So I have the Kronecker delta which is denoted as $\delta_{ij}$=$I$.

Let $a_1, a_2, \cdots, a_n$ be a set of $n$ real numbers, I must show that:

$\sum\limits_{i=1}^n a_i \delta_{ij} = a_j$ and $\sum\limits_{j=1}^n a_j \delta_{ij} = a_i$

i.e. $Ia=a$

Secondly, I must show that:

$\sum\limits_{j=1}^n \delta_{ij} \delta_{jk} = \delta_{ik}$

i.e. $I^2=I$

Any help is appreciated.


Well, by definition

$$\delta_{ij}=\begin{cases}1, & \text{if} & i=j, \\0, & \text{if} & i\neq j,\end{cases}$$ and because of that, it should be clear that

$$a_{i}\delta_{ij}=\begin{cases}a_i, & \text{if} & i=j,\\0, & \text{if} & i\neq j\end{cases}.$$

But since $a_{i}\delta_{ij}=a_i$ just when $i=j$, we can write in this case $a_i\delta_{ij}=a_j$. Now sum the elements, along all the sum $\delta_{ij}$ will be zero, unless when the index $i$ hits $j$. Because of that we have

$$\sum_{i}a_i \delta_{ij} = a_j.$$

The other case can be handled similarly. Give it a try.

EDIT: What I meant is the following, write the sum explicitly

$$\sum_{i}a_i\delta_{ij} = a_1 \delta_{1j}+a_2\delta_{2j}+\cdots+a_{n-1}\delta_{(n-1)j}+a_n\delta_{nj}.$$

From all the terms in the sum, you can see the first index of the $\delta$ varying. Amongst all of them there'll be one such that $i=j$. For this particular one, $\delta_{jj}=1$ and for all the others it is zero. In that case, in the whole sum you have the contribution just from that term, which corresponds to $a_j$.

  • $\begingroup$ Ok thanks. When you say "Now sum the elements, along all the sum $\delta_{ij}$ will be zero, unless when the index $i$ hits $j$. Can you explain this to me? $\endgroup$ – Ryan Oct 31 '13 at 11:20
  • $\begingroup$ I've added an edit. See if it helps. $\endgroup$ – user1620696 Oct 31 '13 at 11:24
  • $\begingroup$ It does indeed. Much appreciated, I understand it now. In regards to showing $\sum\limits_{j=1}^n \delta_{ij} \delta_{jk} = \delta_{ik}$, how would I start this? I can see it is obvious that $I^2=I$ I just have trouble expressing it in a way that is mathematically correct. $\endgroup$ – Ryan Oct 31 '13 at 11:32
  • $\begingroup$ Just a quick suggestion: expand the sum, for the first $\delta$ there will only remain terms when $i=j$ and for the second, only when $j=k$. Now, If $i\neq k$, then the product will be zero, because one of the deltas becomes one and the other not. In that case, that thing will be zero unless $i=k$ when it'll be one. But that's the definition of $\delta_{ik}$. $\endgroup$ – user1620696 Oct 31 '13 at 11:40
  • $\begingroup$ Right I got you. So I have expanded the sum as $\delta_{1i} \delta_{1k}, \cdots, \delta_{ni} \delta_{nk}$ and stated that the only termes remaining are when $i=j$ and $j=k$. All other terms are $0$. However if $i$ does not equal $k$ then the product will be $0$ because when $i=j$ and $j=k$, the other delta term will be $0$. So $\sum\limits_{j=1}^n \delta_{ij} \delta_{jk} = \delta_{ik}$ when $i=k$ $\endgroup$ – Ryan Oct 31 '13 at 11:55

A quick suggestion (that should be a comment): put the $a_{j}$s in a "diagonal" matrix and use (the formal definition of) matrix multiplication to get what you're after.

  • $\begingroup$ What are you referring to as $a_js$ ? $\endgroup$ – Ryan Oct 31 '13 at 11:09
  • $\begingroup$ The elements of $\{a_{1}, \dots , a_{n}\}$. $\endgroup$ – Shaun Oct 31 '13 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.