Square of Vector Dot Product
Mia Lopez 
It is given in my textbook that, given two vectors x and z:
is equal to
How does this expansion happen
$\endgroup$2 Answers
$\begingroup$Let $x,z \in \mathbb{R}^n$, so we can write $x = (x_1,x_2,...,x_n)$ and $z=(z_1,...,z_n)$. Hence,
$$(x\cdot z)^2= (x_1z_1 + ....+x_nz_n)(x_1z_1 + ....+x_nz_n) $$
Notice that as we expand the sum, we get in the first monomial, second monomial, and so on the following
$$ (x_1z_1)^2 + x_1z_1 x_2 z_2 +x_1z_1 x_3 z_3 + ... + x_1z_1 x_n z_n $$
$$ x_2z_2 x_1 z_1 + (x_2z_2)^2 + x_2z_2 x_3 z_3 + ... + x_2z_1 x_n z_n $$
......
$$ x_nz_n x_1 z_1 + x_nz_n x_2 z_2 + x_nz_n x_3 z_3 + ... + (x_n z_n)^2 $$
Add the diagonals first and we obtain
$$ \sum_{i=1}^n x^2_iz^2_i $$
now, observe that the lower and upper triangular part of the array above are equal and so we are addings terms in the forsm $2 x_iz_i x_j z_j $. Can you finish it?
$\endgroup$ 1 $\begingroup$$$X.Z = x_1z_1 + x_2 z_2 +....+x_n z_n$$
$$(X.Z)^2 = ( x_1z_1 + x_2 z_2 +....+x_n z_n)^2$$
$$= (x_1^2z_1^2 +x_2^2z_2^2 +...+x_2^2z_n^2) +$$$$ 2(x_1z_1x_2 z_2+x_1z_1x_3 z_3+....+$$$$x_{(n-1)} z_{(n-1)} x_n z_n)$$
$\endgroup$
