Day 19 - Cauchy-Schwarz Inequality

The Cauchy–Schwarz Inequality uses the Dot Product to establish a fundamental inequality relating two vectors, for any vectors ($v$ and $w$) in an inner product space.

$$|v\cdot w| \le \|v\|\|w\|$$

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First, we derive this inequality utilize the property of the cosine function:

$$|\cos\theta| \le 1$$

Using the dot product formula, we can rewrite it as such.

$$\left| \frac{v\cdot w}{\|v\|\|w\|} \right| \le 1$$

Multiplying both sides by the magnitude of both vectors gives us the inequality.

$$|v\cdot w| \le \|v\|\,\|w\|$$

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The Triangle Inequality#

A direct and important application of the Cauchy–Schwarz Inequality is the Triangle Inequality, which states that for all vectors $v$ and $w$:

$$\|v + w\| \le \|v\| + \|w\|$$

The inequality originates from classical geometry, where it states that the length of one side of a triangle cannot exceed the sum of the other two sides.

Some “people” believe this to be a piece of useless mathematics; however it serves as a fundamental bridge between geometry and linear algebra.

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We start by expanding the square of the norm using the dot product:

$$\begin{aligned} \|v + w\|^{2} &= (v + w)\cdot(v + w)\\ &= v\cdot v + 2(v\cdot w) + w\cdot w \\ &= \|v\|^2 + 2(v\cdot w) + \|w\|^{2}\\ \end{aligned}$$

Using the Schwarz Inequality in the equation:

$$\begin{aligned} \|v + w\|^{2} &\le \|v\|^2 + 2(\|v\| \|w\|) + \|w\|^{2}\\ &=(\|v\| + \|w\|)^{2} \end{aligned}$$

Taking the square roots of both sides (noting that magnitude can not be negative) yields the final inequality.

$$\|v + w\| \le \|v\| + \|w\|$$

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• Strang, Gilbert. Introduction to Linear Algebra. 5th ed. Wellesley–Cambridge Press, 2016. Chapter 1.2.