Complex Numbers – Lagrange’s Identity in Complex Form

Question

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
$$
\lvert a_i\bar{b}_j – a_j\bar{b}_i\rvert^2 = \lvert a_i\rvert^2\lvert b_j\rvert^2 + \lvert a_j\rvert^2\lvert b_i\rvert^2 – a_i\bar{a}_jb_i\bar{b}_j – a_j\bar{a}_ib_j\bar{b}_i.
$$

Answer 1

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 in two sums, one where the indexes agree 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}