Problem 79 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove the Cauchy-Schwarz Inequality for two-dimensional vectors: $$ |\mathbf{u} \cdot \mathbf{v}| \leq\|\mathbf{u}\|\|\mathbf{v}\| $$

Step-by-Step Solution

Answer

Cauchy-Schwarz Inequality is proven by showing \((u_1v_2 - u_2v_1)^2 \geq 0\).

Step 1: Understanding the vectors and operations

For two-dimensional vectors \( \mathbf{u} = (u_1, u_2) \) and \( \mathbf{v} = (v_1, v_2) \), we need to understand the dot product and the magnitudes. The dot product is given by \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). The magnitudes are \( \| \mathbf{u} \| = \sqrt{u_1^2 + u_2^2} \) and \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2} \).

Step 2: Expressing the inequality with vectors

The Cauchy-Schwarz Inequality we need to prove is \( |\mathbf{u} \cdot \mathbf{v}| \leq \|\mathbf{u}\|\|\mathbf{v}\| \). This means the absolute value of the dot product should be less than or equal to the product of the magnitudes of the vectors.

Step 3: Algebraic expression reformulation

The inequality can be rewritten as: \[ (u_1v_1 + u_2v_2)^2 \leq (u_1^2 + u_2^2)(v_1^2 + v_2^2) \] This translates to comparing the square of the dot product with the product of the magnitudes squared.

Step 4: Expanding both sides

Expand both sides: The left hand side becomes: \[ (u_1v_1 + u_2v_2)^2 = u_1^2v_1^2 + 2u_1u_2v_1v_2 + u_2^2v_2^2 \] The right hand side turns into: \[ (u_1^2 + u_2^2)(v_1^2 + v_2^2) = u_1^2v_1^2 + u_1^2v_2^2 + u_2^2v_1^2 + u_2^2v_2^2 \]

Step 5: Rearrange the inequality

Subtract the left hand side from the right hand side to put everything on one side to check if non-negative:\[ (u_1^2v_2^2 + u_2^2v_1^2) + (u_1u_2v_1v_2 - u_1u_2v_1v_2) - (u_1u_2v_1v_2) \geq 0 \] This simplifies to: \[ u_1^2v_2^2 + u_2^2v_1^2 - u_1u_2v_1v_2 \geq 0 \]

Step 6: Final Verification

Notice that the simplified inequality can also be written as: \[ (u_1v_2 - u_2v_1)^2 \geq 0 \] Since a square of any real number is always non-negative, this verifies the inequality. Thus, the Cauchy-Schwarz Inequality holds true.

Key Concepts

Two-Dimensional Vectors

In mathematics, a two-dimensional vector is simply a pair of numbers. These numbers can represent a wide range of things, like force, velocity, or any quantity that has both a magnitude and direction in a plane. For instance, if we have two two-dimensional vectors, \( \mathbf{u} = (u_1, u_2) \) and \( \mathbf{v} = (v_1, v_2) \), they exist within a plane, such as the space of your desk or a piece of paper. The components can either be positive or negative depending on their direction. When working with these vectors, we use operations like addition, subtraction, and especially the dot product, to find various characteristics of the vectors. Keeping it simple, understanding these basics is foundational when dealing with vector operations.

Dot Product

The dot product is one way to multiply two vectors, which results in a scalar value. For our two-dimensional vectors, \( \mathbf{u} = (u_1, u_2) \) and \( \mathbf{v} = (v_1, v_2) \), the dot product is calculated by multiplying their corresponding components and then summing those results:

• \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \)

This scalar value is crucial in many applications, such as finding the angle between vectors or understanding vector projections. An interesting fact about the dot product is that it relates directly to the projection of one vector onto another. If the dot product ends up being zero, you know the vectors are perpendicular, or orthogonal, to each other.

Vector Magnitude

The magnitude of a vector is essentially its length – how far it "stretches" in space. For a vector \( \mathbf{u} = (u_1, u_2) \), the magnitude, or the "norm," is computed using the Pythagorean theorem:

• \( \| \mathbf{u} \| = \sqrt{u_1^2 + u_2^2} \)

Similarly, \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2} \) for vector \( \mathbf{v} \). This length is always non-negative and denotes the size of the vector regardless of its direction. Understanding magnitudes is essential not only for the Cauchy-Schwarz Inequality but also for assessing how large or small a vector really is, in terms of physical interpretation.

Inequality Proof

The Cauchy-Schwarz Inequality is a powerful tool in linear algebra and is central to many proofs and solutions in geometry and physics. The inequality, which is written as \( |\mathbf{u} \cdot \mathbf{v}| \leq \|\mathbf{u}\|\|\mathbf{v}\| \), asserts that the absolute value of the dot product of two vectors is never greater than the product of their magnitudes. To prove this inequality, we reformulate both the dot product and magnitudes squared, obtaining:

• \((u_1v_1 + u_2v_2)^2 \leq (u_1^2 + u_2^2)(v_1^2 + v_2^2)\)

By expanding both sides and rearranging the inequality, it becomes evident that the equation \((u_1v_2 - u_2v_1)^2 \geq 0\) holds.Given that squares of real numbers are always non-negative, the expression confirms the inequality is valid. This shows the elegant relationship between the geometry of the vectors and their algebraic interactions.