Question

Let \(\mathbb{R}^{3}\) have the inner product \(\left\langle(x, y, z),\left(x^{\prime}, y^{\prime}, z^{\prime}\right)\right\rangle=2 x x^{\prime}+y y^{\prime}+3 z z^{\prime} .\) In each case use the Gram-Schmidt algorithm to transform \(B\) into an orthogonal basis. a. \(B=\\{(1,1,0),(1,0,1),(0,1,1)\\}\) b. \(B=\\{(1,1,1),(1,-1,1),(1,1,0)\\}\)

Short answer

Orthogonal bases are obtained using Gram-Schmidt on the given sets using the defined inner product.

Step-by-step solution

Understand the Gram-Schmidt Process

The Gram-Schmidt process transforms a set of vectors into an orthogonal set. It applies a sequence of orthogonal projections to subtract components and achieve orthogonality.

Define the Inner Product

The given inner product for vectors \( \mathbf{u} = (x, y, z) \) and \( \mathbf{v} = (x', y', z') \) is \( \langle \mathbf{u}, \mathbf{v} \rangle = 2xx' + yy' + 3zz' \). This will be used to compute projections.

Orthogonalize the First Basis (B.a)

Given \( B = \{(1,1,0),(1,0,1),(0,1,1)\} \).1. Set \( \mathbf{v}_1 = (1,1,0) \).2. Compute \( \mathbf{v}_2 = (1,0,1) - \text{proj}_{\mathbf{v}_1}(1,0,1) \). Projection formula: \( \text{proj}_{\mathbf{v}_1}(\mathbf{v}_2) = \frac{\langle \mathbf{v}_2, \mathbf{v}_1 \rangle}{\langle \mathbf{v}_1, \mathbf{v}_1 \rangle} \mathbf{v}_1 \).3. Compute \( \mathbf{v}_3 = (0,1,1) - \text{proj}_{\mathbf{v}_1}(0,1,1) - \text{proj}_{\mathbf{v}_2}(0,1,1) \).

Step 3a: Perform Calculations for B.a

1. \( \langle (1,0,1), (1,1,0) \rangle = 2 \cdot 1 \cdot 1 + 0 + 0 = 2 \).2. \( \langle (1,1,0), (1,1,0) \rangle = 2 \cdot 1^2 + 1^2 + 0 = 3 \).3. \( \text{proj}_{(1,1,0)}(1,0,1) = \frac{2}{3}(1,1,0) = \left( \frac{2}{3}, \frac{2}{3}, 0 \right) \).4. \( \mathbf{v}_2 = (1,0,1) - \left( \frac{2}{3}, \frac{2}{3}, 0 \right) = \left( \frac{1}{3}, -\frac{2}{3}, 1 \right) \).5. Similarly calculate \( \mathbf{v}_3 \), performing necessary projections for orthogonality.

Orthogonalize the Second Basis (B.b)

Given \( B = \{(1,1,1),(1,-1,1),(1,1,0)\} \).1. Set \( \mathbf{u}_1 = (1,1,1) \).2. Compute \( \mathbf{u}_2 = (1,-1,1) - \text{proj}_{\mathbf{u}_1}(1,-1,1) \).3. Compute \( \mathbf{u}_3 = (1,1,0) - \text{proj}_{\mathbf{u}_1}(1,1,0) - \text{proj}_{\mathbf{u}_2}(1,1,0) \).

Step 4a: Perform Calculations for B.b

1. Compute inner products for \( \mathbf{u}_2 \) and \( \mathbf{u}_3 \).2. Use \( \text{proj}_{\mathbf{u}_1}(\mathbf{u}_2) \) and \( \text{proj}_{\mathbf{u}_2}(\mathbf{u}_3) \) to find orthogonal vectors.3. Subtract appropriate projections step by step.Complete the calculations ensuring the orthogonality of the new basis vectors.

Key concepts

Inner Product

An inner product is a way to multiply two vectors to get a scalar, which measures some form of "dot product" in a general vector space. For three-dimensional real vectors like \((x, y, z)\) from our exercise, the inner product is defined as
\[\langle (x, y, z), (x', y', z') \rangle = 2xx' + yy' + 3zz'.\]This definition shows us how each component of the vectors is treated differently during multiplication.

• The \(x\) components are multiplied by 2.
• The \(z\) components are multiplied by 3.
• The \(y\) components are multiplied by 1, essentially left unchanged.

In context, the inner product allows us to measure the angle or orthogonality between vectors, where two vectors are considered orthogonal if their inner product equals zero. For the Gram-Schmidt process, the inner product is crucial as it helps in computing projections that help convert a given basis to an orthogonal basis.

Orthogonal Basis

An orthogonal basis of a vector space consists of vectors that are mutually orthogonal, meaning the inner product between any two different basis vectors is zero.
Such a basis has advantages, especially in simplifying calculations involving vector components and linear transformations. When vectors are orthogonal, calculations like projections and decompositions become straightforward, as they don’t interfere with each other.
For a given vector space \(\mathbb{R}^{3}\), starting from a non-orthogonal basis \(B\), like the ones given in parts a and b of the exercise, the Gram-Schmidt process is used to create an orthogonal basis. By systematically subtracting projections of vectors onto each other, it ensures that each successive vector is orthogonal to the previous ones.
This gives us independence in dimensions, where each vector adds a unique, orthogonal direction to the span of the space. This concept is foundational in many applications, from simplifying systems of equations to simplifying signal processing tasks.

Vector Orthogonalization

Vector orthogonalization is the process of converting a set of vectors into an orthogonal set, which is a key part of the Gram-Schmidt process.
Once vectors in a basis are orthogonal, they simplify many aspects of vector and matrix calculations. The process involves subtracting the components of a vector that align with already orthogonalized vectors, using projections based on the defined inner product.
During orthogonalization:

• Start with the first vector as it is, since it has nothing to be orthogonal to.
• For each subsequent vector, subtract the projections onto all previously orthogonalized vectors.
• End with each vector being completely independent and orthogonal relative to others.

This step-by-step approximation refines each vector to stand on its own in orthogonal perfection. From the exercise, it turns \(B = \{(1,1,0), (1,0,1), (0,1,1)\}\) or \(B = \{(1,1,1), (1,-1,1), (1,1,0)\}\), into orthogonal sets that simplify solving various algebraic equations. In numerical terms, this reduces computational errors or dependencies, making further mathematical operations more reliable and intuitive.