# Squaring Norms solved by Algebra

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.

• Since the original statement has the norm squared, your exponent of $1/2$ shouldn't be there. Then, if you rewrite the right-hand side, it will be correct (remember that the square-root function isn't linear) Mar 1, 2015 at 17:03

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).

Note that, in two dimensions,

$$\|a-b\|^2 = \left((a_x-b_x)^2+(a_y-b_y)^2\right)$$

not $\left((a_x+b_x)^2+(a_y+b_y)^2\right)$ as you have written.

In general, you can write $$\|a - b\|^2 = \left<a - b, a- b\right> = \left<a, a- b\right> - \left<b, a- b\right> \\ = \left<a, a\right> - 2\left<a, b\right> + \left<b, b\right>$$

• Thank you for this answer, it has pointed out an important flaw in my trying to understand the calculation. I have marked the following answer because it pointed out the crux of my misunderstanding $\|c\|^2=\langle c,c\rangle=c^Tc$. Mar 2, 2015 at 10:35