I'm having trouble understanding why this condition is true. If anyone could help me understand the following statement (whether by proof or by simple reasoning), I would greatly appreciate it.
"For general rotations, $\sum_i a_{ij}a_{ik} = \delta_{j,k}$"
Here, $\delta_{j,k}$ is the Kronecker-Delta function.
Thanks in advance
On
A rotation, expressed as a matrix, is a member of $SO(3)$, that is, it is a $3\times $ orthogonal matrix (thus the $O(3)$) with determinant $1$ (thus the $S$ standing for Special.
For an orthogonal matrix $A$, the inverse and transpose are the same $$ A^{-1} = A^T $$ which if you interpret the three columns of $R$ as the new coordinate frame axes, merely insist that these three axes be orthogonal to one another and of unit length. Multiplying by $A$ on the right, that also says $$ AA^T = I $$
Now let's say that in indices, with $A$ elements being $a_{ji}$ (and the $(k,i)$ elements of $A^T$ being $a_{ik}$). $$ \sum_i a_{ji} a_{ik} = \delta_{jk} $$
Rotating two orthogonal vectors gives you two orthogonal vectors, right?
Well, the vectors $e_j$ and $e_k$ (which are all zero except for their $j$th (resp. $k$th) entry which is a $1$) are orthogonal for $j \ne k$.
Applying a rotation $A$ to $e_j$ produces a vector with entries $a_{ij}$, $i = 1, \ldots, n$. A similar statement applies to $Ae_k$. What's the dot product of these two resulting vectors? It's exactly the thing in your sum.
For for $j \ne k$, the dot product must be 0, since they're orthogonal vectors.
For the case $j = k$, both $e_j$ adn $e_k$ are unit vectors, indeed, the same unit vector. So $Ae_j$ is a unit vector, which means that $$ 1 = \| Ae_j \|^2 = Ae_j \cdot Ae_j = Ae_j \cdot Ae_k, $$ so in the case $j = k$, your sum ends up being 1.