Tensor tranformation between basis?

Question

If I am the basis vector $e_i$ into another basis to get $e'_j$ I use: $$e'_j=S_{ij}e_i$$ My text book says that $S_{ij}$ is the ith component of the vector $e'_j$ with respect to the unprimed basis. Please can you explain what this last sentence means what does this mean (in simpler terms)? Thanks

Answer 1

The last sentence means exactly what's written in the equation: If you write $e'_j = S_{ij} e_i = S_{1j} e_1 + S_{2j} e_2 + \cdots + S_{nj} e_n$ over the $\{e_n\}$ basis, then the coefficient of $e_i$ is exactly $S_{ij}$. (Since the $e_i$ are a basis, the $S_{ij}$ are well-defined and unique.)

It might be throwing you off that the book is presumably using the Einstein summation convention here: repeated free indices (i.e., here, indices only on one side of the equation) are summed over. Thus $$e'_j = S_{ij} e_i$$ actually means $$e'_j = \sum_{i = 1}^n S_{ij} e_i.$$ It's a common convention when dealing with tensor calculus, especially in physics, to simplify all the bookkeeping.

Answer 2

Here the formula $$e'_j=S_{ij}e_i$$ really means that $$e'_j=\sum_{i=1}^nS_{ij}e_i.$$ Now it is clear that $S_{ij}$ is the ith component of the vector $e'_j$ with respect to the unprimed basis.

Answer 3

As others have pointed out, you have an implicit summation going over the index $i$, called Einstein Summation. I would like to do an explicit example.

Consider two basis $\beta = \{ e_1, e_2, e_3 \}$ and $\beta' = \{ e_1', e_2', e_3' \}$ of $\Bbb R^3$ where $$\left\{ \begin{align} e_1 & = (1,1,0), \\ e_2 & = (1,0,1), \\ e_3 & = (0,0,1), \end{align} \right.$$ and $$\left\{ \begin{align} e_1' & = (2,0,3), \\ e_2' & = (4,0,0), \\ e_3' & = (0,1,0). \end{align} \right.$$ That relation says that $$\left\{ \begin{align} e_1' & = \sum_{i=1}^3 S_{i1}e_i = S_{11} e_1 + S_{21} e_2 + S_{31} e_3, \\ e_2' & = \sum_{i=1}^3 S_{i2}e_i = S_{12} e_1 + S_{22} e_2 + S_{32} e_3, \\ e_3' & = \sum_{i=1}^3 S_{i3}e_i = S_{13} e_1 + S_{23} e_2 + S_{33} e_3. \end{align} \right.$$ Just as an example, writing the first in matrix form: $$\begin{bmatrix} 2 \\ 0 \\ 3 \end{bmatrix} = \begin{bmatrix} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}.$$ Write those sums in coordinates, you'll arrive at: $$\left\{ \begin{align} (2,0,3) & = (S_{11} + S_{21}, S_{11}, S_{21} + S_{31}), \\ (4,0,0) & = (S_{12} + S_{22}, S_{12}, S_{22} + S_{32}), \\ (0,1,0) & = (S_{13} + S_{23}, S_{13}, S_{23} + S_{33}). \end{align} \right.$$ This is straightforward to solve and you'll find $$[S]_{\beta}^{\beta'} = \begin{bmatrix} 0 & 0 & 1 \\ 2 & 4 & -1 \\ 1 & -4 & 1 \end{bmatrix}.$$