Understanding Inner Products

Oct 13, 2025 by magicalceylontours.com 13 views

Introduction

In the realm of inner products, the Cauchy-Schwarz inequality is a fundamental concept that plays a crucial role in various mathematical and physical applications. The inequality states that for any vectors $x$ and $y$ in a normed space $X$ equipped with an inner product $\langle \cdot,\cdot\rangle$ , the following inequality holds:

\langle x-y,y-z\rangle \leq 0

However, proving this inequality can be a challenging task, especially for those who are new to the concept of inner products. In this article, we will delve into the world of inner products and provide a step-by-step guide on how to prove the Cauchy-Schwarz inequality.

Understanding Inner Products

Before we dive into the proof, let's take a moment to understand what an inner product is. An inner product is a way of combining two vectors to produce a scalar value. It is a fundamental concept in linear algebra and is used to define the length and angle between vectors.

In a normed space $X$ , an inner product $\langle \cdot,\cdot\rangle$ satisfies the following properties:

Positive definiteness: $\langle x,x\rangle \geq 0$ for all $x\in X$ , and $\langle x,x\rangle = 0$ if and only if $x=0$ .
Linearity in the first argument: $\langle ax+by,z\rangle = a\langle x,z\rangle + b\langle y,z\rangle$ for all $x,y,z\in X$ and scalars $a,b$ .
Conjugate symmetry: $\langle x,y\rangle = \overline{\langle y,x\rangle}$ for all $x,y\in X$ .

The Cauchy-Schwarz Inequality

Now that we have a good understanding of inner products, let's focus on the Cauchy-Schwarz inequality. The inequality states that for any vectors $x$ and $y$ in a normed space $X$ equipped with an inner product $\langle \cdot,\cdot\rangle$ , the following inequality holds:

\langle x-y,y-z\rangle \leq 0

To prove this inequality, we will use the properties of inner products and some clever algebraic manipulations.

Proof of the Cauchy-Schwarz Inequality

Let $x,y,z\in X$ be arbitrary vectors. We can rewrite the inequality as:

\langle x-y,y-z\rangle = \langle x,y-z\rangle - \langle y,y-z\rangle

Using the linearity property of inner products, we can expand the first term as:

\langle x,y-z\rangle = \langle x,y\rangle - \langle x,z\rangle

Similarly, we can expand the second term as:

\langle y,y-z\rangle = \langle y,y\rangle - \langle y,z\rangle

Substituting these expressions back into the original inequality, we get:

\langle x,y\rangle - \langle x,z\rangle - \langle y,y\rangle + \langle y,z\rangle \leq 0

Now, let's focus on the first two terms. We can rewrite them as:

\langle x,y\rangle - \langle x,z\rangle = \langle x,y-z\rangle

Using the positive definiteness property of inner products, we know that $\langle x,x\rangle \geq 0$ for all $x\in X$ . We can rewrite this as:

\langle x,x\rangle = \langle x,y-z\rangle + \langle x,z-y\rangle

Using the linearity property of inner products, we can expand the second term as:

\langle x,z-y\rangle = \langle x,z\rangle - \langle x,y\rangle