Proof of Lagrange Identity I need to prove Lagrange Identity for complex case, i.e.
$$
\left( \sum_{i=1}^n|a_i|^2 \right)\left( \sum_{i=1}^n |b_i|^2 \right)-\left| \sum_{i=1}^na_ib_i \right|^2=\sum_{1\leq i<j\leq n}|\bar{a}_ib_j-\bar{a}_jb_i|^2
$$
The proof should use summation directly and without something like vectors or induction. 
 A: \begin{align}
\left( \sum_{i=1}^n|a_i|^2 \right)\left( \sum_{i=1}^n |b_i|^2 \right)-\left| \sum_{i=1}^na_ib_i \right|^2&=\sum_{i,j=1}^n|a_i|^2|b_j|^2 -\sum_{i,j=1}^n\bar{a}_i\bar{b}_ia_jb_j
\\
&=\sum_{i,j=1, i\neq j}^n|a_i|^2|b_j|^2 +\sum_{i=1}^n|a_i|^2|b_i|^2 -\sum_{i=1}^n|a_i|^2|b_i|^2
\\
& \hspace{5 mm}-\sum_{i,j=1, i\neq j}^n\bar{a}_i\bar{b}_ia_jb_j
\\
&=\sum_{i,j=1, i\neq j}^n|a_i|^2|b_j|^2 -\sum_{i,j=1, i\neq j}^n\bar{a}_i\bar{b}_ia_jb_j
\\
&=\dfrac{1}{2}\left(\sum_{i,j=1, i\neq j}^n |a_i|^2|b_j|^2+\sum_{i,j=1, i\neq j}^n |a_j|^2|b_i|^2\right) -
\\&\quad\quad \dfrac{1}{2}\left(\sum_{i,j=1, i\neq j}^n \bar{a}_i\bar{b}_ia_jb_j+\sum_{i,j=1, i\neq j}^n \bar{a}_j\bar{b}_ja_ib_i\right)\:\text{*}
\\
&=\dfrac{1}{2}\sum_{i,j=1, i\neq j}^n(|a_i|^2|b_j|^2+|a_j|^2|b_i|^2 -\bar{a}_i\bar{b}_ia_jb_j-\bar{a}_j\bar{b}_ja_ib_i)
\\
&=\dfrac{1}{2}\sum_{i,j=1, i\neq j}^n(\bar{a}_ib_j-\bar{a}_jb_i)(a_i\bar{b}_j-a_j\bar{b}_i)
\\
&=\dfrac{1}{2}\sum_{i,j=1, i\neq j}^n|\bar{a}_ib_j-\bar{a}_jb_i|^2
\\
&=\dfrac{1}{2}\left(\sum_{1\leqslant i<j\leqslant n}|\bar{a}_ib_j-\bar{a}_jb_i|^2+\sum_{1\leqslant j<i\leqslant n}|\bar{a}_ib_j-\bar{a}_jb_i|^2\right)
\\
&=\dfrac{1}{2}\left(\sum_{1\leqslant i<j\leqslant n}|\bar{a}_ib_j-\bar{a}_jb_i|^2+\sum_{1\leqslant i<j\leqslant n}|\bar{a}_jb_i-\bar{a}_ib_j|^2\right)
\\
&=\sum_{1\leqslant i<j\leqslant n}|\bar{a}_ib_j-\bar{a}_jb_i|^2
\\
\end{align}
* Since
$$
\quad \sum \limits_{i,j=1, i\neq j}^n\bar{a}_i\bar{b}_i a_j b_j=\left| \sum_{i=1}^na_ib_i \right|^2-\sum_{i=1}^n|a_i|^2|b_i|^2
$$ 
It is real. So
$$
\overline{\sum \limits_{i,j=1, i\neq j}^n\bar{a}_i\bar{b}_ia_jb_j}=\sum \limits_{i,j=1, i\neq j}^n a_ib_i\bar{a}_j\bar{b}_j=\sum \limits_{i,j=1, i\neq j}^n\bar{a}_i\bar{b}_ia_jb_j
$$
