# Scalar product and the Kronecker delta symbol

1. Aug 3, 2015

### Karol

From a text book. 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.

2. Aug 3, 2015

### Orodruin

Staff Emeritus
The relation is true regardless of whether the first or the second index is summed over.

3. Aug 4, 2015

### Karol

I know, but why?
When the second index is summed over the expression $\lambda_{ij} \lambda_{kj}$ makes sense, the 3 are the direction cosines of different, orthogonal, lines $x'_i$:
$$\sum_j \lambda_{ij} \lambda_{kj}=\left\{ \begin{array}{l} \cos^2(x'_1,x_1)+\cos^2(x'_1,x_2)+\cos^2(x'_1,x_3)=1 \\ \cos(x'_1,x_1)\cos(x'_2,x_1)+\cos(x'_1,x_2)\cos(x'_2,x_2)+\cos(x'_1,x_3)\cos(x'_2,x_3)=0 \\ \cos(x'_1,x_1)\cos(x'_3,x_1)+\cos(x'_1,x_2)\cos(x'_3,x_2)+\cos(x'_1,x_3)\cos(x'_3,x_3)=0 \\ \cos(x'_2,x_1)\cos(x'_1,x_1)+\cos(x'_2,x_2)\cos(x'_1,x_2)+\cos(x'_2,x_3)\cos(x'_1,x_3)=0 \\ \cos^2(x'_2,x_1)+\cos^2(x'_2,x_2)+\cos^2(x'_2,x_3)=1 \\ \cos(x'_2,x_1)\cos(x'_3,x_1)+\cos(x'_2,x_2)\cos(x'_3,x_2)+\cos(x'_2,x_3)\cos(x'_3,x_3)=0 \\ \cos(x'_3,x_1)\cos(x'_1,x_1)+\cos(x'_3,x_2)\cos(x'_1,x_2)+\cos(x'_3,x_3)\cos(x'_1,x_3)=0 \\ \cos(x'_3,x_1)\cos(x'_2,x_1)+\cos(x'_3,x_2)\cos(x'_2,x_2)+\cos(x'_3,x_3)\cos(x'_2,x_3)=0 \\ \cos^2(x'_3,x_1)+\cos^2(x'_3,x_2)+\cos^2(x'_3,x_3)=1 \end{array} \right.$$
But in $\sum_i \lambda_{ij} \lambda_{ik}$ the multiplication in each member is between the cosine direction of the same line, which doesn't have a meaning:
$$\sum_i \lambda_{ij} \lambda_{ik} =\left\{ \begin{array}{l} \cos^2(x'_1,x_1)+\cos^2(x'_2,x_1)+\cos^2(x'_3,x_1)=? \\ \cos(x'_1,x_1)\cos(x'_1,x_2)+\cos(x'_2,x_1)\cos(x'_2,x_2)+\cos(x'_3,x_1)\cos(x'_3,x_2)=? \\ \cos(x'_1,x_1)\cos(x'_1,x_3)+\cos(x'_2,x_1)\cos(x'_2,x_3)+\cos(x'_3,x_1)\cos(x'_3,x_3)=? \\ \cos(x'_1,x_2)\cos(x'_1,x_1)+\cos(x'_2,x_2)\cos(x'_2,x_1)+\cos(x'_3,x_2)\cos(x'_3,x_1)=? \\ \cos^2(x'_1,x_2)+\cos^2(x'_2,x_2)+\cos^2(x'_3,x_2)=? \\ \cos(x'_1,x_2)\cos(x'_1,x_3)+\cos(x'_2,x_2)\cos(x'_2,x_3)+\cos(x'_3,x_2)\cos(x'_3,x_3)=? \\ \cos(x'_1,x_3)\cos(x'_1,x_1)+\cos(x'_2,x_3)\cos(x'_2,x_1)+\cos(x'_3,x_3)\cos(x'_3,x_1)=? \\ \cos(x'_1,x_3)\cos(x'_1,x_2)+\cos(x'_2,x_3)\cos(x'_2,x_2)+\cos(x'_3,x_3)\cos(x'_3,x_2)=? \\ \cos^2(x'_1,x_3)+\cos^2(x'_2,x_3)+\cos^2(x'_3,x_3)\cos(x'_3,x_1)=? \\ \end{array} \right.$$
This problem can be visualized. the multiplication in $\sum_j \lambda_{ij} \lambda_{kj} =\delta_{jk}$ is between 2 rows of the matrices while in $\sum_i \lambda_{ij} \lambda_{ik}$ the multiplication is between 2 columns.

4. Aug 4, 2015

### Orodruin

Staff Emeritus
Are you familiar with the fact that any matrix commutes with its inverse? In this case, the inverse is the transpose (which follows from the relation you are familiar with).

Naturally, this relation also follows from the fact that it does not matter which system you decided to call the primed system and which system you decided to call unprimed, the completeness relation must be true regardless.
This is wrong, it has a meaning. It is just the decomposition of the unprimed basis in the primed eigenvectors instead of vice versa!

5. Aug 8, 2015

### Karol

Thanks Orodruin