Problem 35 - Differential Equations and Linear Algebra

Question

Let \(V\) be a real inner product space. (a) Prove that for all \(\mathbf{v}, \mathbf{w} \in V\) \(\|\mathbf{v}+\mathbf{w}\|^{2}=\|\mathbf{v}\|^{2}+2\langle\mathbf{v}, \mathbf{w}\rangle+\|\mathbf{w}\|^{2}\) [Hint: \(\left.\|\mathbf{v}+\mathbf{w}\|^{2}=\langle\mathbf{v}+\mathbf{w}, \mathbf{v}+\mathbf{w}\rangle .\right]\) (b) Two vectors \(\mathbf{v}\) and \(\mathbf{w}\) in an inner product space \(V\) are called orthogonal if \(\langle\mathbf{v}, \mathbf{w}\rangle=0 .\) Use (a) to prove the general Pythagorean Theorem: If \(\mathbf{v}\) and \(\mathbf{w}\) are orthogonal in an inner product space \(V,\) then $$\|\mathbf{v}+\mathbf{w}\|^{2}=\|\mathbf{v}\|^{2}+\|\mathbf{w}\|^{2}$$

Step-by-Step Solution

Answer

In a real inner product space \(V\), to prove the expression \(\|\mathbf{v}+\mathbf{w}\|^{2}=\|\mathbf{v}\|^{2}+2\langle\mathbf{v},\mathbf{w}\rangle+\|\mathbf{w}\|^{2}\), we first compute the squared norm of the sum of the vectors \(\mathbf{v}\) and \(\mathbf{w}\), using the given hint. Then we expand the inner product and use the properties of real inner product spaces to simplify the expression, finally reaching the desired result. If \(\mathbf{v}\) and \(\mathbf{w}\) are orthogonal vectors (meaning \(\langle\mathbf{v}, \mathbf{w}\rangle=0\)), then we can use this result to prove the general Pythagorean theorem, \(\|\mathbf{v}+\mathbf{w}\|^{2}=\|\mathbf{v}\|^{2}+\|\mathbf{w}\|^{2}\), in an inner product space.

Step 1: Part (a)

To prove this, we start by finding the squared norm of the sum of the vectors \(\mathbf{v}\) and \(\mathbf{w}\). We are given the hint: $$\|\mathbf{v}+\mathbf{w}\|^2 = \langle\mathbf{v}+\mathbf{w},\mathbf{v}+\mathbf{w}\rangle$$ Now, use the properties of inner products to expand the expression: $$\langle\mathbf{v}+\mathbf{w},\mathbf{v}+\mathbf{w}\rangle = \langle\mathbf{v},\mathbf{v}\rangle + \langle\mathbf{v},\mathbf{w}\rangle + \langle\mathbf{w},\mathbf{v}\rangle + \langle\mathbf{w},\mathbf{w}\rangle$$ Since it is a real inner product space, we have \(\langle\mathbf{w},\mathbf{v}\rangle = \langle\mathbf{v},\mathbf{w}\rangle\). Thus, the expression above simplifies to: $$\|\mathbf{v}+\mathbf{w}\|^2 = \langle\mathbf{v},\mathbf{v}\rangle + 2\langle\mathbf{v},\mathbf{w}\rangle + \langle\mathbf{w},\mathbf{w}\rangle$$ Recall that \(\|\mathbf{v}\|^2 = \langle\mathbf{v},\mathbf{v}\rangle\) and \(\|\mathbf{w}\|^2 = \langle\mathbf{w},\mathbf{w}\rangle\). Substituting this into the last equality, we get the desired result: $$\|\mathbf{v}+\mathbf{w}\|^2 = \|\mathbf{v}\|^2 + 2\langle\mathbf{v},\mathbf{w}\rangle + \|\mathbf{w}\|^2$$

Step 2: Part (b)

Given that \(\mathbf{v}\) and \(\mathbf{w}\) are orthogonal, so \(\langle\mathbf{v}, \mathbf{w}\rangle=0\). We want to prove the general Pythagorean theorem: $$\|\mathbf{v}+\mathbf{w}\|^2 = \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2$$ From part (a), we have: $$\|\mathbf{v}+\mathbf{w}\|^2 = \|\mathbf{v}\|^2 + 2\langle\mathbf{v},\mathbf{w}\rangle + \|\mathbf{w}\|^2$$ Since \(\langle\mathbf{v}, \mathbf{w}\rangle=0\), the expression becomes: $$\|\mathbf{v}+\mathbf{w}\|^2 = \|\mathbf{v}\|^2 + 2(0) + \|\mathbf{w}\|^2$$ So we have: $$\|\mathbf{v}+\mathbf{w}\|^2 = \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2$$ Thus, we have proved the general Pythagorean theorem for vectors in an inner product space.

Key Concepts

Vector Norm

The concept of a vector norm is foundational in understanding inner product spaces. A vector norm is a function that assigns a positive length or size to the vectors in a vector space, and it is often denoted by \( \|\mathbf{v}\| \). In mathematical terms, for any vector \( \mathbf{v} \) in a vector space \( V \), the norm \( \|\mathbf{v}\| \) is defined as: \[ \|\mathbf{v}\| = \sqrt{\langle\mathbf{v}, \mathbf{v}\rangle} \]The properties of vector norms include:

• Positive Definiteness: \(\|\mathbf{v}\| \geq 0\) and \(\|\mathbf{v}\| = 0\) if and only if \( \mathbf{v} = \mathbf{0} \).
• Homogeneity (or Scaling): \(\|c\mathbf{v}\| = |c|\|\mathbf{v}\| \) for any scalar \( c \).
• Triangle Inequality: \(\|\mathbf{v} + \mathbf{w}\| \leq \|\mathbf{v}\| + \|\mathbf{w}\| \) for all vectors \( \mathbf{v}, \mathbf{w} \).

A clear understanding of vector norms helps in grasping how distances and sizes are computed in inner product spaces. It's important to realize that vector norms are essential in defining the measurement of vector magnitude within the context of an inner product space.

Orthogonality

In the world of inner product spaces, understanding the concept of orthogonality is key. Two vectors \( \mathbf{v} \) and \( \mathbf{w} \) in an inner product space are said to be orthogonal if their inner product \( \langle \mathbf{v}, \mathbf{w} \rangle \) equals zero. This is often interpreted as the vectors being "perpendicular" to each other. The condition for orthogonality can be written mathematically as:\[ \langle \mathbf{v}, \mathbf{w} \rangle = 0 \]Orthogonality has several important properties that make it useful:

• Zero Inner Product: When \( \mathbf{v} \) and \( \mathbf{w} \) are orthogonal, they do not "interfere" with each other, meaning they are independent in the geometric sense.
• Basis Construction: Orthogonal vectors can be used to construct an orthogonal basis, simplifying many calculations, including those related to projections.
• Simplified Calculations: In many cases, calculations involving orthogonal vectors are simplified because of their zero inner product.

Recognizing orthogonal vectors is crucial, especially when solving problems that involve simplifying complex vector operations. This comes in very handy when applying the Pythagorean Theorem in inner product spaces.

Pythagorean Theorem in Inner Product Spaces

The Pythagorean theorem is a well-known geometric theorem extended into the realm of inner product spaces. In its simplest form, it states that, for two orthogonal vectors \( \mathbf{v} \) and \( \mathbf{w} \) in an inner product space, the squared norm of their sum is equal to the sum of their squared norms. Mathematically, it is expressed as:\[ \|\mathbf{v} + \mathbf{w}\|^2 = \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 \]This relationship boils down to the absence of the cross term \( 2\langle\mathbf{v}, \mathbf{w}\rangle \) when \( \mathbf{v} \) and \( \mathbf{w} \) are orthogonal, simplifying the general formula given by:\[ \|\mathbf{v} + \mathbf{w}\|^2 = \|\mathbf{v}\|^2 + 2\langle\mathbf{v}, \mathbf{w}\rangle + \|\mathbf{w}\|^2 \]Since orthogonality makes \( \langle \mathbf{v}, \mathbf{w} \rangle = 0 \), the Pythagorean theorem emerges naturally. The importance of this theorem:

• Verifies orthogonality: If given vectors satisfy the theorem, they must be orthogonal.
• Applies in higher dimensions: The theorem holds regardless of the dimensionality of the space.
• Stays at the core of many applications: Critical in fields such as signal processing, linear algebra, and physics.

Overall, the Pythagorean theorem in inner product spaces bridges geometric intuition and algebraic formalism, making it an indispensable tool in understanding vector relations.