Page 276 - Matrix Analysis & Applied Linear Algebra

272 Chapter 5 Norms, Inner Products, and Orthogonality

Cauchy–Bunyakovskii–Schwarz (CBS) Inequality

∗ n×1
|x y|≤ x y for all x, y ∈C . (5.1.3)
∗
Equality holds if and only if y = αx for α = x y/x x.
∗
2
∗
∗
∗
Proof. Set α = x y/x x = x y/ x (assume x = 0 because there is nothing
to prove if x = 0) and observe that x (αx − y)=0, so
∗
2 ∗
∗
0 ≤ αx − y =(αx − y) (αx − y)= ¯αx (αx − y) − y (αx − y)
∗
2 2
y x − (x y)(y x) (5.1.4)
∗
∗
∗
∗
∗
= −y (αx − y)= y y − αy x = .
2
x
2
Since y x = x y, it follows that (x y)(y x)= |x y| , so
∗
∗
∗
∗
∗
2 2 2
∗
y x −|x y|
0 ≤ .
2
x
2 2 2 2
∗
Now, 0 < x implies 0 ≤ y x −|x y| , and thus the CBS inequality
is obtained. Establishing the conditions for equality is Exercise 5.1.9.
One reason that the CBS inequality is important is because it helps to
establish that the geometry in higher-dimensional spaces is consistent with the
3
2
geometry in the visual spaces and . In particular, consider the situation
depicted in Figure 5.1.3.
x + y
y
||x + y|| ||y||
||x|| x
Figure 5.1.3
Imagine traveling from the origin to the point x and then moving from x to the
point x+y. Clearly, you have traveled a distance that is at least as great as the
direct distance from the origin to x+y along the diagonal of the parallelogram.
In other words, it’s visually evident that x + y ≤ x + y . This observation