I am trying to show Lagrange's identity in the complex form; that is, $$ \Bigl\lvert\sum_{i = 1}^na_ib_i\Bigr\rvert^2 = \sum_{i = 1}^n\lvert a_i\rvert^2\sum_{i = 1}^n\lvert b_i\rvert^2 - \sum_{1\leq i\lt j\leq n} \lvert a_i\bar{b}_j - a_j\bar{b}_i\rvert^2 $$ Then \begin{align} \Bigl\lvert\sum_{i = 1}^na_ib_i\Bigr\rvert^2 &= \Bigl(\sum_{i = 1}^na_ib_i\Bigr)\Bigl(\sum_{j = 1}^n\bar{a}_j\bar{b}_j\Bigr)\\ &= \sum_{i,j=1}^na_i\bar{a}_jb_i\bar{b}_j\\ &= \begin{aligned} a_1\bar{a}_1b_1\bar{b}_1 &+ \cdots + a_n\bar{a}_1b_n\bar{b}_1+\\ \vdots\phantom{....} & \phantom{...}\ddots\phantom{......}\vdots\phantom{.....}+\\ a_1\bar{a}_nb_1\bar{b}_n &+ \cdots + a_n\bar{a}_nb_n\bar{b}_n \end{aligned} \end{align} I see that down the diagonals I will have $\sum_{i=1}^n\lvert a_i\rvert^2\sum_{i=1}^n\lvert b_i\rvert^2$ which leaves me with $$ \sum_{1\leq i\leq j\leq n}a_i\bar{a}_jb_i\bar{b}_j+a_j\bar{a}_ib_j\bar{b}_i\tag{1} $$ but I don't see how equation $(1)$ is equal to $$ \lvert a_i\bar{b}_j - a_j\bar{b}_i\rvert^2 = \lvert a_i\rvert^2\lvert b_j\rvert^2 + \lvert a_j\rvert^2\lvert b_i\rvert^2 - a_i\bar{a}_jb_i\bar{b}_j - a_j\bar{a}_ib_j\bar{b}_i. $$


2 Answers 2


When you sum the terms in the diagonal you don't get $$\sum_{i=1}^n |a_i|^2\sum_{j=1}^n |b_j|^2.$$

Instead, what you get is $$\sum_{i=1}^n |a_ib_i|^2.$$

Let's write $[n] = \{1,\ldots,n\}$. Here it helps to separate the sum into two sums, one where the indexes are equal and one where the indexes are different. Notice that

$$ \sum_{i=1}^n |a_i|^2\sum_{j=1}^n |b_j|^2 = \sum_{i,j\in[n]} |a_ib_j|^2 = \sum_{i=j} |a_ib_j|^2 + \sum_{i\neq j} |a_ib_j|^2 = \sum_{i=1}^n |a_ib_i|^2 + \sum_{i\neq j} |a_ib_j|^2. $$

But if $i\neq j$, of course $j\neq i$, so $$\sum_{i\neq j} |a_ib_j|^2 = \sum_{1\leq i\lt j\leq n} |a_ib_j|^2 + |a_jb_i|^2.$$

Following your line of thought \begin{align} \left|\sum_{i = 1}^na_ib_i\right|^2 &=\left(\sum_{i = 1}^na_ib_i\right)\left(\sum_{j = 1}^n\bar{a}_j\bar{b}_j\right)\\ &= \sum_{i,j\in[n]}a_i\bar{a}_jb_i\bar{b}_j\\ &= \sum_{i=1}^n |a_ib_i|^2 + \sum_{1\leq i\lt j\leq n}a_i\bar{a}_jb_i\bar{b}_j+a_j\bar{a}_ib_j\bar{b}_i \\ &= \sum_{i=1}^n |a_ib_i|^2 + \sum_{i\neq j} |a_ib_j|^2 - \sum_{i\neq j} |a_ib_j|^2 + \sum_{1\leq i\lt j\leq n}a_i\bar{a}_jb_i\bar{b}_j+a_j\bar{a}_ib_j\bar{b}_i \\ &= \sum_{i=1}^n |a_i|^2\sum_{j=1}^n |b_j|^2 - \left( \sum_{i\neq j} |a_ib_j|^2 - \sum_{1\leq i\lt j\leq n}a_i\bar{a}_jb_i\bar{b}_j+a_j\bar{a}_ib_j\bar{b}_i \right) \\ &= \sum_{i=1}^n |a_i|^2\sum_{j=1}^n |b_j|^2 - \sum_{1\leq i\lt j\leq n} |a_ib_j|^2 + |a_jb_i|^2 - a_i\bar{a}_jb_i\bar{b}_j-a_j\bar{a}_ib_j\bar{b}_i \\ &= \sum_{i=1}^n |a_i|^2\sum_{j=1}^n |b_j|^2-\sum_{1\leq i\lt j\leq n}\lvert a_i\bar{b}_j - a_j\bar{b}_i\rvert^2 \end{align}


I think you are some wrong,in fact, we have $$\left|\sum_{i=1}^{n}a_{i}b_{i}\right|^2=\Re{\left(\sum_{j=1}^{n}\sum_{k=1}^{n}a_{j}b_{j}\overline{a_{k}b_{k}}\right)}=\dfrac{1}{2}\sum_{j=1}^{n}\sum_{k=1}^{n}2\Re{(a_{j}b_{j}\overline{a_{k}b_{k}})}$$ by $2\Re(z)=z+\overline{z}$ and $|z|^2=z\overline{z}$. and doing suitable factorisations. we can get the well known indentity: $$2\Re{(a_{j}b_{j}\overline{a_{k}b_{k}})}=|a_{j}|^2|b_{k}|^2+|a_{k}|^2|b_{j}|^2-|a_{j}\overline{b_{k}}-a_{k}\overline{b_{j}}|^2$$ so \begin{align*}\left|\sum_{i=1}^{n}a_{i}b_{i}\right|^2&=\dfrac{1}{2}\sum_{j=1}^{n}\sum_{k=1}^{n}\left(|a_{j}|^2|b_{k}|^2+|a_{k}|^2|b_{j}|^2-|a_{j}\overline{b_{k}}-a_{k}\overline{b_{j}}|^2\right)\\ &=\sum_{j=1}^{n}\sum_{k=1}^{n}|a_{j}|^2|b_{k}|^2-\sum_{1\le k<j\le n}\left(|a_{j}\overline{b_{k}}-a_{k}\overline{b_{j}}|^2\right)\\ &=\left(\sum_{i=1}^{n}|a_{i}|^2\right)\left(\sum_{i=1}^{n}|b_{i}|^2\right)-\sum_{1\le k<j\le n}\left(|a_{j}\overline{b_{k}}-a_{k}\overline{b_{j}}|^2\right) \end{align*}

  • $\begingroup$ I don't see why your summation is taking the real part. We have that $\lvert z\rvert^2 = z\bar{z}$. Let $z = \sum a_ib_i$. Then we have $\bigl(\sum a_ib_i\bigr)\bigl(\sum\bar{a}_j\bar{b}_j\bigr)$. $\endgroup$
    – dustin
    Commented Jan 23, 2015 at 17:38

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