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From Schaum's Outline to Tensor Calculus Chapter 1, Example 1.8 —
$a_{\large{ij}}(x_{\large{i}} + y_{\large{j}}) \neq a_{\large{ij}}x_{\large{i}} + a_{\large{ij}}y_{\large{j}}$.

I was trying to grok why Einstein summation turned this into a NONidentity without writing out the sums explicitly. I failed to do this so I tried writing out $ i= j = 2$.

LHS = $\sum_{i = 1}^2\sum_{j = 1}^2 a_{\large{ij}}(x_{\large{i}} + y_{\large{j}})$ = $\sum_{i = 1}^2 a_{\large{i1}}(x_{\large{i}} + y_{\large{1}}) + a_{\large{i2}}(x_{\large{i}} + y_{\large{2}}) $
$= \left[a_{\large{11}}(x_{\large{1}} + y_{\large{1}}) + a_{\large{11}}(x_{\large{1}} + y_{\large{1}})\right]_{\large{i = 1}} + \left[ a_{\large{21}}(x_{\large{2}} + y_{\large{1}}) + a_{\large{22}}(x_{\large{2}} + y_{\large{2}})\right]_{\large{i = 2}} $

RHS = $\sum_{i = 1}^2\sum_{j = 1}^2 a_{\large{ij}}x_{\large{i}} + a_{\large{ij}}y_{\large{j}}$ = $\sum_{i = 1}^2 a_{\large{i1}}x_{\large{i}} + a_{\large{i1}}y_{\large{1}} + a_{\large{i2}}x_{\large{i}} + a_{\large{i2}}y_{\large{2}}$ $= \left[a_{\large{11}}(x_{\large{1}} + y_{\large{1}}) + a_{\large{12}}(x_{\large{1}} + y_{\large{2}})\right]_{\large{i = 1}} + \left[ a_{\large{21}}(x_{\large{2}} + y_{\large{1}}) + a_{\large{22}}(x_{\large{2}} + y_{\large{2}})\right]_{\large{i = 2}} $ = LHS?

What went amiss? Further, how would you fathom that Einstein summation razes equality without writing out anything?

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    I think the RHS is interpreted as $\left(\sum_i a_{ij}x_i\right) + \left(\sum_j a_{ij}y_j\right)$. But I never understood the Einstein summation notation. (If I think right, that's one reason why I never understood it.) – Daniel Fischer Jul 07 '13 at 15:13

1 Answers1

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If you define

$$b_{ij}:=x_i+y_j,$$

the l.h.s. of your relation is equivalent to

$$a_{ij}b_{ij}:=\sum_{i,j}a_{ij}b_{ij},$$

due to Einstein summation convention. In other words, the l.h.s. of the given expression is a scalar. The r.h.s. is

$$a_{ij}x_i+a_{ij}y_j=\sum_{i}a_{ij}x_i+\sum_{j}a_{ij}y_j=(\text{we can change the name of repeated indices})=\sum_{k}a_{kj}x_k+\sum_{l}a_{il}y_l :=q_{ji},$$

i.e. the $(j,i)$-component of a matrix (called $q$), instead.

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