Einstein Summation: How do I show $a_{ij} (x_i + y_j) \not= a_{ij}x_i + a_{ij}y_j $?

Question

Einstein Summation: How do I show $a_{ij} (x_i + y_j) \not= a_{ij}x_i + a_{ij}y_j $?

Answer 1

Parsing the much-maligned notation yields $$\sum_{i,j} a_{ij} (x_i + y_j) \ne \sum_i a_{ij}x_i + \sum_j a_{ij}y_j$$ which is an evident inequality.

Answer 2

I will assume $x=(x_i)$ to be a $1\times N$ (row) vector, $y=(y_i)$ an $N \times 1$ (column) vector, and $A=(a_{ij})$ an $N\times N$ matrix. Then the terms on the RHS correspond to the matrix multiplications $x_i a_{ij} =(x A)_j$ and $a_{ij}y_j = (A y)_j$. Note that these are the components of a row vector and a column vector respectively.

What about the LHS? At the level of algebra, the equality seems correct since we can factor out $a_{ij}$ from both terms. But this amounts to adding the $j$th components of a row vector and a column vector; they do not correspond to addition of two vectors. Hence the issue boils down to $x_i+y_j$ not being a meaningful vector quantity.