Question

We often write vectors in \(\mathbb{R}^{n}\) as row \(\mathrm{n}\) -tuples. Obtain orthonormal bases of \(\mathbb{R}^{3}\) by normalizing the following. $$ \begin{aligned} &\text { a. }\\{(1,-1,2),(0,2,1),(5,1,-2)\\}\\\ &\text { b. }\\{(1,1,1),(4,1,-5),(2,-3,1)\\} \end{aligned} $$

Short answer

a: \( \left\{ \frac{1}{\sqrt{6}}(1,-1,2), \frac{1}{\sqrt{5}}(0,2,1), \frac{1}{\sqrt{30}}(5,1,-2) \right\} \) b: \( \left\{ \frac{1}{\sqrt{3}}(1,1,1), \frac{1}{\sqrt{42}}(4,1,-5), \frac{1}{\sqrt{14}}(2,-3,1) \right\} \)

Step-by-step solution

Understanding Problem Requirements

We need to find orthonormal bases from provided vectors by normalizing vectors and ensuring they are orthogonal.

Gram-Schmidt Orthogonalization (Part a)

Apply Gram-Schmidt process to vectors \((1,-1,2)\), \((0,2,1)\), \((5,1,-2)\).1. Let \(v_1 = (1,-1,2)\).2. Project \((0,2,1)\) onto \(v_1\): \[ \text{proj}_{v_1}(0,2,1) = \frac{(0,2,1)\cdot(1,-1,2)}{(1,-1,2)\cdot(1,-1,2)}(1,-1,2) = \frac{0+(-2)+2}{1+1+4}(1,-1,2) = \frac{0}{6}(1,-1,2) = (0,0,0) \] So \(u_2 = (0,2,1) - (0,0,0) = (0,2,1)\).3. Now consider \((5,1,-2)\) and make it orthogonal: \[ \text{proj}_{v_1}(5,1,-2) = \frac{(5,1,-2)\cdot(1,-1,2)}{6}(1,-1,2) = \frac{5-1-4}{6}(1,-1,2) = 0\] \[ \text{proj}_{u_2}(5,1,-2) = \frac{(5,1,-2)\cdot(0,2,1)}{(0,2,1)\cdot(0,2,1)}(0,2,1) = \frac{0+2-2}{4+1}(0,2,1) = (0,0,0)\] \(u_3 = (5,1,-2) - (0,0,0)\) remains \((5,1,-2)\).

Normalize Vectors (Part a)

Normalize the vectors \(v_1 = (1,-1,2)\), \(u_2 = (0,2,1)\), \(u_3 = (5,1,-2)\).1. Normalizing \(v_1\): \[ e_1 = \frac{1}{\sqrt{6}}(1,-1,2) \]2. Normalizing \(u_2\): \[ e_2 = \frac{1}{\sqrt{5}}(0,2,1) \]3. Normalizing \(u_3\): \[ e_3 = \frac{1}{\sqrt{30}}(5,1,-2) \]Thus, the orthonormal basis is \(\left\{ \frac{1}{\sqrt{6}}(1,-1,2), \frac{1}{\sqrt{5}}(0,2,1), \frac{1}{\sqrt{30}}(5,1,-2) \right\}\).

Gram-Schmidt Orthogonalization (Part b)

Apply Gram-Schmidt process to vectors \((1,1,1)\), \((4,1,-5)\), \((2,-3,1)\).1. Let \(v_1 = (1,1,1)\).2. Project \((4,1,-5)\) onto \(v_1\): \[ \text{proj}_{v_1}(4,1,-5) = \frac{(4,1,-5)\cdot(1,1,1)}{3}(1,1,1) = \frac{0}{3}(1,1,1) = (0,0,0) \] So \(u_2 = (4,1,-5) - (0,0,0) = (4,1,-5)\).3. Now consider \((2,-3,1)\) and make it orthogonal: \[ \text{proj}_{v_1}(2,-3,1) = \frac{(2,-3,1)\cdot(1,1,1)}{3}(1,1,1) = 0 \] \[ \text{proj}_{u_2}(2,-3,1) = \frac{(2,-3,1)\cdot(4,1,-5)}{42}(4,1,-5) = 0\] \(u_3 = (2,-3,1)\).

Normalize Vectors (Part b)

Normalize the vectors \(v_1 = (1,1,1)\), \(u_2 = (4,1,-5)\), \(u_3 = (2,-3,1)\).1. Normalizing \(v_1\): \[ e_1 = \frac{1}{\sqrt{3}}(1,1,1) \]2. Normalizing \(u_2\): \[ e_2 = \frac{1}{\sqrt{42}}(4,1,-5) \]3. Normalizing \(u_3\): \[ e_3 = \frac{1}{\sqrt{14}}(2,-3,1) \]Thus, the orthonormal basis is \(\left\{ \frac{1}{\sqrt{3}}(1,1,1), \frac{1}{\sqrt{42}}(4,1,-5), \frac{1}{\sqrt{14}}(2,-3,1) \right\}\).

Key concepts

Gram-Schmidt Process

The Gram-Schmidt Process is a method used in linear algebra to convert a set of linearly independent vectors into an orthogonal set. This orthogonal set is then normalized to form an orthonormal set. This process is particularly useful when you want to find orthonormal bases in spaces like \(\mathbb{R}^{3}\), which means sets of vectors that are both orthogonal to each other and are unit vectors.

Here's how it works:

• Start with a set of linearly independent vectors.
• Take each vector and "project" it onto each of the previous vectors in the set, subtracting this projection from the vector to make it orthogonal with respect to the previous vectors.
• Continue this step-by-step for each vector in the set until the complete set is orthogonal.

Once your vectors are orthogonal, you can easily make them orthonormal by normalizing each one. The final set is an orthonormal basis for your vector space. Understanding this process involves careful attention to both the algebraic manipulation of vector coordinates and the geometric intuition behind projecting one vector onto another. It also helps to visualize how orthogonality ensures that no vector in the set is pointing in the same general direction as another.

Vector Normalization

Vector Normalization is the process of adjusting the length of a vector to make it a unit vector, which means its length becomes 1. It is a critical step when creating orthonormal bases, such as those found using the Gram-Schmidt Process.

To normalize a vector, follow these steps:

• Find the magnitude of the vector. This is done by taking the square root of the sum of the squares of its components.
• Divide each component of the vector by its magnitude. This scales the vector to a new length of 1, maintaining its direction but ensuring it is now a unit vector.

For example, given a vector \( \mathbf{v} = (x, y, z) \), the magnitude is calculated as \( \sqrt{x^2 + y^2 + z^2} \). The normalized vector \( \mathbf{e} \) is \( \mathbf{e} = \frac{1}{\sqrt{x^2 + y^2 + z^2}}(x, y, z) \). With normalized vectors, you can be sure they form the desired orthonormal basis when paired with the orthogonal vectors from Gram-Schmidt. Normalizing is not only about scaling but about ensuring that the underlying geometry of your vector space is preserved and usable for further mathematical operations.

Linear Algebra

Linear Algebra is a branch of mathematics concerned with vectors, vector spaces (also known as linear spaces), and linear mappings between these spaces. It involves the study of lines, planes, and subspaces, but it also extends to higher dimensions. This subject forms the foundation for numerous mathematical theories and practical applications in science and engineering, including computer graphics, machine learning, and quantum mechanics.

Key concepts in linear algebra, such as vector spaces, convergence of vectors, and transformations, are essential for understanding the ideas behind orthonormal bases, Gram-Schmidt Process, and more. In particular, linear algebra provides the language and frameworks needed to work with systems of linear equations and transformations that are at the heart of the Gram-Schmidt process and normalization of vectors.

• Vectors: Basic objects in linear algebra, represented as directed segments with both magnitude and direction.
• Vector Spaces: Collections of vectors where vector addition and scalar multiplication are defined.
• Systems of Equations: Linear systems that can be solved using matrices and determinants, common in linear algebra.

Mastering linear algebra involves understanding these fundamental components and recognizing their interconnectedness within wider mathematical applications.