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This module provides both statement and proof of the Cauchy-Schwarz inequality and discusses its practical implications with regard to the matched filter detector.

Introduction

Any treatment of linear algebra as relates to signal processing would not be complete without a discussion of the Cauchy-Schwarz ineqaulity, a relation that enables a wide array of signal procesing applications related to pattern matching through a method called the matched filter. Recall that in standard Euclidean space, the angle θ between two vectors x , y is given by

cos ( θ ) = x , y | | x | | | | y | | .

Since cos ( θ ) 1 , it follows that

| x , y | 2 x , x y , y .

Furthermore, equality holds if and only if cos ( θ ) = 0 , implying that

| x , y | 2 = x , x y , y

if and only if y = a x for some real a . This relation can be extended to all inner product spaces over a real or complex field and is known as the Cauchy-Schwarz inequality, which is of great importance to the study of signals.

The cauchy-schwarz inequality

Statement of the cauchy-schwarz inequality

The general statement of the Cauchy-Schwarz inequality mirrors the intuition for standard Euclidean space. Let V be an inner product space over the field of complex numbers C with inner product · , · . For every pair of vectors x , y V the inequality

| x , y | 2 x , x y , y

holds. Furthermore, the equality

| x , y | 2 = x , x y , y

holds if and only if y = a x for some a C . That is, equality holds if and only if x and y are linearly dependent.

Proof of the cauchy-schwarz inequality

Let V be a vector space over the real or complex field F , and let x , y V be given. In order to prove the Cauchy-Schwarz inequality, it will first be proven that | x , y | 2 = x , x y , y if y = a x for some a F . It will then be shown that | x , y | 2 < x , x y , y if y a x for all a F .

Consider the case in which y = a x for some a F . From the properties of inner products, it is clear that

| x , y | 2 = | x , a x | 2 = | a ¯ x , x | 2 .

Hence, it follows that

| x , y | 2 = | a ¯ | 2 | x , x | 2 = | a | 2 x , x 2 .

Similarly, it is clear that

x , x y , y = x , x a x , a x = x , x a a ¯ x , x = | a | 2 x , x 2 .

Thus, it is proven that | x , y | 2 = x , x y , y if x = a y for some a F .

Next, consider the case in which y a x for all a F , which implies that y 0 so y , y 0 . Thus, it follows by the properties of inner products that, for all a F , x - a y , x - a y > 0 . This can be expanded using the properties of inner products to the expression

x - a y , x - a y = x , x - a y - a y , x - a y = x , x - a ¯ x , y - a y , x + | a | 2 y , y

Choosing a = x , y y , y ,

x - a y , x - a y = x , x - y , x y , y x , y - x , y y , y y , x + x , y y , x y , y 2 y , y = x , x - x , y y , x y , y

Hence, it follows that x , x - x , y y , x y , y > 0 . Consequently, x , x y , y - x , y x , y ¯ > 0 . Thus, it can be concluded that | x , y | 2 < x , x y , y if y a x for all a F .

Therefore, the inequality

| x , y | 2 x , x y , y

holds for all x , y V , and equality

| x , y | 2 = x , x y , y

holds if and only if y = a x for some a F .

Additional mathematical implications

Consider the maximization of x | | x | | , y | | y | | where the norm | | · | | = · , · is induced by the inner product. By the Cauchy-Schwarz inequality, we know that x | | x | | , y | | y | | 2 1 and that x | | x | | , y | | y | | 2 = 1 if and only if y | | y | | = a x | | x | | for some a C . Hence, x | | x | | , y | | y | | attains a maximum where y | | y | | = a x | | x | | for some a C . Thus, collecting the scalar variables, x | | x | | , y | | y | | attains a maximum where y = a x . This result will be particulaly useful in developing the matched filter detector techniques.

Questions & Answers

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Joy Reply
Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
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To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
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8m/s²
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Someone
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Source:  OpenStax, Signals and systems. OpenStax CNX. Aug 14, 2014 Download for free at http://legacy.cnx.org/content/col10064/1.15
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