I have two questions that refer to the same subject: index notation. Both of them refer to the use of index notation to represent a matrix times a vector. When we have generally a matrix multiplicating a vector, we have, in index notation: $$v'_i = a_{ij} v_j$$ (I've come to this formula just by performing the multiplication for a $2 \times 2$ matrix.
But, when we talk about a rotation matrix: $$S_{ij} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$ Then our multiplication is $$v_i' = S_{ji} v_j$$
Why has $i$ and $j$ changed places? I can see with when we construct the new basis vectors $\mathbf{\hat{e}}_1$ and $\mathbf{\hat{e}}_2$ we arrive to this, but why is the matrix not equal to the transformation $L_{ij} = S^{T}_{ij}$? I dont understand, since i see the matrix $S_{ij}$ as a change of basis matrix.
Another problem with this notation appears when i impose that the norm of a vector must remain constant. So $$v_i v_i = v'_i v'_i$$ And since $v'_i = L_{ij}v_j$, we could substitute this in the equation above. But the result actually is $$v_i v_i = L_{ij}L_{ik}v_jv_k$$
Why do we write $L_{ij}L_{ik}v_jv_k$ instead of $L_{ij}v_jL_{ij}v_j$?
When we say that the matrix $$S = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$ represents a rotation by an angle $\theta$ in the counterclockwise direction within a plane, what we really mean is that if you have a vector $\mathbf v$ in the plane whose coordinates $(v_1,v_2)$ are given according to a certain orthonormal basis $B = \{\mathbf b_1, \mathbf b_2\},$ and you want to know the coordinates in the same basis $B$ of a vector $\mathbf v'$ of the same magnitude as $\mathbf v$ but in a direction turned at an angle $\theta$ counterclockwise from the direction of $\mathbf v,$ the answer is $$ \begin{bmatrix} v'_1 \\ v'_2 \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} (\cos\theta) v_1 - (\sin\theta) v_2 \\ (\sin\theta) v_1 + (\cos\theta) v_2 \end{bmatrix} = \begin{bmatrix} S_{11} v_1 + S_{12} v_2 \\ S_{21} v_1 + S_{22} v_2 \end{bmatrix}. $$
In other words, the coordinates of the new vector are given by the formula $$v'_i = S_{ij}vj.$$ We do not reverse the order of the subscripts from the usual order for matrix multiplication.
On the other hand, suppose we do not want to rotate the vector $\mathbf v,$ but instead we want to know the coordinates of the same vector in a new orthonormal basis $B' = \{\mathbf b'_1, \mathbf b'_2\},$ where the new basis $B'$ is rotated by an angle $\theta$ counterclockwise from the old basis $B.$ If we simply copied the coordinates of $\mathbf v$ that we found over the basis $B,$ and said that these were the coordinates of a vector over the basis $B',$ we would be describing a vector $\mathbf v'$ that is rotated an angle $\theta$ counterclockwise from the original vector $\mathbf v.$ But we do not want a new vector; we want new coordinates of the old vector.
In order to describe our old vector over the new basis, we can take the new vector $\mathbf v'$ and rotate it backward (by an angle $\theta$ clockwise) so that we get back to the original vector $\mathbf v.$ That is, we want to "undo" the undesired rotation of the vector that occurs due to using the old coordinates over a new basis. We cannot do this by multiplying by the matrix $S$; that would rotate the vector an additional angle $\theta$ counterclockwise, so we end up twice as far from the desired direction. Instead, we have to multiply by $S^{-1},$ the matrix that reverses the rotation that $S$ performs. The inverse of a rotation matrix is the transpose of the matrix, so the new coordinates of the old vector are
$$ S^{-1} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = S^T \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} S_{11} & S_{21} \\ S_{12} & S_{22} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} S_{11} v_1 + S_{21} v_2 \\ S_{12} v_1 + S_{22} v_2 \end{bmatrix}, $$ and that's how we might encounter the formula $v'_i = S_{ji}v_j.$ The coordinates of $S$ are "reversed" because we do not multiply by $S$ for this calculation, but rather multiply by a (usually) different matrix whose $ij$ entry is the $ji$ entry of $S.$
As for the inner-product question, it's important to remember that the notation $v'_i = L_{ij}v_j$ represents a summation over the variable $j,$ that is, when we do this with vectors in a plane what we really have is $$ v'_i = L_{i1} v_1 + L_{i2} v_2 $$ for $i = 1$ and for $i = 2.$
When you write $v_i v_i = v'_i v'_i,$ clearly you do not mean that $v_1v_1 = v'_1v'_1$ and $v_2v_2 = v'_2v'_2,$ because in the general case those equations are not true; instead you mean that $v_1v_1 + v_2v_2 = v'_1 v'_1 + v'_2 v'_2.$
Now let's look at the $i$th term of the right-hand sum, $v'_i v'_i.$ We already know that $v'_i = L_{i1} v_1 + L_{i2} v_2.$ Therefore $$ v'_i v'_i = (L_{i1} v_1 + L_{i2} v_2)(L_{i1} v_1 + L_{i2} v_2). $$ If we expand out the right-hand side of this, we not only get the terms $L_{i1} v_1 L_{i1} v_1$ and $L_{i2} v_2 L_{i2} v_2,$ we also get the terms $L_{i1} v_1 L_{i2} v_2$ and $L_{i2} v_2 L_{i1} v_1.$ The pattern $L_{ij} v_j L_{ij} v_j$ matches only the first two terms, so it cannot give the correct sum in general. The pattern $L_{ij} v_j L_{ik} v_k$ represents all the necessary terms as long as you remember to consider all possible combinations of $(j,k)$ and not just the ones where $j=k.$ And of course since $L_{ij},$ $v_j,$ $L_{ik},$ and $v_k$ all are numbers, the multiplication commutes, and $$L_{ij} v_j L_{ik} v_k = L_{ij} L_{ik} v_j v_k.$$