{:check ["true"]}
Index
Vector Spaces
Algebra
Linear Algebra¶
import numpy as np
import matplotlib.pyplot as pl
from mpl_toolkits.mplot3d import Axes3D
Vectors¶
Vectors are points in some space.
$u = \left[\begin{array}{c} 1 \\ 3 \end{array}\right]\in\mathbb{R}^2$
$v = \left[\begin{array}{c} -2 \\1 \\ 3 \end{array}\right]\in\mathbb{R}^3$
u = np.array([1,3])
v = np.array([-2, 1, 3])
pl.figure(figsize=(6,6))
pl.scatter(u[0], u[1], s=100)
pl.xlim(-5,5)
pl.ylim(-5,5)
pl.grid(True)
fig = pl.figure(figsize=(6,6))
ax = fig.gca(projection='3d')
ax.scatter(v[0], v[1], v[2], s=100)
ax.set_xlim(-5, 5)
ax.set_ylim(-5, 5)
ax.set_zlim(-5, 5);
Lines¶
An infinite line is given by:
$$ L = \{\mathbf{u} + t\cdot\mathbf{v}: t\in\mathbb{R}\} $$A finite line segment is given by
$$ L = \{\mathbf{u} + t\cdot\mathbf{v}: t\in[0,1]\} $$ts = np.linspace(0, 1, 20)
u = np.array([1, -1.5])
v = np.array([-1.5, 1])
L = np.array([u + t * v for t in ts])
fig = pl.figure(figsize=(10,5))
pl.subplot(1,2,1)
pl.grid(True)
pl.scatter(L[:,0], L[:,1])
pl.xlim(-2, 2)
pl.ylim(-2, 2);
pl.arrow(0, 0, *v, width=0.1)
pl.scatter(*u, color='r', s=200);
Planes¶
Planes are sets of points formed as linear combinations of basis vectors.
In 3D, a plane has two basis vectors.
$$ P = \{s\mathbf{u} + t\mathbf{v}: s, t\in\mathbb{R}\}$$For higher dimensions, we will need more basis vectors to describe a plane... unless we use a normal vector.
u = np.array([1, 2, 0.])
v = np.array([1, 2, 1])
S = np.linspace(-1, 1, 10)
T = np.linspace(-1, 1, 10)
P = np.array([s*u + t*v for s in S for t in T])
fig = pl.figure(figsize=(10, 10))
ax = fig.gca(projection='3d')
for point in P:
ax.scatter(*point, color='blue')
Normal Vectors of Planes¶
It turns out that a plane can be described by a single vector in any dimension.
Given a vector $\mathbf{w}\in\mathbb{R}^3$, we can define the plane as any vector that is orthogonal to $\mathbf{w}$.
$$P = \{ \mathbf{x}\in\mathbb{R}^3: \left<\mathbf{x}, \mathbf{w}\right> = 0\} $$Getting the normal vector¶
From the basis vectors of $P$, we can get its normal vector using the cross product:
$$ \mathbf{w} = \mathbf{u}\times \mathbf{v} $$Getting the basis vectors¶
If we have the normal vector $\mathbf{w}\mathbb{R}^n$, we can compute the basis vectors by solving the equations:
$\left<\mathbf{w},\mathbf{u}\right>$ = 0
$\left<\mathbf{w},\mathbf{v}\right>$ = 0
w = np.cross(u, v)
w
# Let's check points in P
P @ w