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I am looking at the Schaum's Outlines "Tensor Calculus" by David C. Kay, and on page 3, the following non-identity and identity are presented:

$$ \begin{align} a_{ij}(x_i + y_j) &\neq a_{ij} x_i + a_{ij} y_j\tag{1} \\ a_{ij}(x_j + y_j) &= a_{ij} x_j + a_{ij} y_j \tag{2} \end{align} $$

Let's assume that $i,j$ both run over $\{1,2\}$ for simplicity.

For (2), I write:

$$\begin{align} a_{ij}(x_j + y_j) &= a_{i1}(x_1+y_1)+a_{i2}(x_2+y_2) \\ &= a_{i1}x_1+a_{i2}x_2+a_{i1}y_1+a_{i2}y_2 \\ &= a_{ij}x_j+a_{ij}y_j \end{align}$$ where $i$ is a free index, and thus only $j$ is summed over.

For (1), I write: $$\begin{align} a_{ij}(x_i + y_j) &= a_{11}(x_1+y_1)+a_{12}(x_1+y_2)+a_{21}(x_2+y_1)+a_{22}(x_2+y_2) \\ &= (a_{11}+a_{12})x_1+(a_{21}+a_{22})x_2+(a_{11}+a_{21})y_1+(a_{12}+a_{22})y_2\\ &=\ (a_{i1}+a_{i2})x_i+(a_{1j}+a_{2j})y_j \end{align}$$ Is that correct? Can we go further, or is that as much as we can derive?

EDIT it seems that in (1) the inequality results from the RHS being illegal: in one term on the RHS, i is free, whereas in the other, j is free. The LHS however is legit, and the derivation I propose seems to hold, but cannot proceed any further. In (2), the situation is much better, because the index i is free in both the LHS and RHS.

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In (1) the sum $x_i + y_j$ simply doesn't make sense. If you have two vectors $x,y$ you can surely add them, but then you get $(x+y)_i = x_i + y_i$. This is true in general: you can only add tensors with matching indices. That's all there is to it.

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  • $\begingroup$ Good point - I had not seen that, being oblivious to the "vector" semantics here. $\endgroup$ – Frank Jul 9 at 15:27

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