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I just want to ask if $$\sum_{i=1}^n \sum_{j=1}^i |a_i \bar b_j-a_j\bar b_i|^2 = \sum_{i=1}^n \sum_{j=1}^n (a_i \bar b_j-a_j\bar b_i)(\bar a_ib_j + \bar a_j b_i)$$

is true and possibly an explanation on why the summations change from $\sum^n$ to to $\sum^i$

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  • $\begingroup$ I didn't spot the sign change in the second factor of the summand on the RHS. The summands are not equal. $\endgroup$
    – John Bentin
    Feb 10, 2015 at 21:04
  • $\begingroup$ @JohnBentin if there was not a sign change would it be equal? $\endgroup$
    – user214010
    Feb 10, 2015 at 21:33

1 Answer 1

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The answer is no. As a counterexample, take $n=2$, $a_2=b_1=1$, and $a_1=b_2=0$. Then the left-hand side equals $1$ while the right-hand side equals $0$.

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