import numpy as np import numpy.linalg as la
x = np.array([1.,2,3])
First, let's compute the 2-norm by hand:
Next, let's use
numpy machinery to compute it:
Both of the values above represent the 2-norm: $\|x\|_2$.
The $\infty$ norm represents a special case, because it's actually (in some sense) the limit of $p$-norms as $p\to\infty$.
Recall that: $\|x\|_\infty = \max(|x_1|, |x_2|, |x_3|)$.
Where does that come from? Let's try with $p=100$:
array([1.00000000e+00, 1.26765060e+30, 5.15377521e+47])
Compare to last value in vector: the addition has essentially taken the maximum:
Numpy can compute that, too:
Once you know the set of vectors for which $\|x\|=1$, you know everything about the norm, because of semilinearity. The graphical version of this is called the 'unit ball'.
We'll make a bunch of vectors in 2D (for visualization) and then scale them so that $\|x\|=1$.
alpha = np.linspace(0, 2*np.pi, 2000, endpoint=True) x = np.cos(alpha) y = np.sin(alpha) vecs = np.array([x,y]) p = 5 norms = np.sum(np.abs(vecs)**p, axis=0)**(1/p) norm_vecs = vecs/norms import matplotlib.pyplot as pt %matplotlib inline pt.grid() pt.gca().set_aspect("equal") pt.plot(norm_vecs, norm_vecs) pt.xlim([-1.5, 1.5]) pt.ylim([-1.5, 1.5])