- #1
Karol
- 1,380
- 22
From a textbook. proof that the scalar product ##A\centerdot B## is a scalar:
Vectors A' and B' are formed by rotating vectors A and B:
$$A'_i=\sum_j \lambda_{ij} A_j,\; B'_i=\sum_j \lambda_{ij} B_j$$
$$A' \centerdot B'=\sum_i A'_i B'_i =\sum_i \left( \sum_j \lambda_{ij} A_j \right)\left( \sum_k \lambda_{ik} B_k \right)$$
$$=\sum_{i,k} \left( \sum_{i} \lambda_{ij} \lambda_{ik} \right) A_j B_k=\sum_{j} \left( \sum_{k} \delta_{jk} A_j B_k \right) $$
But:
$$\sum_j \lambda_{ij} \lambda_{kj} =\delta_{jk} $$
The order of the indexes in ##\sum_i \lambda_{ij} \lambda_{ik}## is inverse.
$$\lambda_{ij}=\cos(x'_i,x_j)$$
And, for example, for i=2:
$$\lambda^2_{21}+\lambda^2_{22}+\lambda^2_{23}=1$$
The identical index i=2 comes first, while in ##\sum_{i} \lambda_{ij} \lambda_{ik}## the (should be, or not to be) identical indexes j and k are second and the first index, i, changes.
Vectors A' and B' are formed by rotating vectors A and B:
$$A'_i=\sum_j \lambda_{ij} A_j,\; B'_i=\sum_j \lambda_{ij} B_j$$
$$A' \centerdot B'=\sum_i A'_i B'_i =\sum_i \left( \sum_j \lambda_{ij} A_j \right)\left( \sum_k \lambda_{ik} B_k \right)$$
$$=\sum_{i,k} \left( \sum_{i} \lambda_{ij} \lambda_{ik} \right) A_j B_k=\sum_{j} \left( \sum_{k} \delta_{jk} A_j B_k \right) $$
But:
$$\sum_j \lambda_{ij} \lambda_{kj} =\delta_{jk} $$
The order of the indexes in ##\sum_i \lambda_{ij} \lambda_{ik}## is inverse.
$$\lambda_{ij}=\cos(x'_i,x_j)$$
And, for example, for i=2:
$$\lambda^2_{21}+\lambda^2_{22}+\lambda^2_{23}=1$$
The identical index i=2 comes first, while in ##\sum_{i} \lambda_{ij} \lambda_{ik}## the (should be, or not to be) identical indexes j and k are second and the first index, i, changes.