# Kronecker delta symbol

1. Dec 18, 2009

### tony873004

$$\begin{array}{l} \delta _{jk} A_k \\ \\ \delta _{jk} A_k = \left( {\delta _{1,1} + \delta _{1,2} + \delta _{1,3} + \delta _{2,1} + \delta _{2,2} + \delta _{2,3} + \delta _{3,1} + \delta _{3,2} + \delta _{3,3} } \right)A_k \\ \,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {1 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 1} \right)A_k \\ \,\,\,\,\,\,\,\,\,\,\,\,\, = 3A_k \\ \end{array}$$

But the answer should be $$A_j$$. Where did I go wrong?

2. Dec 18, 2009

### Pengwuino

As far as I know, the sum is only over k, not j.

3. Dec 18, 2009

### diazona

That's true, and you also need to recognize that $A_k$ has different values for different terms in the sum.

4. Dec 18, 2009

### Pengwuino

and to further add, the delta will only be 1 when k=j in the sum.

5. Dec 18, 2009

### tony873004

Sorry, I don't understand your explanations. We don't have a textbook for this, only class notes. So I don't even know what "the sum is over k" means. Is it possible to work out the example? Thanks!

6. Dec 19, 2009

### phsopher

The sum is always only over a repeated index, the other indices are fixed. For example suppose A is a 2x2 matrix and x is a 2-vector:

$$A_{ij} x_j = \sum_{j=1}^2 A_{ij} x_j = A_{i1} x_1 + A_{i2} x_2$$

7. Dec 19, 2009

### Phyisab****

really your only problem is that you summed over j

8. Dec 19, 2009

### tiny-tim

Hi tony873004!

See http://en.wikipedia.org/wiki/Einstein_summation_convention" [Broken]

Last edited by a moderator: May 4, 2017
9. Dec 20, 2009

### HallsofIvy

No. As others have said, the sum is over k, the repeated index, not i.
$$\delta_{ik}A_k= (A_{i1}A_1+ A_{i2}A_2+ A_{i3}A_3[/itex] for every i. That is [tex]\delta_{1k}A_k= A_{11}A_1+ A_{12}A_2+ A_{13}A_3= A_1$$
$$\delta_{2k}A_k= A_{21}A_1+ A_{22}A_2+ A_{23}A_3= A_2$$
$$\delta_{3k}A_3= A_{21}A_1+ A_{32}A_2+ A_{33}A_3= A_3$$
That is, it is the vector $<A_1, A_2, A_3>$ which can be written as $A_i$ or $A_j$ as they mean the same thing.