Clarification on Einstein Summation Convention I was doing a MOOC quiz question on sun's rays hitting an object and casting a shadow on the ground in a 3D space. The variables are listed below and visualized in an illustration beneath that:

*

*Space: 3D, with unit vectors $\hat{e}_1$ to $\hat{e}_3$

*Vector $\hat{s}$: Unit vector of sun's rays (should be a column vector with 3 elements)

*Vector $r$: Any point on an object in the space.

*Vector $r'$: $r$ projected onto the ground due to shadow cast when sun's rays hits object.


The question asked for this (equation (1)) to be represented in Einstein's Summation notation:
$$r' = r - \frac{\hat{s}\left(r \cdot \hat{e}_3\right)}{\hat{s}\cdot \hat{e}_3} \tag 1$$
These are 3 options (equations (2),(3, and (4)) which were listed to be correct (indices are the $i$'s and $j$'s outside the brackets):
\begin{align} r'_i &= r_i - \frac{s_i (\hat{e}_3)_j r_j}{(\hat{s}\cdot \hat{e}_3)} \tag 2 \\
 r'_i &= \left(I_{ij} - \frac{s_i[\hat{e}_3]_j }{ \hat{s}\cdot \hat{e}_3}\right)r_j \tag 3 \\
 r'_i &= \left(I_{ij} - \frac{s_i(I)_{i=3,j} }{ \hat{s}\cdot \hat{e}_3}\right)r_j \tag 4
\end{align}
Where $I$ is an identity matrix.
Questions:
(1) Can someone help to explain how they are equivalent representations?
(2) What is $[\hat{e}_3]_j$? Isn't it a $3\times 1$ column matrix - doesn't that mean $j=1$?
 A: 
A bit of a preamble



*

*If $\{e_i\}_i$ forms a base, then you can write any vector ${\bf x}$ as 


$$
x = x_i \hat{\bf e}_i \tag{a}
$$


*

*If the base is orthonormal you have


$$
\hat{\bf e}_i \cdot \hat{\bf e}_j = \delta_{ij} = \begin{cases}1 & i =j \\ 0 & {\rm otherwise}\end{cases} \equiv I_{ij} \tag{b}
$$


*

*If the base is orthonormal and ${\bf x}, {\bf y}$ are vectors


$$
{\bf x}\cdot {\bf y} \stackrel{(a)}{=} (x_i \hat{\bf e}_i)\cdot (y_j \hat{\bf e}_j) \stackrel{(b)}{=} x_i y_j \delta_{ij} = x_i y_i \tag{c}
$$


*

*In an orthonormal basis you can calculate the component of vector ${\bf x}$ along the direction $j$ fairly easy


$$
{\bf x}\cdot \hat{\bf e}_j = (x_i \hat{\bf e}_i) \cdot \hat{\bf e}_j = x_i \delta_{ij} = x_j \tag{d}
$$


*

*As a corollary the $j$-th component of the vector $\hat{\bf e}_i$ is simply 
$$
\hat{\bf e}_i \cdot \hat{\bf e}_j = \delta_{ij} \tag{e}
$$

Now your problem

\begin{eqnarray}
{\bf r}' &=& {\bf r} - \hat{\bf s} \frac{{\bf r}\cdot \hat{\bf e}_3}{\hat{\bf s}\cdot \hat{\bf e}_3} \\
r'_i &=& r_i - s_i \frac{{\bf r}\cdot \hat{\bf e}_3}{\hat{\bf s}\cdot \hat{\bf e}_3} \\
&\stackrel{(c)}{=}& r_i - s_i \frac{r_j [e_3]_j}{\hat{\bf s}\cdot \hat{\bf e}_3} \\
&=& r_i -  \frac{s_i [e_3]_j}{\hat{\bf s}\cdot \hat{\bf e}_3}r_j \tag{2} \\
&\stackrel{(e)}{=}& r_j \delta_{ij} - \frac{s_i \delta_{3j}}{\hat{\bf s}\cdot \hat{\bf e}_3}r_j \tag{3}
\end{eqnarray}
As for your last question $[e_3]_j$ means the $j$-th component of the vector $\hat{\bf e}_3$ in the base $\{\hat{\bf e}_i \}_i$ which is just $\delta_{3j}$ (see Eq (e) above)
