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Chapter 6: Problem 31

Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are vectors. Show that $$ |\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}| $$ [This result is called the Cauchy-Schwarz Inequality. Although this problem asks for a proof only in the setting of vectors in the plane, a similar inequality is true in many other settings and has important uses throughout mathematics.]

Short Answer

Expert verified
Using the definition of dot product, \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}| \cos(\theta)\), and knowing that \(-1 \leq \cos(\theta) \leq 1\), we can derive the inequality \(-|\mathbf{u}||\mathbf{v}| \leq \mathbf{u} \cdot \mathbf{v} \leq |\mathbf{u}||\mathbf{v}|\). Therefore, by definition of absolute value, we conclude that \(|\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|\), proving the Cauchy-Schwarz Inequality for vectors in the plane.

Step by step solution

01

Write down the definition of dot product

The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is defined as: \[ \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}| \cos(\theta) \] where \(|\mathbf{u}|\) and \(|\mathbf{v}|\) are the magnitudes of the vectors \(\mathbf{u}\) and \(\mathbf{v}\), and \(\theta\) is the angle between them.
02

Use properties of cosine to derive inequality

Since the cosine function is always between -1 and 1: \[ -1 \leq \cos(\theta) \leq 1 \] Multiply both sides of the inequality by \(|\mathbf{u}||\mathbf{v}|\) (magnitudes are always non-negative, so the inequality remains valid): \[ -|\mathbf{u}||\mathbf{v}| \leq |\mathbf{u}||\mathbf{v}| \cos(\theta) \leq |\mathbf{u}||\mathbf{v}| \] Since \(|\mathbf{u}||\mathbf{v}| \cos(\theta)\) is equivalent to \(\mathbf{u} \cdot \mathbf{v}\): \[ -|\mathbf{u}||\mathbf{v}| \leq \mathbf{u} \cdot \mathbf{v} \leq |\mathbf{u}||\mathbf{v}| \]
03

Conclude the proof

From Step 2, we have shown that \[ -|\mathbf{u}||\mathbf{v}| \leq \mathbf{u} \cdot \mathbf{v} \leq |\mathbf{u}||\mathbf{v}| \] Therefore, by definition of absolute value, we can conclude that: \[ |\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}| \] Thus, the Cauchy-Schwarz Inequality for vectors in the plane is proven.

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