TITLE: Squaring Norms solved by Algebra
QUESTION [6 upvotes]: I found the following in a paper and am not sure how it is correct:
$\Vert a - b \Vert^2$
was expanded to:
$\Vert a \Vert^2 - 2a^Tb + \Vert b \Vert^2$
The paper was on gps location algorithms so $a$ and §b§ are and $\Vert a\Vert $ is the second norm of a
My question is how this assertion can be made? I'm stuck on even multiplying out the terms individually to arrive at equalt roots:
$\left((a_x+b_x)^2+(a_y+b_y)^2\right)^\frac{1}{2} = \left(a_x^2+a_y^2\right)^\frac{1}{2} +2 \left(a_xb_x+a_yb_y\right) +\left(b_x^2+b_y^2\right)^\frac{1}{2}$
Is this stamement correct?
Additionally: What Can I do if i have more complex terms including matrices for rotation or the like.

REPLY [8 votes]: The reason for this statement is that one way to write the norm for (column) vectors is
$\|c\|^2=\langle c,c\rangle=c^Tc$
Then
$\|a-b\|=\langle a-b,a-b\rangle$.
Using the linearity of the inner product, you get that 
$\langle a-b,a-b\rangle=\langle a,a\rangle+\langle a,-b\rangle+\langle -b,a\rangle+\langle -b,-b\rangle$.
Since $\langle -b,a\rangle=\langle a,-b\rangle=-\langle a,b\rangle=-a^Tb$, you have the desired result (combining all facts).