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Chapter 6: Problem 17

Determine which sets of vectors are orthonormal. If a set is only orthogonal, normalize the vectors to produce an orthonormal set. \(\left[\begin{array}{c}{1 / 3} \\ {1 / 3} \\ {1 / 3}\end{array}\right],\left[\begin{array}{c}{-1 / 2} \\ {0} \\ {1 / 2}\end{array}\right]\)

Short Answer

Expert verified
The vectors are orthogonal; the orthonormal set is \( \left[ \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right] \) and \( \left[ -\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \right] \)."

Step by step solution

01

Checking Orthogonality

To determine if the vectors are orthogonal, we compute their dot product. Let \( \mathbf{a} = \left[ \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right] \) and \( \mathbf{b} = \left[ -\frac{1}{2}, 0, \frac{1}{2} \right] \). The dot product is computed as follows: \[ \mathbf{a} \cdot \mathbf{b} = \frac{1}{3}(-\frac{1}{2}) + \frac{1}{3}(0) + \frac{1}{3}\left(\frac{1}{2}\right) = -\frac{1}{6} + 0 + \frac{1}{6} = 0. \] Since the dot product is zero, the vectors are orthogonal.
02

Normalizing the Vectors

Next, we need to normalize each vector to produce an orthonormal set. A vector \( \mathbf{v} \) is normalized by dividing by its magnitude: \[ \mathbf{v}_{norm} = \frac{\mathbf{v}}{\|\mathbf{v}\|}. \] First, calculate the magnitude of each vector.
03

Computing the Magnitude of Vector a

The magnitude of vector \( \mathbf{a} = \left[ \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right] \) is: \[ \| \mathbf{a} \| = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2} = \sqrt{\frac{1}{9} + \frac{1}{9} + \frac{1}{9}} = \sqrt{\frac{3}{9}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}. \]
04

Normalizing Vector a

Normalize vector \( \mathbf{a} \) by dividing by its magnitude: \[ \mathbf{a}_{norm} = \frac{1}{3} \left( \frac{1}{\sqrt{3}} \right) \left[ 1, 1, 1 \right] = \left[ \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right]. \]
05

Computing the Magnitude of Vector b

The magnitude of vector \( \mathbf{b} = \left[ -\frac{1}{2}, 0, \frac{1}{2} \right] \) is: \[ \| \mathbf{b} \| = \sqrt{\left(-\frac{1}{2}\right)^2 + 0 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}. \]
06

Normalizing Vector b

Normalize vector \( \mathbf{b} \) by dividing by its magnitude: \[ \mathbf{b}_{norm} = \frac{1}{2} \left( \frac{1}{\sqrt{2}} \right) \left[ -1, 0, 1 \right] = \left[ -\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \right]. \]
07

Final Orthonormal Set

The orthonormal set is composed of the normalized versions of the original vectors: \[ \left[ \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right] \text{ and } \left[ -\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \right]. \]

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Key Concepts

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

Orthogonality
Orthogonality is a core concept in vector algebra involving two vectors being perpendicular to each other. Vectors are orthogonal when their dot product equals zero. You can think of it as a multidimensional version of the right angle concept you see in geometry. Consider two vectors, \( \mathbf{a} \) and \( \mathbf{b} \). Their dot product is computed by multiplying corresponding components and summing up the result. If the dot product is zero, it means that the vectors are at right angles to each other.
To check if vectors are orthogonal, calculate their dot product using the formula \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). If this equals zero, the vectors do not influence each other, which is crucial in calculations where independence is important.
Vector Normalization
Vector normalization is the process of converting a vector into a unit vector. This means the vector has a magnitude of 1 and retains its original direction. Normalization is essential when you want to maintain direction but scale down to a standard size.
The formula for normalizing a vector \( \mathbf{v} \) is \( \mathbf{v}_{norm} = \frac{\mathbf{v}}{\| \mathbf{v} \|} \), where \( \|\mathbf{v}\| \) is the magnitude of the vector. After normalization, the vector provides directionality without imposing any scale, making it suitable for representing directions or as unit basis vectors in calculations.
Dot Product
The dot product, also known as the scalar product, is a way to multiply two vectors, resulting in a scalar. This operation combines both the direction and magnitude of vectors to give insight into their relationship. If the dot product of two vectors is zero, they are orthogonal.
Computing the dot product is straightforward: for vectors \( \mathbf{a} \) and \( \mathbf{b} \), use the formula \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). The dot product essentially tells you how much of one vector goes in the direction of the other. It's fundamental for tasks like finding projections and in physics, where work done is calculated using the force vector and displacement vector.
Magnitude of a Vector
The magnitude of a vector, also known as its length, measures how long the vector is from the origin to its endpoint. It is crucial in determining how strong a vector's effect is in any given direction.
You calculate the magnitude of a vector \( \mathbf{v} = [v_1, v_2, v_3] \) using the formula \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \). This formula is a three-dimensional extension of the Pythagorean theorem. The magnitude provides context on how much impact a vector has, independent of its direction, aiding in applications such as physics where force and velocity are central.

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Most popular questions from this chapter

Let \(\mathbb{P}_{3}\) have the inner product given by evaluation at \(-3,-1\) \(1,\) and \(3 .\) Let \(p_{0}(t)=1, p_{1}(t)=t,\) and \(p_{2}(t)=t^{2}\) a. Compute the orthogonal projection of \(p_{2}\) onto the sub- space spanned by \(p_{0}\) and \(p_{1}\) . b. Find a polynomial \(q\) that is orthogonal to \(p_{0}\) and \(p_{1},\) such that \(\left\\{p_{0}, p_{1}, q\right\\}\) is an orthogonal basis for \(\operatorname{Span}\left\\{p_{0}, p_{1}, p_{2}\right\\} .\) Scale the polynomial \(q\) so that its vec- tor of values at \((-3,-1,1,3)\) is \((1,-1,-1,1)\)

Let \(\mathbf{u}=\left[\begin{array}{r}{2} \\ {-5} \\ {-1}\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{r}{-7} \\ {-4} \\ {6}\end{array}\right] .\) Compute and compare \(\mathbf{u} \cdot \mathbf{v},\|\mathbf{u}\|^{2},\|\mathbf{v}\|^{2},\) and \(\|\mathbf{u}+\mathbf{v}\|^{2} .\) Do not use the Pythagorean Theorem.

Given \(a \geq 0\) and \(b \geq 0,\) let \(\mathbf{u}=\left[\begin{array}{c}{\sqrt{a}} \\ {\sqrt{b}}\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{c}{\sqrt{b}} \\\ {\sqrt{a}}\end{array}\right]\) Use the Cauchy-Schwarz inequality to compare the geometric mean \(\sqrt{a b}\) with the arithmetic mean \((a+b) / 2\)

Let \(\overline{x}=\frac{1}{n}\left(x_{1}+\cdots+x_{n}\right)\) and \(\overline{y}=\frac{1}{n}\left(y_{1}+\cdots+y_{n}\right) .\) Show that the least-squares line for the data \(\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)\) must pass through \((\overline{x}, \overline{y}) .\) That is, show that \(\overline{x}\) and \(\overline{y}\) satisfy the linear equation \(\overline{y}=\hat{\beta}_{0}+\hat{\beta}_{1} \overline{x}\) .[Hint: Derive this equation from the vector equation \(\mathbf{y}=X \hat{\boldsymbol{\beta}}+\boldsymbol{\epsilon} .\) Denote the first column of \(X\) by \(\mathbf{1}\) . Use the fact that the residual vector \(\epsilon\) is orthogonal to the column space of \(X\) and hence is orthogonal to \(\mathbf{1} . ]\)

Find a formula for the least-squares solution of \(A \mathbf{x}=\mathbf{b}\) when the columns of \(A\) are orthonormal.
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