2.2: Orthonormal bases in ℝ² - Math 124 Course Notes

import numpy as np
import plotly.graph_objects as go
from pathlib import Path
import sys

_notes_root = next(p for p in [Path.cwd(), *Path.cwd().parents] if (p / 'myst.yml').exists())
if str(_notes_root) not in sys.path:
    sys.path.insert(0, str(_notes_root))
from plot_style import style_plotly

BLUE, ORANGE, PINK = '#3d81f6', 'orange', '#d81a60'
ANNOTATION_SIZE = 24


def base2(xrange=(-5, 6), yrange=(-5, 5), width=700, height=580, tick=1):
    fig = style_plotly(go.Figure(), renderer='png')
    fig.update_layout(width=width, height=height, showlegend=False,
                      font=dict(size=17), margin=dict(l=55, r=30, t=30, b=50))
    fig.update_xaxes(title='x', range=xrange, dtick=tick, zeroline=True, zerolinewidth=1.5,
                     constrain='domain', scaleanchor='y', scaleratio=1)
    fig.update_yaxes(title='y', range=yrange, dtick=tick, zeroline=True, zerolinewidth=1.5,
                     constrain='domain')
    return fig


def line2(fig, start, end, color=BLUE, dash='solid', width=2):
    fig.add_trace(go.Scatter(x=[start[0], end[0]], y=[start[1], end[1]], mode='lines',
                            line=dict(color=color, width=width, dash=dash),
                            hoverinfo='skip', showlegend=False))


def label2(fig, xy, label, dx=0, dy=0, color='black'):
    # MathJax needs its own size command when exporting math labels to PNG.
    if label.startswith('$') and label.endswith('$'):
        label = r'$\Large ' + label[1:]
    fig.add_annotation(x=float(xy[0])+dx, y=float(xy[1])+dy, text=label, showarrow=False,
                       font=dict(family='Palatino', color=color, size=ANNOTATION_SIZE))


def point2(fig, xy, label=None, dx=0.35, dy=0.35):
    fig.add_trace(go.Scatter(x=[xy[0]], y=[xy[1]], mode='markers',
                            marker=dict(color='black', size=7), hoverinfo='skip'))
    if label: label2(fig, xy, '$' + label.replace('−', '-') + '$', dx, dy)


def vec2(fig, end, label=None, color=BLUE, start=(0, 0), dx=0.3, dy=0.3, width=4):
    # Match Chapter 1.4: a colored shaft, arrowhead, and midpoint math label.
    line2(fig, start, end, color=color, width=width)
    fig.add_annotation(x=float(end[0]), y=float(end[1]), ax=float(start[0]), ay=float(start[1]),
                       xref='x', yref='y', axref='x', ayref='y', showarrow=True,
                       text='', arrowhead=3, arrowsize=1.2, arrowwidth=3, arrowcolor=color)
    if label:
        midpoint = (np.asarray(start, dtype=float) + np.asarray(end, dtype=float)) / 2
        label2(fig, midpoint, label, dx, dy, color=color)


def square2(fig, corner, along, perp, size=0.3):
    corner=np.asarray(corner,dtype=float)
    u=np.asarray(along,dtype=float); u=u/np.linalg.norm(u)*size
    v=np.asarray(perp,dtype=float); v=v/np.linalg.norm(v)*size
    line2(fig,corner+u,corner+u+v,'gray',width=1.5)
    line2(fig,corner+v,corner+u+v,'gray',width=1.5)

In Chapter 1.4, we wrote vectors as linear combinations of the standard basis vectors. Could we use a different pair of directions? In this section, we’ll see why perpendicular unit vectors make this especially convenient.

The standard basis¶

Recall the standard basis vectors from Chapter 1.4:

\vec e_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad \vec e_2= \begin{bmatrix} 0\\ 1 \end{bmatrix}.

Every vector in $\mathbb R^2$ is a linear combination of these two vectors. Its entries give the coefficients.

fig=base2((-0.5,1.7),(-0.5,1.7),width=520,height=520,tick=0.5)
vec2(fig,(1,0),r"$\vec e_1$",BLUE,dx=0.1,dy=0.16,width=4)
vec2(fig,(0,1),r"$\vec e_2$",ORANGE,dx=0.17,dy=0.08,width=4)
square2(fig,(0,0),(1,0),(0,1),0.13)
fig.show(scale=2)

For example,

\begin{bmatrix} 5\\ 7 \end{bmatrix} = \begin{bmatrix} 5\\ 0 \end{bmatrix} + \begin{bmatrix} 0\\ 7 \end{bmatrix} = 5 \begin{bmatrix} 1\\ 0 \end{bmatrix} + 7 \begin{bmatrix} 0\\ 1 \end{bmatrix} = 5\vec e_1+7\vec e_2.

To do geometry in $\mathbb{R}^2$ , it is useful to consider other such pairs of vectors. The key properties of $(\vec e_1,\vec e_2)$ are:

length one,
orthogonal to each other.

These two properties will let us find coefficients using dot products, even when the directions are tilted.

Recall from Chapter 1.5 that a unit vector has length one. We can turn any nonzero vector $\vec w$ into a unit vector by dividing by its length: $\vec w/\lVert\vec w\rVert$ .

Building an orthonormal basis¶

First, let’s check the lengths of two vectors.

The vector

\begin{bmatrix} \frac45\\[2pt] \frac35 \end{bmatrix}

has length one since

\sqrt{\left(\frac45\right)^2+\left(\frac35\right)^2} = \sqrt{\frac{16}{25}+\frac{9}{25}} = 1.

A unit vector can point in a different direction. For example, the vector

\begin{bmatrix} \frac{1}{\sqrt{2}}\\[2pt] -\frac{1}{\sqrt{2}} \end{bmatrix}

has length one since

\sqrt{ \left(\frac{1}{\sqrt{2}}\right)^2+ \left(-\frac{1}{\sqrt{2}}\right)^2 } = \sqrt{\frac12+\frac12} = 1.

fig=base2((-1.4,1.5),(-1.4,1.5),width=580,height=580,tick=0.5)
t=np.linspace(0,2*np.pi,240)
fig.add_trace(go.Scatter(x=np.cos(t),y=np.sin(t),mode='lines',line=dict(color='#9ca3af',width=2,dash='dash'),hoverinfo='skip'))
vec2(fig,(4/5,3/5),r"$\vec u$",BLUE,dx=-0.08,dy=0.24,width=4)
vec2(fig,(1/np.sqrt(2),-1/np.sqrt(2)),r"$\vec v$",ORANGE,dx=0.1,dy=-0.22,width=4)
fig.show(scale=2)

To make an orthonormal basis, we need two unit vectors that are also perpendicular. Let

\vec u_1= \begin{bmatrix} \frac45\\[2pt] \frac35 \end{bmatrix}, \qquad \vec u_2= \begin{bmatrix} -\frac35\\[2pt] \frac45 \end{bmatrix}.

Both vectors have length one, and their dot product is

\vec u_1\cdot\vec u_2=\frac45\left(-\frac35\right)+\frac35\left(\frac45\right)=0.

By Chapter 1.6, they are orthogonal. This pair is an orthonormal basis of $\mathbb R^2$ .

fig=base2((-1.5,1.5),(-1.3,1.5),width=560,height=560,tick=0.5)
u1=np.array([4/5,3/5]); u2=np.array([-3/5,4/5])
vec2(fig,u1,r"$\vec u_1$",BLUE,dx=-0.05,dy=0.24,width=4)
vec2(fig,u2,r"$\vec u_2$",ORANGE,dx=-0.24,dy=0.0,width=4)
square2(fig,(0,0),u1,u2,0.14)
fig.show(scale=2)

We can choose a different pair of perpendicular unit directions. Let

\vec w_1= \begin{bmatrix} -\frac{1}{\sqrt{2}}\\[2pt] \frac{1}{\sqrt{2}} \end{bmatrix}, \qquad \vec w_2= \begin{bmatrix} -\frac{1}{\sqrt{2}}\\[2pt] -\frac{1}{\sqrt{2}} \end{bmatrix}.

Each vector has squared length $1/2+1/2=1$ , and

\vec w_1\cdot\vec w_2=\frac12-\frac12=0.

So $(\vec w_1,\vec w_2)$ is also an orthonormal basis. There are many choices of orthonormal basis, just as there are many choices of direction vector for a line.

fig=base2((-1.5,1.4),(-1.5,1.4),width=560,height=560,tick=0.5)
w1=np.array([-1,1])/np.sqrt(2); w2=np.array([-1,-1])/np.sqrt(2)
vec2(fig,w1,r"$\vec w_1$",BLUE,dx=0.1,dy=0.24,width=4)
vec2(fig,w2,r"$\vec w_2$",ORANGE,dx=-0.05,dy=-0.28,width=4)
square2(fig,(0,0),w1,w2,0.14)
fig.show(scale=2)

Finding coefficients using dot products¶

An orthonormal basis gives us two perpendicular directions. Every vector can be written uniquely as a sum of multiples of those directions. How can we find the two coefficients?

Let $\vec u_1,\vec u_2$ be an orthonormal basis of $\mathbb{R}^2$ .

Let $\vec v$ be any vector in $\mathbb{R}^2$ .

Then

\vec v=a\vec u_1+b\vec u_2

for some scalars $a,b$ . The figure below shows this decomposition by completing a rectangle.

fig=base2((-3,5.5),(-1.5,7),width=680,height=660)
u1=np.array([4/5,3/5]); u2=np.array([-3/5,4/5])
a,b=5,3
p=a*u1; q=b*u2; v=p+q
line2(fig,p,v,'gray','dash'); line2(fig,q,v,'gray','dash')
vec2(fig,p,r"$a\vec u_1$",BLUE,dx=0.5,dy=-0.3)
vec2(fig,q,r"$b\vec u_2$",ORANGE,dx=-0.55,dy=-0.1)
vec2(fig,v,r"$\vec v$",PINK,dx=-0.45,dy=0.2)
square2(fig,(0,0),u1,u2,0.35)
vec2(fig,u1,r"$\vec u_1$",BLUE,dx=0.35,dy=-0.3)
vec2(fig,u2,r"$\vec u_2$",ORANGE,dx=-0.4,dy=0.15)
fig.show(scale=2)

But how do we find $a,b$ ?

The distributive and scalar-multiplication properties from Chapter 1.6 let us isolate one coefficient at a time. Take the dot product with $\vec u_1$ :

\vec v\cdot \vec u_1 = (a\vec u_1+b\vec u_2)\cdot \vec u_1 = a(\vec u_1\cdot \vec u_1) + b(\vec u_2\cdot \vec u_1).

Since $\vec u_1$ is a unit vector,

\vec u_1\cdot \vec u_1=1.

Since $\vec u_1$ and $\vec u_2$ are orthogonal,

\vec u_2\cdot \vec u_1=0.

Therefore

\vec v\cdot \vec u_1=a.

Similarly,

b=\vec v\cdot \vec u_2.

So

\vec v = (\vec v\cdot \vec u_1)\vec u_1 + (\vec v\cdot \vec u_2)\vec u_2.

Representing the same vector in different bases¶

Let

\vec v= \begin{bmatrix} 7\\ 5 \end{bmatrix}.

Using the standard basis¶

For the standard basis

(\vec e_1,\vec e_2) = \left( \begin{bmatrix} 1\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1 \end{bmatrix} \right),

we have

\vec v\cdot \vec e_1 = \begin{bmatrix} 7\\ 5 \end{bmatrix} \cdot \begin{bmatrix} 1\\ 0 \end{bmatrix} = 7,

and

\vec v\cdot \vec e_2 = \begin{bmatrix} 7\\ 5 \end{bmatrix} \cdot \begin{bmatrix} 0\\ 1 \end{bmatrix} = 5.

Therefore

(\vec v\cdot \vec e_1)\vec e_1 + (\vec v\cdot \vec e_2)\vec e_2 = 7\vec e_1+5\vec e_2 = \begin{bmatrix} 7\\ 0 \end{bmatrix} + \begin{bmatrix} 0\\ 5 \end{bmatrix} = \begin{bmatrix} 7\\ 5 \end{bmatrix}.

Using a different orthonormal basis¶

Now take the orthonormal basis

(\vec u_1,\vec u_2) = \left( \begin{bmatrix} \frac45\\[2pt] \frac35 \end{bmatrix}, \begin{bmatrix} -\frac35\\[2pt] \frac45 \end{bmatrix} \right).

Then

\vec v\cdot \vec u_1 = \begin{bmatrix} 7\\ 5 \end{bmatrix} \cdot \begin{bmatrix} \frac45\\[2pt] \frac35 \end{bmatrix} = \frac{28}{5}+\frac{15}{5} = \frac{43}{5}.

Also,

\vec v\cdot \vec u_2 = \begin{bmatrix} 7\\ 5 \end{bmatrix} \cdot \begin{bmatrix} -\frac35\\[2pt] \frac45 \end{bmatrix} = -\frac{21}{5}+\frac{20}{5} = -\frac15.

So

\vec v = \frac{43}{5}\vec u_1 - \frac15 \vec u_2.

The coefficient $-1/5$ means the second component points opposite to $\vec u_2$ . The vector itself is still $\begin{bmatrix}7\\5\end{bmatrix}$ .

Activity 1

Continuing the example, suppose

(\vec w_1,\vec w_2) = \left( \begin{bmatrix} -\frac{1}{\sqrt{2}}\\[2pt] \frac{1}{\sqrt{2}} \end{bmatrix}, \begin{bmatrix} -\frac{1}{\sqrt{2}}\\[2pt] -\frac{1}{\sqrt{2}} \end{bmatrix} \right).

Let

\vec v= \begin{bmatrix} 7\\ 5 \end{bmatrix}.

Find the coefficients of $\vec v$ in this orthonormal basis. What changes, and what stays the same?

Solution

\vec v\cdot \vec w_1 = \begin{bmatrix} 7\\ 5 \end{bmatrix} \cdot \begin{bmatrix} -\frac{1}{\sqrt{2}}\\[2pt] \frac{1}{\sqrt{2}} \end{bmatrix} = -\frac{7}{\sqrt{2}}+\frac{5}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}.

Also,

\vec v\cdot \vec w_2 = \begin{bmatrix} 7\\ 5 \end{bmatrix} \cdot \begin{bmatrix} -\frac{1}{\sqrt{2}}\\[2pt] -\frac{1}{\sqrt{2}} \end{bmatrix} = -\frac{7}{\sqrt{2}}-\frac{5}{\sqrt{2}} = -\frac{12}{\sqrt{2}} = -6\sqrt{2}.

So

\vec v = -\sqrt{2}\,\vec w_1 - 6\sqrt{2}\,\vec w_2.

The vector is the same in all three cases. Its coefficients change because we are describing it using different pairs of directions.