Question

Show that \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal, if and only if \(\|\mathbf{u}+\mathbf{v}\|^{2}=\|\mathbf{u}\|^{2}+\|\mathbf{v}\|^{2}\).

Short answer

\( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal if and only if \( \|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 \), as verified by setting \( \mathbf{u} \cdot \mathbf{v} = 0 \).

Step-by-step solution

Understanding Orthogonal Vectors

Two vectors \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal if their dot product is zero, i.e., \( \mathbf{u} \cdot \mathbf{v} = 0 \). This property is fundamental to the relationship given in the problem, connecting it with the expression \( \|\mathbf{u} + \mathbf{v}\|^2 \).

Expanding Norm Squared

The square of the norm of a vector \( \mathbf{a} \) is given by \( \|\mathbf{a}\|^2 = \mathbf{a} \cdot \mathbf{a} \). Therefore, \( \|\mathbf{u} + \mathbf{v}\|^2 = (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) \).

Use Distributive Property

Apply the distributive property for the dot product: \[ (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = \mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v} \]. The cross terms \( \mathbf{u} \cdot \mathbf{v} \) and \( \mathbf{v} \cdot \mathbf{u} \) are equal.

Condition for Orthogonality

Remember that if \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal, \( \mathbf{u} \cdot \mathbf{v} = 0 \). Therefore, the expression is simplified to \( \mathbf{u} \cdot \mathbf{u} + 2(\mathbf{u} \cdot \mathbf{v}) + \mathbf{v} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v} \).

Verify Both Conditions

Start from the given condition \( \|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 \). With expanded form: \( \mathbf{u} \cdot \mathbf{u} + 2(\mathbf{u} \cdot \mathbf{v}) + \mathbf{v} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v} \), the middle term must be zero, hence \( \mathbf{u} \cdot \mathbf{v} = 0 \). This confirms orthogonality. Conversely, if vectors are orthogonal, the equality holds.

Key concepts

Dot Product

The dot product is a fundamental operation in vector algebra that combines two vectors to yield a scalar. For two vectors \( \mathbf{u} = (u_1, u_2, ..., u_n) \) and \( \mathbf{v} = (v_1, v_2, ..., v_n) \), the dot product is calculated as:\[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + ... + u_n v_n \]When vectors are orthogonal, their dot product is zero. This means that they are perpendicular to each other. Visualize this in two dimensions as two lines intersecting at a right angle.It's important to understand that the dot product is:- Commutative: \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \)- Distributive over vector addition, which will help as we explore further concepts.In terms of practical application, dot product calculations can determine angles between vectors, projections, and can also be used to check for orthogonality.

Vector Norm

The vector norm, also known as the Euclidean norm, measures the 'length' of a vector. For a vector \( \mathbf{a} = (a_1, a_2, ..., a_n) \), its norm is calculated as:\[ \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2 + ... + a_n^2} \]To find the square of the norm, simply square the value of the norm:\[ \| \mathbf{a} \|^2 = a_1^2 + a_2^2 + ... + a_n^2 \]However, using the dot product, this can also be written as \( \| \mathbf{a} \|^2 = \mathbf{a} \cdot \mathbf{a} \).Understanding vector norms is crucial in contexts like physics, machine learning, and computer graphics. It provides a method to quantitatively compare vectors and their magnitudes.For the exercise at hand, understanding that \( \| \mathbf{u} + \mathbf{v} \|^2 = (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) \), helps in breaking down the problem using algebraic expansion.

Distributive Property

The distributive property is a vital principle that applies to operations involving addition and multiplication, and it is no different in the context of vectors and dot products. This property allows you to expand expressions like \( (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) \).Using the distributive property for dot products:\[ (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = \mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v} \]Note that \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \) due to commutativity of the dot product. This results in those middle terms doubling up: \( 2(\mathbf{u} \cdot \mathbf{v}) \).For vectors to be orthogonal, as discussed, the dot product \( \mathbf{u} \cdot \mathbf{v} \) must be zero. This effectively removes the middle terms in the expanded equation, confirming the condition given in the exercise. Mastery of this property simplifies many operations in linear algebra and calculus, making it easier to tackle complex vector equations.