Question

In each case, use the Gram-Schmidt algorithm to convert the given basis \(B\) of \(V\) into an orthogonal basis. a. \(V=\mathbb{R}^{2}, B=\\{(1,-1),(2,1)\\}\) b. \(V=\mathbb{R}^{2}, B=\\{(2,1),(1,2)\\}\) c. \(V=\mathbb{R}^{3}, B=\\{(1,-1,1),(1,0,1),(1,1,2)\\}\) d. \(V=\mathbb{R}^{3}, B=\\{(0,1,1),(1,1,1),(1,-2,2)\\}\)

Short answer

Orthogonal bases: (a) \(\{(1,-1),(1.5,1.5)\}\); (b) \(\{(2,1),(-0.6,1.2)\}\); (c) \(\{(1,-1,1),(\frac{1}{3},\frac{2}{3},\frac{1}{3}),(0,0,1)\}\); (d) \(\{(0,1,1),(1,0,0),(0,-2,2)\}\).

Step-by-step solution

Understanding the Gram-Schmidt Process

The Gram-Schmidt process takes a set of vectors and produces an orthogonal set of vectors. Starting with the first vector, we progressively subtract components that are in the direction of previously orthogonalized vectors, ensuring the new vector is orthogonal to all previously determined vectors.

Orthogonalizing the First Set

For case (a), use the given vectors \( (1,-1) \) and \( (2,1) \). Set \( u_1 = (1,-1) \). Then, apply\[ u_2 = (2,1) - \text{proj}_{u_1}(2,1), \] where \( \text{proj}_{u_1}(2,1) = \frac{(2,1) \cdot (1,-1)}{(1,-1)\cdot(1,-1)} \cdot (1,-1).\)Compute the dot products and simplify:\[ u_2 = (2,1) - \frac{2 imes 1 + 1 imes (-1)}{1^2 + (-1)^2} (1,-1) = (2,1) - \frac{1}{2} (1,-1). \]evaluate\[ u_2 = (2,1) - (0.5,-0.5) = (1.5, 1.5). \] So, the orthogonal basis is \( \{ (1,-1), (1.5, 1.5) \} \).

Orthogonalizing the Second Set

For case (b), use the given vectors \( (2,1) \) and \( (1,2) \). Set \( u_1 = (2,1) \). Then,\[ u_2 = (1,2) - \text{proj}_{u_1}(1,2), \] where \( \text{proj}_{u_1}(1,2) = \frac{(1,2) \cdot (2,1)}{(2,1)\cdot(2,1)} \cdot (2,1).\)Compute\[ u_2 = (1,2) - \frac{1 imes 2 + 2 imes 1}{2^2 + 1^2} (2,1) = (1,2) - \frac{4}{5} (2,1). \]Calculate\[ u_2 = (1,2) - (1.6,0.8) = (-0.6, 1.2). \] Thus, the orthogonal basis is \( \{ (2,1), (-0.6, 1.2) \} \).

Orthogonalizing the Third Set

For case (c), use the given vectors \( (1,-1,1), (1,0,1), (1,1,2) \).1. Set \( u_1 = (1,-1,1) \).2. Compute\[ u_2 = (1,0,1) - \text{proj}_{u_1}(1,0,1), \] where \( \text{proj}_{u_1}(1,0,1) = \frac{(1,0,1) \cdot (1,-1,1)}{(1,-1,1)\cdot(1,-1,1)}\cdot(1,-1,1).\)Evaluate\[ u_2 = (1,0,1) - \frac{2}{3} (1,-1,1) = (1,0,1) - \left(\frac{2}{3},-\frac{2}{3},\frac{2}{3}\right). \]Reduce\[ u_2 = \left(\frac{1}{3}, \frac{2}{3}, \frac{1}{3}\right). \]3. Compute\[ u_3 = (1,1,2) - \text{proj}_{u_1}(1,1,2) - \text{proj}_{u_2}(1,1,2). \]Find\[ \text{proj}_{u_1} = \frac{4}{3} (1,-1,1),\text{proj}_{u_2} = 1 \left(\frac{1}{3}, \frac{2}{3}, \frac{1}{3}\right). \]Finally, solve\[ u_3 = (1,1,2) - \left(\frac{4}{3} - \frac{1}{3}, \frac{4}{3} + \frac{2}{3}, \frac{4}{3} - \frac{1}{3}\right). \]Thus, the orthogonal basis: \( \{(1,-1,1), \left(\frac{1}{3},\frac{2}{3},\frac{1}{3}\right), \left(\frac{0}{1}, 0, 1\right) \} \).

Orthogonalizing the Fourth Set

For case (d), use *(0,1,1), (1,1,1), (1,-2,2)*.1. Set \( u_1 = (0,1,1) \).2. Compute\[ u_2 = (1,1,1) - \text{proj}_{u_1}(1,1,1), \] where \( \text{proj}_{u_1}\) is\[ \frac{2}{2} (0,1,1).\]Chanve\[ u_2 = (1,1,1) - (0,1,1) = (1,0,0). \]3. Compute\[ u_3 = (1,-2,2) - \text{proj}_{u_1}(1,-2,2) - \text{proj}_{u_2}(1,-2,2). \]Calculate\[ \text{proj}_{u_1} = 0 (0,1,1), \text{proj}_{u_2} = 1 (1,0,0). \]Finally\[ u_3 = (1,-2,2) - (1,0,0) = (0,-2,2). \]Thus, the orthogonal basis is \( \{ (0,1,1), (1,0,0), (0,-2,2) \} \).

Key concepts

Orthogonal Basis

In the world of linear algebra, an orthogonal basis is a set of vectors in a vector space that are all perpendicular to each other. This implies that the dot product (or inner product) of any pair of distinct vectors in the set is zero. Orthogonal sets are particularly desirable in computations because they simplify a lot of matrix operations.

• An orthogonal basis maintains the "shape" of the space, meaning geometrical properties are preserved during transformations.
• Each vector's length is unaffected by its peers, making equations easier to handle.
• To make a basis orthonormal, each vector is normalized, meaning each vector is turned into a unit vector.

When dealing with problems, using the Gram-Schmidt process is a common way to convert any basis into an orthogonal one. The process involves subtracting components aligned with previously determined orthogonal vectors from each subsequent vector until the entire set is orthogonal. This strategy is beneficial as it keeps calculations neat and reduces errors.

Linear Algebra

Linear algebra is a branch of mathematics that is concerned with vectors, vector spaces (also called linear spaces), linear transformations, and systems of linear equations. It is a fundamental part of modern mathematics and has wide applications across various domains including physics, computer science, engineering, and more. Here are a few essential components of linear algebra that are relevant to understanding orthogonal bases:

• Vectors: Objects that have both a magnitude and a direction; they are represented as coordinates in space.
• Vector Spaces: Collections of vectors that can be added together and multiplied by scalars to produce another vector within the space.
• Basis: A set of vectors in a vector space such that every vector in the space can be expressed as a linear combination of them.
• Linear Transformations: Functions that map one vector space to another, adhering to the rules of scalar multiplication and vector addition.

Understanding linear algebra provides the algebraic framework necessary for solving problems involving vector spaces and finding orthogonal bases.

Vector Spaces

A vector space is a set of vectors where vector addition and scalar multiplication are defined and satisfy certain axioms such as distributivity, associativity, and the existence of an additive identity and inverses. Vector spaces play a vital role in understanding the concepts of linear transformations and span many areas of applied mathematics. Here's what you need to know:

• Subspace: A subset of a vector space that itself forms a vector space.
• Linear Combination: An expression constructed from a set of terms by multiplying each term by a constant and adding the results.
• Dimension: The number of vectors in a basis of the vector space. It is a measure of the "size" or "degree of freedom" of the space.
• Span: The set of all possible vectors that can be created as a linear combination of a given set of vectors.

The concept of vector spaces allows for a deeper comprehension of when and how the Gram-Schmidt algorithm can be applied, especially in transforming a given basis into an orthogonal basis. Understanding the structural properties of these spaces supports solving complex problems in linear algebra.