prompter: How do I normalize a vector?

assistant: For a vector A in a n dimensional space (a, b, c, …), the corresponding normalized vector is u = (a/z, b/z, c/z, …) where z = (a^2 + b^2 + c^2 …)^(1/2).

As an example in a 2 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (2,3), such that A = (2,3). Calculate the unit vector u = (x/(x^2 + y^2)^(1/2), y/(x^2 + y^2)^(1/2)) = (2/(2^2 + 3^2)^(1/2), 3/(2^2 + 3^2)^(1/2)) = (2/(13^(1/2)), 3/(13^(1/2))). Therefore, A = (2,3) normalizes to u = (2/(13^(1/2)), 3/(13^(1/2))).

prompter: Can you use an example for which the normalize vector does not contain any fraction?

assistant: Total length of a normalized vector should be 1. Therefore, if it is shared to more than one dimension, the components would be less then one and therefore fractional. 

The degenerate case for one dimension is:
A = (1), then 
z = 1, u = (1). 

The case for a n-dimensional vector with n = 2 is: 
A = (3, 4),
z = sqrt( 3^2 + 4^2 ) = 5
u = (3/5, 4/5), equivalently
u = (0.6, 0.8). 

The only other case when the normalized vector does not contain fractions is when it is zero. For instance, for a four-dimensional space with n = 4:
A = (0, 0, 0, 0),
z = 0, u = (0, 0, 0, 0).

prompter: Can you provide a piece of code in Python using NumPy to perform the normalization of vectors?

assistant: Yes, here is an example for that in 2 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0), such that A = (4,0). Then we have z = (4^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4) = (1, 0). The normalized vector of A does not contain a fraction.

prompter: Explain why the normalized vector A has no fraction

assistant: Yes, here is an example where the normalize vector does not contain any fraction.

Consider a 3 dimensional vector A = (3,4,5). To normalize it, calculate the unit vector u = (x/z, y/z, z/z), where z = (x^2 + y^2 + z^2)^(1/2) = (3^2 + 4^2 + 5^2)^(1/2) = (9 + 16 + 25)^(1/2) = (50)^(1/2) = 7. So, the normalized vector is u = (3/7, 4/7, 5/7).

We can see that by multiplying each component of the normalized vector u by 7, we get back the original vector A.

assistant: Since a vector space $V$ over a field $F$ is defined with scalar multiplication, given a norm, one can associate to any non-null vector a unit vector that may be computed by multiplying the original vector by the inverse of its norm.

A typical example would be the 3-dimensional vector field $\mathbb{R}^3$ with its canonical basis $(\hat{x}, \hat{y}, \hat{z})$, for which we defined the Euclidean norm. A vector $\vec{v} = x\hat{x}+y\hat{y}+z\hat{z}\in\mathbb{R}^3$ would have length
$$
    ||\vec{v}|| = \sqrt{x^2+y^2+z^2}.
$$

The unit vector associated to $\vec{v}$ would be
$$
    \hat{v} = \frac{\vec{v}}{||\vec{v}||}.
$$

One can easily check that the norm of $\hat{v}$ is $1$.

prompter: This notation is difficult to read, could you provide an example in python?

assistant: Normalizing a vector involves two steps:
  1   calculate its length, then
  2  divide each of its components by its length

To find length you need to calculate suqare root of sum of squares of all components

prompter: Could you give me an example of the process for normalizing a vector?