, thank you both for your help.

14. Feb 24, 2016

### blue_leaf77

That the Kronecker delta is defined as $\delta_{ik}=1$ for $i=k$ and zero for $i\neq k$, doesn't it remind you of the elements of the identity matrix?

15. Feb 24, 2016

### FluidStu

Yes, it does. I can see that the Kronecker Delta in matrix form is simply an identity matrix. But how does this help me to prove:
?

16. Feb 24, 2016

### blue_leaf77

$$\delta_{ij}\delta_{jk} = \delta_{i1}\delta_{1k} + \delta_{i2}\delta_{2k} + \delta_{i3}\delta_{3k} = \delta_{ik}$$
for the matrix equation $AB=C$. If A=B=C=I, you will only need to replace $a$, $b$, and $c$ with $\delta$ in that quoted equation.

17. Feb 24, 2016

### FluidStu

I see that replacing each with δ gives the required proof. But is it "ok" to replace a, b and c with δ? Aren't a, b and c simply elements of matrices? How can we just replace them with δ?

18. Feb 24, 2016

### zinq

At the risk of repetition: To put the essential point in one place:

Let A = (aij), B = (bij), and C = (cij) be n-dimensional matrices such that

AB = C.​

Then you know from the definition of matrix multiplication that

cij = Σ aik bkj

for any i, j in the range 1 ≤ i, j ≤ n, where the sum is over k = 1,...,n.

Specializing to the case of n = 3, for the case of A = B = C = I3 (the 3D identity matrix), and then using the repeated-index summation convention, you get

δik δkj = δij

for any i, j in the range 1 ≤ i, j ≤ 3.

19. Feb 24, 2016

### blue_leaf77

Note that the last sentence in post #16 is a conditional statement, look what the condition is in order for the conclusion to be true.

20. Feb 24, 2016

### PeroK

Here's an alternative proof (for which I'll abandon the summation convention):

$\Sigma_{j} \delta_{ij} \delta_{jk} = \delta_{ii} \delta_{ik} = \delta_{ik} \ \$ (as $\delta_{ij} = 0 \ (i \ne j)$)

Or, more laboriously:

$\Sigma_{j} \delta_{ij} \delta_{jk} = \delta_{ii} \delta_{ik} + \Sigma_{j \ne i} \delta_{ij} \delta_{jk} = \delta_{ik} + 0 \ \$ (as $\delta_{ii} = 1, \ \delta_{ij} = 0 \ (i \ne j)$)