Prove that $\delta_{ij}\delta_{jk}=\delta_{ik}$

Question

Firstly, I should mention that I have just started learning about tensors, the problem is that I need to understand why the result $$\fbox{$\delta_{ij}\delta_{jk}=\delta_{ik}$}\tag{1}$$ is true in order to be able to follow certain proofs given in my lecture course.

Before starting tensors I was taught that $$\delta_{ij} = \begin{cases} 0, & \mbox{if } i \ne j \\ 1, & \mbox{if } i=j \end{cases}\tag{2}$$

But now that I have been introduced to tensors I have been told that the Kronecker delta can actually be thought of as a tensor and that

$$\delta_{ij}=\begin{bmatrix}\delta _{11}&\delta _{12}&\delta _{13}\\\delta _{21}&\delta _{22}&\delta _{23}\\\delta _{31}&\delta _{32}&\delta _{33}\end{bmatrix}$$ for $i=1,2,3$ and $j=1,2,3$.

Using the definition in $\mathrm(2)$ then $$\delta_{ij}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I$$

I know that in $\mathrm(1)$ the Einstein summation convention is being used such that $$\delta_{ij}\delta_{jk}=\sum_{j=1}^{3}\delta_{ij}\delta_{jk}$$

But if I try to prove $(1)$ as follows

$$\delta_{ij}\delta_{jk}=\begin{bmatrix}\delta _{11}&\delta _{12}&\delta _{13}\\\delta _{21}&\delta _{22}&\delta _{23}\\\delta _{31}&\delta _{32}&\delta _{33}\end{bmatrix}\begin{bmatrix}\delta _{11}&\delta _{12}&\delta _{13}\\\delta _{21}&\delta _{22}&\delta _{23}\\\delta _{31}&\delta _{32}&\delta _{33}\end{bmatrix}$$ for $i=1,2,3$ and $j=1,2,3$ and $k=1,2,3$.

It doesn't bother me to multiply this whole thing out if it helps my understanding, but frankly, I don't see the point as

$$\delta_{ij}\delta_{jk}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I^2=I \ne \delta_{ik}$$

I have already watched this: Why does $\delta_{ij}\delta_{jk}=\delta_{ik}\,$ An explanation to make sense! which does not explain why $\delta_{ij}\delta_{jk}=\delta_{ik}$

I have also looked at a similar question on this site but the answer is too complicated for me to understand.

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So to summarize, is there a simple way of showing that $$\delta_{ij}\delta_{jk}=\delta_{ik}?$$

Answer 1

Everyone struggles to gain intuition with this. The way I came to realize it was to think about the terms in $\delta_{ij}\delta_{jk}$.

$\delta_{ij}$ is zero unless $i=j$, and $\delta_{jk}$ is zero unless $j=k$. So for any $j$, $\delta_{ij}\delta_{jk}$ (no summation) is $1$ if $i$, $j$, and $k$ are equal, and $0$ otherwise.

Now $i$, $j$, and $k$ are all being drawn from the same pool of cartesian indices. So when we do sum over all $j$'s, exactly one of those $j$'s will agree with $i$, and exactly one will agree with $k$. If $j$ agrees with $i$ at the same time it agrees with $k$ (that is, all three indices are equal), we will pick up a $1$. For all the rest, we're summing up zeroes. In other words, the sum is $1$ if $i=k$, and $0$ otherwise. That tensor is $\delta_{ik}$.

Answer 2

$\delta_{ij}\delta_{jk}$ denotes a multiplication on the index j. $\delta_{ij}\delta_{jk}=\delta_{i1}\delta_{1k}+\delta_{i2}\delta_{2k}+\delta_{i3}\delta_{3k}$. If $i\neq 1$ and $k\neq 1$ for the first term, then the first term will be zero. Similarly If $i\neq 2$ and $k\neq 2$ for the second term, then the second term will be zero. Therefore $\delta_{ij}\delta_{jk}=\delta_{11}\delta_{11}+\delta_{22}\delta_{22}+\delta_{33}\delta_{33}=\delta_{ik}$.