Square of Vector Dot Product

Question

It is given in my textbook that, given two vectors x and z:

is equal to

How does this expansion happen

Answer 1

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?

Answer 2

$$X.Z = x_1z_1 + x_2 z_2 +....+x_n z_n$$

$$(X.Z)^2 = ( x_1z_1 + x_2 z_2 +....+x_n z_n)^2$$

$$= (x_1^2z_1^2 +x_2^2z_2^2 +...+x_2^2z_n^2) +$$ $$ 2(x_1z_1x_2 z_2+x_1z_1x_3 z_3+....+$$ $$x_{(n-1)} z_{(n-1)} x_n z_n)$$