Question

Prove the Cauchy-Schwarz Inequality \(|\mathbf{u} \cdot \mathbf{v}| \leq\|\mathbf{u}\|\|\mathbf{v}\| .\)

Short answer

The Cauchy-Schwarz inequality \( |\mathbf{u} \cdot \mathbf{v}|^2 \leq \|\mathbf{u}\|\|\mathbf{v}\|^2 \) is proved by considering a polynomial \( P(t) = \|\mathbf{v} - t\mathbf{u}\|^2 \), finding its minimum, and using this to demonstrate the inequality.

Step-by-step solution

Setup

Let \( \mathbf{u} \) and \( \mathbf{v} \) be two vectors in an inner product space, and suppose \( \mathbf{u} \) and \( \mathbf{v} \) are not orthogonal (if they were the inequality is true as the dot product would be 0). Without loss of generality, we can assume \( \|\mathbf{u}\| = 1 \). This is done to simplify calculations without changing the overall concept.

Define a Polynomial

Consider the following polynomial: \( P(t) = \|\mathbf{v} - t\mathbf{u}\|^2 \). This polynomial is non-negative for all real numbers t as the square of a vector's magnitude is always greater than or equal to zero.

Expand the Polynomial

We can expand the polynomial \( P(t) \) as follows: \( P(t) = (\mathbf{v} - t\mathbf{u}) \cdot (\mathbf{v} - t\mathbf{u}) = \|\mathbf{v}\|^2 - 2t(\mathbf{v} \cdot \mathbf{u}) + t^2\|\mathbf{u}\|^2 = t^2 - 2t(\mathbf{v} \cdot \mathbf{u}) + \|\mathbf{v}\|^2 \). Since \( \|\mathbf{u}\| = 1 \), \(\|\mathbf{u}\|^2 = 1 \) and so it simplifies to avoid multiplication by one.

Find the Minimum of Polynomial

We can find the minimum of \( P(t) \) by taking the derivative and setting it equal to zero. We find that the derivative \( P'(t) = 2t - 2(\mathbf{v} \cdot \mathbf{u}) \). Setting \( P'(t) = 0 \) gives \( t = (\mathbf{v} \cdot \mathbf{u}) \). This shows that \( P(t) \) achieves it's minimum at \( t = (\mathbf{v} \cdot \mathbf{u}) \).

Proving Cauchy-Schwarz inequality

Using the minimum of \( P(t) \) found in the previous step and the fact that \( P(t) \geq 0 \) for all \( t \), we get \( 0 \leq P((\mathbf{v} \cdot \mathbf{u})) = \|\mathbf{v}\|^2 - 2(\mathbf{v} \cdot \mathbf{u})^2 + (\mathbf{v} \cdot \mathbf{u})^2 = \|\mathbf{v}\|^2 - (\mathbf{v} \cdot \mathbf{u})^2 \). Re-arranging gives the desired result: \( |\mathbf{u} \cdot \mathbf{v}|^2 \leq \|\mathbf{u}\|\|\mathbf{v}\|^2 \). Which proves the Cauchy-Schwarz inequality.

Key concepts

Vector Magnitude

Understanding vector magnitude is crucial for grasping the essence of the Cauchy-Schwarz Inequality. In the context of vectors, the magnitude represents the 'length' or 'size' of a vector. It can be calculated using the square root of the sum of the squares of all the vector's components. For a vector \( \mathbf{a} = (a_1, a_2, ..., a_n) \) in an n-dimensional space, the magnitude is expressed as \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + ... + a_n^2} \).

The significance of vector magnitude is highlighted during the proof of the Cauchy-Schwarz Inequality, where it serves as a standard to compare the dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) against the product of their magnitudes.

Inner Product Space

An inner product space is a vector space endowed with an additional structure called an inner product. This inner product allows for the measurement of angles and lengths within the space, introducing geometric concepts to the realm of vector calculus.

The dot product, a common example of an inner product for real-valued vectors, is given by \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + ... + u_nv_n \), for vectors \( \mathbf{u} \) and \( \mathbf{v} \) with components \( u_i \) and \( v_i \) respectively. The Cauchy-Schwarz Inequality uses this concept to show that the absolute value of the dot product does not exceed the product of the vectors' magnitudes.

Polynomial Minimization

Polynomial minimization involves finding the smallest value (minimum) that a given polynomial function can take. In the proof of the Cauchy-Schwarz Inequality, a polynomial \( P(t) \) is constructed reflecting the squared magnitude of the difference between a vector \( \mathbf{v} \) and \( t \) times another vector \( \mathbf{u} \).

As shown in the step-by-step solution, minimizing this polynomial and evaluating it at the derived value shows that the result must be non-negative, reinforcing the inequality's validity. This part of the proof is essentially using calculus to find where the function dips to its lowest point, revealing properties about the relationship between \( \mathbf{u} \) and \( \mathbf{v} \) indirectly.