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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.

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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 $a_j$.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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