Finding the Formula of the Product of $e_{i,j}$ and $e_{k,l}$ to Return Zero Matrix My teacher for calculus this year gave a handout on the first day with an excerpt from Rings, Fields, and Vector Spaces by B.A. Sethuraman. The reason for this is in the beginning of Sethuraman's book illustrates what he wants and expects out of his students for his class.
"How should you read this book? The answer, which applies to every book on mathematics, can be given in one word - actively. You may never have heard this before, but it can never be overstressed - you can only learn mathematics by doing mathematics..."
The first example is to show that multiplication of matrices is not commutative:
$$ \begin{pmatrix}
    1 & 0 \\
    0 & 0 \\
\end{pmatrix} \cdot 
\begin{pmatrix}
    0 & 1 \\
    0 & 0 \\
 \end{pmatrix} \neq \begin{pmatrix}
    0 & 1 \\
    0 & 0 \\
\end{pmatrix} \cdot 
\begin{pmatrix}
    1 & 0 \\
    0 & 0 \\
 \end{pmatrix}$$
Well, clearly the left hand side equals 
\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}
And the right hand side equals,
\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix}
"Mathematical insight comes from mathematical experience and you cannot expect to gain mathematical experience if you merely accept somebody else's word that the product on the left side of the equation does not equal the product on the right side."
Later on, it asks to find a pattern how matrices with nonzero numbers can multiply out to zero.

To help you describe this pattern, you will let $e_{i,j}$ stand for the matrix with 1 in the (i,j)-th slot and zeroes everywhere else, and you will try to discover a formula for the product of $e_{i,j}$ and $e_{k,l}$ where i,j,k, and l can each be any element of the set {1,2}.

If $i = 1$, and $j = 1$
$$e_{i,j} =
\begin{pmatrix}
    1 & 0 \\
    0 & 0 \\
 \end{pmatrix}
$$
If $l = 1$, and $k = 1$, what are the elements of $e_{k,l}$ or does that not even matter? 
$$e_{k,l} =
\begin{pmatrix}
    a & b \\
    c & d \\
 \end{pmatrix}
$$
Is the question looking to find the product of elements or is it looking for the product of the 2 matricies? Admittedly, this is my first time seeing this notation and I am a bit confused. I do know that $i$ and $l$ are the row indicies, and $j$ and $k$ are the column indicies.
 A: The notation of $e_{i,j}$ and $e_{k,l}$ is the same notation, they just use different letters for the different indices. In other words, if $i=k$ and $j=l$ then $e_{i,j} = e_{k,l}$. 
The reason why this notation is useful is as follows : given two $n \times n$ matrices $(a_{ij})$ and $(b_{ij})$, their product can be computed as follows : 
$$
(a_{ij})(b_{ij}) = \left( \sum_{q=1}^n a_{iq} b_{qj} \right). 
$$
This is exactly the formula you use to compute products "when you put your finger horizontally along the first matrix's rows and vertically along the second matrix's columns". The terms you come across horizontally are $a_{ik}$ and the terms you come across vertically are $b_{kj}$ ; you multiply those together and then you sum up, so that the $(i,j)^{\text{th}}$ coefficient of the resulting matrix is the sum above. 
Now what is the $(i,j)^{\text{th}}$ coefficient of a matrix $e_{k,l}$? Well a coefficient $a_{ij}$ of this matrix is $1$ if and only if $i=k$ and $j=l$, so if we use the Kronecker $\delta$ : 
$$
\delta_{ik} \overset{def}=
\begin{cases}
1 & \text{ if } i=k \\
0 & \text{ if } i \neq k
\end{cases}
$$
then we get that $a_{ij} = \delta_{ik} \delta_{jl}$. 
Suppose we start with $e_{k,l}$ and $e_{m,p}$ two matrices as above, so that we want to multiply 
$$
e_{k,l} e_{m,p} = (\delta_{ik} \delta_{jl})(\delta_{im} \delta_{jp}) = \sum_{q = 1}^n (\delta_{ik} \delta_{ql})(\delta_{qm} \delta_{jp}) = (\delta_{ik} \delta_{ll} \delta_{lm} \delta_{jp}) = \delta_{lm} (\delta_{ik} \delta_{jp}) = \delta_{lm} e_{k,p}.
$$
This means $e_{k,l} e_{m,p} = 0$ if $l \neq m$ and otherwise $e_{k,l} e_{m,p} = e_{k,p}$. 
Hope that helps,
