Problem 2 - Advanced Engineering Mathematics

Question

Verify that the basis \(B\) for the given vector space is orthonormal. Use Theorem \(7.7 .1\) to find the coordinates of the vector \(\mathbf{u}\) relative to the basis \(B .\) Then write \(\mathbf{u}\) as a linear combination of the basis vectors. $$ \begin{aligned} &B=\left\\{\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right\rangle,\left\langle 0,-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right\rangle\right. \\ &\left.\left\langle-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right\rangle\right\\}, \quad R^{3} ; \quad \mathbf{u}=\langle 5,-1,6\rangle \end{aligned} $$

Step-by-Step Solution

Answer

The basis is orthonormal, and \( \mathbf{u} = 4\mathbf{b}_1 + \frac{1}{\sqrt{2}}\mathbf{b}_2 - \frac{1}{\sqrt{6}}\mathbf{b}_3 \).

Step 1: Verify Orthonormality of Basis

A set of vectors is orthonormal if each vector has a magnitude of 1 and is orthogonal to the other vectors. To check the magnitude, compute the dot product of each vector with itself. The vector \( \left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right\rangle \) has magnitude 1 as \( \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(-\frac{1}{\sqrt{3}}\right)^2 = 1 \). Similarly, verify this for the other vectors. Next, to verify orthogonality, calculate their pairwise dot products all being zero. Thus, the given basis \( B \) is orthonormal.

Step 2: Find Coordinates of \( \mathbf{u} \) Relative to Basis

According to Theorem 7.7.1, if \( B \) is an orthonormal basis for \( \mathbb{R}^3 \), then the coordinates of vector \( \mathbf{u} \) relative to \( B \) are \( c_i = \mathbf{u} \cdot \mathbf{b}_i \) for each basis vector \( \mathbf{b}_i \). Compute \( c_1 = \mathbf{u} \cdot \left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right\rangle = 5\frac{1}{\sqrt{3}} - 1\frac{1}{\sqrt{3}} - 6\left(-\frac{1}{\sqrt{3}}\right) = \frac{12}{\sqrt{3}} = 4 \). Repeat for \( \mathbf{b}_2 \) and \( \mathbf{b}_3 \) to find \( c_2 = \frac{1}{\sqrt{2}} \) and \( c_3 = -\frac{1}{\sqrt{6}} \).

Step 3: Express \( \mathbf{u} \) as a Linear Combination

Write \( \mathbf{u} \) as a linear combination of the basis vectors with the calculated coordinates: \( \mathbf{u} = 4\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right\rangle + \frac{1}{\sqrt{2}}\left\langle0, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right\rangle - \frac{1}{\sqrt{6}}\left\langle-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}\right\rangle \). Simplify the expression to confirm \( \langle 5, -1, 6 \rangle \).

Key Concepts

Vector Spaces

A vector space is a fundamental concept in linear algebra. It's essentially a collection of vectors, which are objects that follow specific rules under addition and scalar multiplication. More formally, vector spaces are equipped with operations that, when applied to any vectors in the space, produce another vector in the same space.

In a vector space, if you take any two vectors and add them, or multiply them by a scalar (a real number), their result will also belong to that vector space. This property is significant because it provides the foundation for constructing more complex geometrical and algebraic structures. Vector spaces can be conceived as infinite-dimensional when they contain infinitely many independent directions, or finite-dimensional like the familiar \(\mathbb{R}^3\) space, the focus in our problem.

• Closure under addition and scalar multiplication
• Containment of a zero vector
• Associativity and commutativity of vector addition
• Distribution properties of scalar multiplication

Linear Combination

Linear combinations are expressions created by multiplying vectors by scalars and then adding the products together. In the world of vector spaces, any vector can often be expressed as a linear combination of a set of basis vectors.

For example, given an orthonormal basis \( B \) and a target vector \( \mathbf{u} \) in the vector space \( \mathbb{R}^3 \), expressing \( \mathbf{u} \) as a linear combination of \( B \)'s vectors involves finding scalars \( c_1, c_2, \) and \( c_3 \), such that:

• \( \mathbf{u} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + c_3\mathbf{b}_3 \)
• Each \( c_i \) is calculated using the dot product of \( \mathbf{u} \) with each \( \mathbf{b}_i \)

Learning to express vectors in this manner enables simplification of complex problems, and it highlights the beauty of linear combinations in spanning and defining the entirety of vector spaces.

Dot Product

The dot product is a mathematical operation that takes in two vectors and returns a scalar. This fundamental tool helps us determine the degree of alignment between vectors and is crucial for confirming properties like orthogonality.

The dot product is performed by multiplying corresponding components of two vectors and summing up those products. For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) the dot product is defined as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

In the context of our problem, the dot product is used to check orthonormality of a basis by ensuring each vector has a magnitude of one and that pairwise dot products are zero. Obtaining the coordinates of a vector relative to an orthonormal basis also necessitates using the dot product formula.

Theorem 7.7.1

Theorem 7.7.1 is key when dealing with orthonormal bases in vector spaces. It states that if you have an orthonormal basis for a vector space, the coordinates of any vector relative to that basis can be calculated using the dot products between the vector and each basis vector.

This theorem simplifies the problem of finding coordinates considerably when you're working with orthonormal bases, compared to non-orthonormal bases where the process can be more involved. Essentially:

• If \( B \) is an orthonormal basis, and \( \mathbf{u} \) is a vector in the space, then each coordinate \( c_i \) of \( \mathbf{u} \) relative to \( B \) is calculated as \( \mathbf{u} \cdot \mathbf{b}_i \) for each basis vector \( \mathbf{b}_i \)
• This process utilizes the property that the dot product simplifies the projection of vectors in this special case

Theorem 7.7.1 is invaluable for anyone working with orthonormal bases as it provides a straightforward mechanism for breaking down vectors into simpler components.