Problem 11 - Advanced Engineering Mathematics

Question

Use the Gram-Schmidt orthogonalization process (4) to transform the given basis \(\boldsymbol{B}=\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) for \(R^{3}\) into an orthogonal basis \(B^{\prime}=\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\). Then form an orthonormal basis \(B^{\prime \prime}=\left\\{\mathbf{w}_{1}, \mathbf{w}_{2}, \mathbf{w}_{3}\right\\}\) $$ B=\left\\{\left\langle\frac{1}{2}, \frac{1}{2}, 1\right\rangle,\left\langle-1,1,-\frac{1}{2}\right\rangle,\left\langle-1, \frac{1}{2}, 1\right\rangle\right\\} $$

Step-by-Step Solution

Answer

The orthogonal basis is \( \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\} \); the orthonormal basis is \( \{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3\} \).

Step 1: Initialize the Process

Start with the given basis for \( R^3 \), \( B=\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\} \), where \( \mathbf{u}_1 = \left\langle \frac{1}{2}, \frac{1}{2}, 1 \right\rangle \), \( \mathbf{u}_2 = \left\langle -1, 1, -\frac{1}{2} \right\rangle \), and \( \mathbf{u}_3 = \left\langle -1, \frac{1}{2}, 1 \right\rangle \).

Step 2: Calculate \( \mathbf{v}_1 \)

Set \( \mathbf{v}_1 = \mathbf{u}_1 \) since the first vector is simply copied from the given basis. So, \( \mathbf{v}_1 = \left\langle \frac{1}{2}, \frac{1}{2}, 1 \right\rangle \).

Step 3: Calculate \( \mathbf{v}_2 \)

Compute \( \mathbf{v}_2 \) by orthogonalizing \( \mathbf{u}_2 \) against \( \mathbf{v}_1 \).\[ \mathbf{v}_2 = \mathbf{u}_2 - \frac{\langle \mathbf{u}_2, \mathbf{v}_1 \rangle}{\langle \mathbf{v}_1, \mathbf{v}_1 \rangle} \mathbf{v}_1 \]Compute \( \langle \mathbf{u}_2, \mathbf{v}_1 \rangle = -1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} - \frac{1}{2} \cdot 1 = -1 \), and \( \langle \mathbf{v}_1, \mathbf{v}_1 \rangle = \frac{1}{2}^2 + \frac{1}{2}^2 + 1^2 = \frac{3}{2} \).Therefore, \[ \mathbf{v}_2 = \left\langle -1, 1, -\frac{1}{2} \right\rangle - \left\langle -\frac{2}{3}, -\frac{2}{3}, -\frac{4}{3} \right\rangle = \left\langle -\frac{1}{3}, \frac{5}{3}, \frac{5}{6} \right\rangle \].

Step 4: Calculate \( \mathbf{v}_3 \)

Compute \( \mathbf{v}_3 \) by orthogonalizing \( \mathbf{u}_3 \) against both \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \).First project \( \mathbf{u}_3 \) onto \( \mathbf{v}_1 \). \[ \langle \mathbf{u}_3, \mathbf{v}_1 \rangle = -1 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + 1 \cdot 1 = \frac{3}{4} \], and we know \( \langle \mathbf{v}_1, \mathbf{v}_1 \rangle = \frac{3}{2} \). Hence, \( \text{proj}_{\mathbf{v}_1}(\mathbf{u}_3) = \left\langle \frac{1}{2}, \frac{1}{2}, 1 \right\rangle \cdot \frac{1}{2} = \frac{3}{2} \times \frac{1}{2} = \left\langle \frac{1}{4}, \frac{1}{4}, \frac{1}{2} \right\rangle \).Now calculate the projection of \( \mathbf{u}_3 \) onto \( \mathbf{v}_2 \). Compute \( \langle \mathbf{u}_3, \mathbf{v}_2 \rangle = \left\langle -\frac{1}{3}, \frac{5}{3}, \frac{5}{6} \right\rangle \), simplifying gives \( \frac{14}{9} \), knowing \( \langle \mathbf{v}_2, \mathbf{v}_2 \rangle = \frac{109}{18} \), \( \text{proj}_{\mathbf{v}_2}(\mathbf{u}_3) = \left\langle -\frac{1}{3}, \frac{5}{3}, \frac{5}{6} \right\rangle \) is required. Calculate \[ \mathbf{v}_3 = \mathbf{u}_3 - \left\langle \frac{1}{4}, \frac{1}{4}, \frac{1}{2} \right\rangle - \left\langle \frac{1}{4}, \frac{5}{12}, \frac{5}{12} \right\rangle \] simplifying \( \mathbf{u}_3 \) gives \[ \mathbf{v}_3 = \left\langle -\frac{3}{2}, -\frac{3}{2}, -1 \right\rangle \].

Step 5: Normalize the Basis to Obtain \( B'' \)

Convert the orthogonal basis into an orthonormal basis by normalizing each vector.Normalize \( \mathbf{v}_1 \): \( \|\mathbf{v}_1\| = \sqrt{\frac{3}{2}} \), so \( \mathbf{w}_1 = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \sqrt{2} \right\rangle \).Normalize \( \mathbf{v}_2 \): \( \|\mathbf{v}_2\| = \sqrt{\frac{109}{18}} \), so \( \mathbf{w}_2 = \left\langle -\frac{1}{3\sqrt{109}}, \frac{5}{3\sqrt{109}}, \frac{5}{6\sqrt{109}} \right\rangle \).Normalize \( \mathbf{v}_3 \): \( \|\mathbf{v}_3\| = \sqrt{\frac{3}{2}} \), so \( \mathbf{w}_3 = \left\langle -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{1}{2} \right\rangle \).

Key Concepts

Orthogonal Basis

When we talk about an orthogonal basis in a vector space, we're referring to a set of vectors that are mutually perpendicular to each other. This means that every pair of distinct vectors in the set has a dot product of zero. This concept is extremely useful in simplifying complex vector problems, especially in higher dimensions.

To visualize this, consider the 2D plane. If two vectors form a right angle with each other, they are orthogonal. Translating this to a set such as the one we encounter in the Gram-Schmidt process for vector spaces in three dimensions, we aim to manipulate a basis set so that they all form right angles with each other.

Here are some of the benefits of using an orthogonal basis:

• Simplifies vector projections: Calculating the projection of a vector onto an orthogonal basis vector becomes straightforward.
• Increases computational efficiency: Since they are perpendicular, the computations involving dot products and cross products become easier.
• Provides a clear geometric interpretation: Help in visualizing vector operations and transformations.

Orthonormal Basis

An orthonormal basis takes the orthogonal basis a step further by normalizing each vector. This means each vector in the basis not only is orthogonal to every other vector but also has a magnitude (or length) of one. Therefore, an orthonormal basis is often preferred for ease of calculation and precision in representing vector quantities in spaces.

Once you have an orthogonal basis using the Gram-Schmidt process, converting it into an orthonormal basis involves normalizing each vector. This is done by dividing each vector by its magnitude. This ensures each vector in the set is a unit vector.

Benefits of an orthonormal basis include:

• Simplification in calculations: Dot products of basis vectors are either 0 or 1, as they are perpendicular and unit-length.
• Ease of use in transforming and rotating vector spaces: It becomes easier to control and manipulate when each vector has the same length.
• Utility in various mathematical and engineering applications: Commonly used in fields like computer graphics and signal processing.

Vector Spaces

A vector space is a foundational concept in linear algebra and provides a framework for discussing vectors and related operations. In a vector space, vectors can be added together and multiplied by scalars to produce new vectors that also belong to the same space. This concept is crucial because it abstracts and generalizes real-world methods of handling quantities like forces, velocities, and more.

For a set to be considered a vector space, it must adhere to certain properties or axioms:

• Closure under addition and scalar multiplication.
• Presence of an additive identity, often a zero vector.
• Existence of additive inverses, meaning each vector has a counterpart that results in a zero vector when added.

In practical terms, vector spaces allow for systematic techniques to solve equations and analyze linear transformations. Within this context, orthogonal and orthonormal bases contribute significantly by providing structured means to represent and manipulate these vector spaces efficiently, especially in higher dimensions.