Question

Challenge Problem Prove the polarization identity, $$ \|\mathbf{u}+\mathbf{v}\|^{2}-\|\mathbf{u}-\mathbf{v}\|^{2}=4(\mathbf{u} \cdot \mathbf{v}) $$

Short answer

\[ \| \mathbf{u}+ \mathbf{v} \|^{2} - \| \mathbf{u} - \mathbf{v} \|^{2} = 4( \mathbf{u} \cdot \mathbf{v}\)\]

Step-by-step solution

Expand the Norm Squares

Start with the given equation \ \ \ \ \[ \ \ \ \ \| \mathbf{u}+ \mathbf{v} \|^{2} - \| \mathbf{u} - \mathbf{v} \|^{2} = 4( \mathbf{u} \cdot \mathbf{v} ) \ \ \ \ \] \ \ \ \ \[\| \mathbf{u}+ \mathbf{v} \|^{2} = ( \mathbf{u}+ \mathbf{v}) \cdot ( \mathbf{u}+ \mathbf{v}) \] \ \ \ \ \ \ \ \ \[\| \mathbf{u}- \mathbf{v} \|^{2} = ( \mathbf{u}- \mathbf{v}) \cdot ( \mathbf{u}- \mathbf{v}) \] \ \ \ \

Apply Distributive Property

Distribute the dot product in both terms: \ \ \ \ \[ \mathbf{u} \cdot \mathbf{u} + ( \mathbf{u} \cdot \mathbf{v}) + ( \mathbf{v} \cdot \mathbf{u}) + ( \mathbf{v} \cdot \mathbf{v}) \] \ \ \ \ for \| \mathbf{u}+ \mathbf{v} \|^{2}, and \ \ \ \ \[ \mathbf{u} \cdot \mathbf{u} - ( \mathbf{u} \cdot \mathbf{v}) - ( \mathbf{v} \cdot \mathbf{u}) + ( \mathbf{v} \cdot \mathbf{v}) \] \ \ \ \ for \| \mathbf{u}- \mathbf{v} \|^{2}.

Combine Like Terms

Now, subtract \| \mathbf{u}- \mathbf{v} \|^{2} from \| \mathbf{u}+ \mathbf{v} \|^{2}: \ \ \ \ \[ ( \mathbf{u} \cdot \mathbf{u}) + ( \mathbf{u} \cdot \mathbf{v}) + ( \mathbf{v} \cdot \mathbf{u}) + ( \mathbf{v} \cdot \mathbf{v}) - [ ( \mathbf{u} \cdot \mathbf{u}) - ( \mathbf{u} \cdot \mathbf{v}) - ( \mathbf{v} \cdot \mathbf{u}) + ( \mathbf{v} \cdot \mathbf{v}) ] \] \ \ \ \ Combine the terms to get: \ \ \ \ \[2( \mathbf{u} \cdot \mathbf{v}) + 2( \mathbf{v} \cdot \mathbf{u}) = 4( \mathbf{u} \cdot \mathbf{v} ) \]

Key concepts

Dot Product

The dot product, also known as the scalar product, is an essential operation in vector algebra. It combines two vectors to produce a scalar value. The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by \( \mathbf{u} \cdot \mathbf{v} \). For vectors in Euclidean space, the dot product is calculated as follows:
\[ \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + \dots + u_nv_n \]
Here, \( u_i \) and \( v_i \) are the components of the vectors \( \mathbf{u} \) and \( \mathbf{v} \) respectively. The dot product is intuitive, representing the product of the magnitudes of the two vectors and the cosine of the angle between them. This relationship is crucial in applications including physics and computer graphics.

Vector Norms

A vector's norm, often referred to as its length or magnitude, measures the vector's size. For a vector \( \mathbf{u} \, \| \mathbf{u} \| \) denotes its norm. The norm is calculated using the square root of the sum of the squared components:
\[ \| \mathbf{u} \| = \sqrt{u_1^2 + u_2^2 + \dots + u_n^2} \]
This formula is similar to the Pythagorean Theorem. Norms help in comparing vector sizes and verifying orthogonality, as orthogonal vectors have a dot product of zero. In the context of the polarization identity, the norms allow us to express the lengths of vector sums and differences.

Distributive Property

The distributive property in vector algebra is a key rule for simplifying expressions involving vectors. When dealing with dot products, this property illustrates how to distribute the operation over addition or subtraction of vectors. For vectors \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w}, \) it is expressed as:
\[ \mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} \]
The distributive property is particularly useful for proving identities like the polarization identity. It allows us to break down complex expressions into simpler components, facilitating easier manipulation and understanding of vector equations.

Vector Algebra

Vector algebra encompasses operations on vectors, including addition, subtraction, and scalar multiplication. Each vector operation has specific rules and properties that help in solving mathematical and physical problems. Vector addition follows the rule:
\[ \mathbf{u} + \mathbf{v} = (u_1 + v_1, u_2 + v_2, \dots, u_n + v_n) \]
Similarly, vector subtraction is defined as:
\[ \mathbf{u} - \mathbf{v} = (u_1 - v_1, u_2 - v_2, \dots, u_n - v_n) \]
Scalars can also multiply vectors, scaling their magnitude without changing their direction. These operations conform to properties like commutativity and distributivity, providing a structured way to manipulate vectors, which is fundamental to understanding and proving mathematical identities.