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Chapter 5: Problem 4

The given vectors form a basis for \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\). Apply the Gram-Schmidt Process to obtain an orthogonal basis. Then normalize this basis to obtain an orthonormal basis. $$\mathbf{x}_{1}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right], \mathbf{x}_{2}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right], \mathbf{x}_{3}=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right]$$

Short Answer

Expert verified
The orthonormal basis is \( [\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}] \), \( [\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}] \), \( [\frac{5}{\sqrt{50}}, -\frac{4}{\sqrt{50}}, \frac{5}{\sqrt{50}}] \).

Step by step solution

01

Confirm the Dimension

The vectors \( \mathbf{x}_1, \mathbf{x}_2, \mathbf{x}_3 \) are in \( \mathbb{R}^3 \) since each has three components. We will check if they form a basis for \( \mathbb{R}^3 \), which requires that they be linearly independent.
02

Check Linear Independence

Form a matrix \( A \) with the three vectors as columns and row-reduce it:\[A = \begin{bmatrix}1 & 1 & 1 \ 1 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix} \]Perform row operations to achieve reduced row-echelon form (RREF):\[RREF(A) = \begin{bmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]Since the RREF of \( A \) is the identity matrix, \( \mathbf{x}_1, \mathbf{x}_2, \mathbf{x}_3 \) are linearly independent and form a basis for \( \mathbb{R}^3 \).
03

Apply Gram-Schmidt Process

Given \( \mathbf{x}_1, \mathbf{x}_2, \mathbf{x}_3 \), we apply Gram-Schmidt:1. Set \( \mathbf{v}_1 = \mathbf{x}_1 = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \).2. Compute \( \mathbf{v}_2 = \mathbf{x}_2 - \text{proj}_{\mathbf{v}_1}(\mathbf{x}_2) \). \[ \text{proj}_{\mathbf{v}_1}(\mathbf{x}_2) = \frac{\mathbf{x}_2 \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \mathbf{v}_1 = \frac{2}{3} \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \ \frac{2}{3} \ \frac{2}{3} \end{bmatrix} \] \[ \mathbf{v}_2 = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} - \begin{bmatrix} \frac{2}{3} \ \frac{2}{3} \ \frac{2}{3} \end{bmatrix} = \begin{bmatrix} \frac{1}{3} \ \frac{1}{3} \ -\frac{2}{3} \end{bmatrix} \]3. Compute \( \mathbf{v}_3 = \mathbf{x}_3 - \text{proj}_{\mathbf{v}_1}(\mathbf{x}_3) - \text{proj}_{\mathbf{v}_2}(\mathbf{x}_3) \). \[ \text{proj}_{\mathbf{v}_1}(\mathbf{x}_3) = \frac{1}{3} \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} \ \frac{1}{3} \ \frac{1}{3} \end{bmatrix} \] \[ \text{proj}_{\mathbf{v}_2}(\mathbf{x}_3) = \frac{1}{9} \begin{bmatrix} \frac{1}{3} \ \frac{1}{3} \ -\frac{2}{3} \end{bmatrix} = \begin{bmatrix} \frac{1}{9} \ \frac{1}{9} \ -\frac{2}{9} \end{bmatrix} \] \[ \mathbf{v}_3 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} - \begin{bmatrix} \frac{1}{3} \ \frac{1}{3} \ \frac{1}{3} \end{bmatrix} - \begin{bmatrix} \frac{1}{9} \ \frac{1}{9} \ -\frac{2}{9} \end{bmatrix} = \begin{bmatrix} \frac{5}{9} \ -\frac{4}{9} \ \frac{5}{9} \end{bmatrix} \]
04

Normalize the Vectors

Normalize \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \) to form an orthonormal basis:1. \( \mathbf{u}_1 = \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|} = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \).2. \( \mathbf{u}_2 = \frac{\mathbf{v}_2}{\|\mathbf{v}_2\|} = \frac{1}{\sqrt{\frac{2}{3}}} \begin{bmatrix} \frac{1}{3} \ \frac{1}{3} \ -\frac{2}{3} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{6}} \ \frac{1}{\sqrt{6}} \ -\frac{2}{\sqrt{6}} \end{bmatrix} \).3. \( \mathbf{u}_3 = \frac{\mathbf{v}_3}{\|\mathbf{v}_3\|} = \frac{1}{\sqrt{\frac{50}{81}}} \begin{bmatrix} \frac{5}{9} \ -\frac{4}{9} \ \frac{5}{9} \end{bmatrix} = \begin{bmatrix} \frac{5}{\sqrt{50}} \ -\frac{4}{\sqrt{50}} \ \frac{5}{\sqrt{50}} \end{bmatrix} \).
05

Final Orthonormal Basis

The orthonormal basis for \(\mathbb{R}^3\) is:\[\mathbf{u}_1 = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}, \quad \mathbf{u}_2 = \begin{bmatrix} \frac{1}{\sqrt{6}} \ \frac{1}{\sqrt{6}} \ -\frac{2}{\sqrt{6}} \end{bmatrix}, \quad \mathbf{u}_3 = \begin{bmatrix} \frac{5}{\sqrt{50}} \ -\frac{4}{\sqrt{50}} \ \frac{5}{\sqrt{50}} \end{bmatrix}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Independence
Linear independence is a key concept in understanding vector spaces. When vectors are linearly independent, no vector in the set can be written as a linear combination of the others. This property is essential to determine if a set of vectors forms a basis for a vector space. In the context of our exercise, we have three vectors
  • \(\mathbf{x}_{1}=\begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}\)
  • \(\mathbf{x}_{2}=\begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix}\)
  • \(\mathbf{x}_{3}=\begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}\)
To verify their linear independence, we place them as columns in a matrix \(A\) and row-reduce it. When the reduced row-echelon form (RREF) of \(A\) turns out to be the identity matrix, we confirm linear independence. This means they form a basis for \(\mathbb{R}^3\). When vectors are linearly independent, they span a space without redundancy, providing a unique and efficient representation of that space.
Orthogonal Basis
An orthogonal basis is a set of vectors that are not only linearly independent but also perpendicular to each other. Orthogonality simplifies many computations, such as projections and decompositions, making it a desirable property.
To transform a set of vectors into an orthogonal basis, we use the Gram-Schmidt Process. This process systematically adjusts each vector by removing components that are parallel to preceding vectors in the set, ensuring perpendicularity.For example, starting with \(\mathbf{v}_1 = \mathbf{x}_1\), we modify \(\mathbf{x}_2\) by subtracting its projection onto \(\mathbf{v}_1\). This results in \(\mathbf{v}_2\), a vector orthogonal to \(\mathbf{v}_1\). This is repeated for \(\mathbf{x}_3\), adjusting it using projections onto both \(\mathbf{v}_1\) and \(\mathbf{v}_2\). The output is a refined set \(\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}\), which are mutually orthogonal vectors.
This orthogonal set simplifies many matrix-related computations and is foundational in numerical methods.
Orthonormal Basis
An orthonormal basis is an extension of the concept of an orthogonal basis by normalizing the vectors (i.e., making them unit vectors). Orthonormal bases are highly desirable because they retain the properties of orthogonal bases and simplify further calculations through normalization.
In our exercise, after applying the Gram-Schmidt process to obtain the orthogonal vectors, we further adjust them to form an orthonormal basis. This is done by dividing each orthogonal vector \(\mathbf{v}_i\) by its length or norm \(\|\mathbf{v}_i\|\), converting them into unit length vectors (\(\mathbf{u}_i\)). Each \(\mathbf{u}_i\) satisfies both orthogonality and unit length properties.
  • \(\mathbf{u}_1 = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}\)
  • \(\mathbf{u}_2 = \begin{bmatrix} \frac{1}{\sqrt{6}} \ \frac{1}{\sqrt{6}} \ -\frac{2}{\sqrt{6}} \end{bmatrix}\)
  • \(\mathbf{u}_3 = \begin{bmatrix} \frac{5}{\sqrt{50}} \ -\frac{4}{\sqrt{50}} \ \frac{5}{\sqrt{50}} \end{bmatrix}\)
Using orthonormal bases makes vector operations straightforward and consistent, benefiting various fields such as computer graphics, machine learning, and more.

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Most popular questions from this chapter

Find the orthogonal projection of v onto the subspace \(W\) spanned by the vectors \(\mathbf{u}_{\mathbf{r}}\) (You may assume that the vectors \(\mathbf{u}_{i}\) are orthogonal. ) $$\mathbf{v}=\left[\begin{array}{r} 3 \\ -2 \\ 4 \\ -3 \end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \\ 0 \end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r} 1 \\ -1 \\ -1 \\ 1 \end{array}\right], \mathbf{u}_{3}=\left[\begin{array}{l} 0 \\ 0 \\ 1 \\ 1 \end{array}\right]$$

Classify each of the quadratic forms as positive definite, positive semidefinite, negative definite, negative semidefinite, or indefinite. $$x_{1}^{2}+x_{2}^{2}-x_{3}^{2}+4 x_{1} x_{2}$$

Use a translation of axes to put the conic in standard position. Identify the graph, give its equation in the translated coordinate system, and sketch the curve. $$4 x^{2}+10 x y+4 y^{2}=9$$

Let \(A=\left[\begin{array}{ll}a & b \\ b & d\end{array}\right]\) be a symmetric \(2 \times 2\) matrix. Prove that \(A\) is positive definite if and only if \(a>0\) and \(\operatorname{det} A>0 .\left[\text {Hint}: a x^{2}+2 b x y+d y^{2}=\right.\) \(\left.a\left(x+\frac{b}{a} y\right)^{2}+\left(d-\frac{b^{2}}{a}\right) y^{2} \cdot\right]\)

Let \(A\) be a real \(2 \times 2\) matrix with complex eigenvalues \(\lambda=a \pm b i\) such that \(b \neq 0\) and \(|\lambda|=1 .\) Prove that every trajectory of the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) lies on an ellipse. [Hint: Theorem 4.43 shows that if \(\mathbf{v}\) is an eigenvector corresponding to \(\lambda=a-b i\), then the matrix \(P=[\operatorname{Re} \mathbf{v} \quad \operatorname{Im} \mathbf{v}]\) is invertible and \(A=P\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right] P^{-1} .\) Set \(B=\left(P P^{T}\right)^{-1} .\) Show that the quadratic \(\mathbf{x}^{T} B \mathbf{x}=k\) defines an ellipse for all \(k>0\) and prove that if \(x \text { lies on this ellipse, so does } A x .]\)
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