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Show that \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}\right\\}\) or \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) is an orthogonal basis for \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3},\) respectively. Then express \(\mathbf{x}\) as a linear combination of the \(\mathbf{u}^{\prime}\)s. \(\mathbf{u}_{1}=\left[\begin{array}{l}{1} \\ {0} \\ {1}\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r}{-1} \\ {4} \\ {1}\end{array}\right], \mathbf{u}_{3}=\left[\begin{array}{r}{2} \\ {1} \\ {-2}\end{array}\right],\) and \(\mathbf{x}=\left[\begin{array}{r}{8} \\ {-4} \\ {-3}\end{array}\right]\)

Short Answer

Expert verified
\(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}\) is an orthogonal basis for \(\mathbb{R}^3\), and \(\mathbf{x} = \frac{5}{2}\mathbf{u}_1 - \frac{11}{18}\mathbf{u}_2 + 2\mathbf{u}_3\).

Step by step solution

01

Confirm Orthogonality for Basis

To begin, check if the set \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}\) is an orthogonal set in \(\mathbb{R}^3\). For orthogonality, the dot product between any two distinct vectors should be zero.Calculate \(\mathbf{u}_1 \cdot \mathbf{u}_2 = [1, 0, 1] \cdot [-1, 4, 1] = (1)(-1) + (0)(4) + (1)(1) = -1 + 0 + 1 = 0\).Calculate \(\mathbf{u}_1 \cdot \mathbf{u}_3 = [1, 0, 1] \cdot [2, 1, -2] = (1)(2) + (0)(1) + (1)(-2) = 2 + 0 - 2 = 0\).Calculate \(\mathbf{u}_2 \cdot \mathbf{u}_3 = [-1, 4, 1] \cdot [2, 1, -2] = (-1)(2) + (4)(1) + (1)(-2) = -2 + 4 - 2 = 0\).All dot products are zero, thus \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}\) is an orthogonal set and forms an orthogonal basis for \(\mathbb{R}^3\).
02

Decompose \(\mathbf{x}\) Using the Basis

Since \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}\) is an orthogonal basis for \(\mathbb{R}^3\), we can express \(\mathbf{x}\) as a linear combination: \(\mathbf{x} = c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + c_3 \mathbf{u}_3\).We use the formula for coefficients in orthogonal bases: for vector \(\mathbf{u}_i\) in an orthogonal set, \(c_i = \frac{\mathbf{x} \cdot \mathbf{u}_i}{\mathbf{u}_i \cdot \mathbf{u}_i}\).Calculate \(c_1:\)\(c_1 = \frac{\mathbf{x} \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} = \frac{[8, -4, -3] \cdot [1, 0, 1]}{[1, 0, 1] \cdot [1, 0, 1]} = \frac{(8)(1) + (-4)(0) + (-3)(1)}{1^2 + 0^2 + 1^2} = \frac{8 - 3}{2} = \frac{5}{2}\).Calculate \(c_2:\)\(c_2 = \frac{\mathbf{x} \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} = \frac{[8, -4, -3] \cdot [-1, 4, 1]}{(-1)^2 + 4^2 + 1^2} = \frac{(-8)(-1) + (-4)(4) + (-3)(1)}{1 + 16 + 1} = \frac{8 - 16 - 3}{18} = \frac{-11}{18}\).Calculate \(c_3:\)\(c_3 = \frac{\mathbf{x} \cdot \mathbf{u}_3}{\mathbf{u}_3 \cdot \mathbf{u}_3} = \frac{[8, -4, -3] \cdot [2, 1, -2]}{2^2 + 1^2 + (-2)^2} = \frac{(8)(2) + (-4)(1) + (-3)(-2)}{4 + 1 + 4} = \frac{16 - 4 + 6}{9} = \frac{18}{9} = 2\).
03

Express \(\mathbf{x}\) as a Linear Combination

With the coefficients calculated, express \(\mathbf{x}\): \(\mathbf{x} = \frac{5}{2}\mathbf{u}_1 - \frac{11}{18}\mathbf{u}_2 + 2\mathbf{u}_3\).

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Key Concepts

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

Dot Product
The dot product is a crucial operation in understanding orthogonality in vector spaces. It's a way to multiply two vectors to get a scalar. To compute the dot product, align the corresponding components of two vectors, multiply each pair, and then sum all these products. For vectors \(\mathbf{a} = [a_1, a_2, a_3]\) and \(\mathbf{b} = [b_1, b_2, b_3]\), the dot product is calculated as \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\).
In the context of checking for an orthogonal basis, a set of vectors in \(\mathbb{R}^n\) is orthogonal when the dot product between every pair of distinct vectors in the set equals zero. This property is essential, as it simplifies many calculations, including vector decompositions. In this exercise, verifying that vectors \(\mathbf{u}_1, \mathbf{u}_2,\) and \(\mathbf{u}_3\) are orthogonal was the first step in ensuring they form an orthogonal basis.
Orthogonal vectors not only simplify the process of breaking down other vectors (decomposition) but also ensure the absence of redundancy within the set, which is beneficial for constructing an efficient and concise basis for a vector space.
Linear Combination
A linear combination involves expressing a vector as a sum of scalar multiples of other vectors. It's a method used to describe vectors within a space in terms' basis vectors. For instance, when you have an orthogonal basis, an arbitrary vector \(\mathbf{x}\) can be expressed as a linear combination of the basis vectors, such that \(\mathbf{x} = c_1\mathbf{u}_1 + c_2\mathbf{u}_2 + c_3\mathbf{u}_3\).
To find the coefficients (\(c_1, c_2, c_3\)), you use the dot product. In orthogonal bases, computing these coefficients is straightforward:
  • Take the dot product of the vector \(\mathbf{x}\) with each basis vector \(\mathbf{u}_i\).
  • Divide by the dot product of \(\mathbf{u}_i\) with itself, \(\mathbf{u}_i \cdot \mathbf{u}_i\).
Thus, each coefficient corresponds to how much of each basis vector is needed to "build" the vector \(\mathbf{x}\). Orthogonal bases simplify the calculation of these coefficients, making the representation clear and arithmetic manageable.
Vector Decomposition
Vector decomposition is a technique that involves breaking a vector down into a sum of vectors, often relative to a particular basis set. This is exceedingly useful for simplifying complex vector operations and solving various mathematical problems.
In the exercise, vector decomposition is carried out with an orthogonal basis, making it especially straightforward. With the vector \(\mathbf{x}\) given and an orthogonal basis \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}\), we decompose \(\mathbf{x}\) using a formula that involves the dot product:
  • Each component of \(\mathbf{x}\) relative to the basis vector \(\mathbf{u}_i\) is determined by \(c_i = \frac{\mathbf{x} \cdot \mathbf{u}_i}{\mathbf{u}_i \cdot \mathbf{u}_i}\).
In practice, the decomposition allows us to reassemble any target vector as a weighted sum of simpler, orthogonal vectors. This method has applications in everything from physics to computer graphics, anytime a complex vector needs to be analyzed or transformed against a given basis.

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Most popular questions from this chapter

Compute the quantities in Exercises \(1-8\) using the vectors $$ \mathbf{u}=\left[\begin{array}{r}{-1} \\ {2}\end{array}\right], \quad \mathbf{v}=\left[\begin{array}{l}{4} \\ {6}\end{array}\right], \quad \mathbf{w}=\left[\begin{array}{r}{3} \\ {-1} \\ {-5}\end{array}\right], \quad \mathbf{x}=\left[\begin{array}{r}{6} \\ {-2} \\ {3}\end{array}\right] $$ \(\|\mathbf{w}\|\)

In Exercises \(7-10,\) let \(W\) be the subspace spanned by the \(\mathbf{u}^{\prime}\) 's, and write \(\mathbf{y}\) as the sum of a vector in \(W\) and a vector orthogonal to \(W\) $$ \mathbf{y}=\left[\begin{array}{l}{3} \\ {4} \\ {5} \\\ {6}\end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{r}{1} \\ {1} \\\ {0} \\ {-1}\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{l}{1} \\\ {0} \\ {1} \\ {1}\end{array}\right], \mathbf{u}_{3}=\left[\begin{array}{r}{0} \\ {-1} \\ {1} \\\ {-1}\end{array}\right] $$

Let \(A=\left[\begin{array}{rr}{3} & {4} \\ {-2} & {1} \\ {3} & {4}\end{array}\right], \mathbf{b}=\left[\begin{array}{r}{11} \\ {-9} \\\ {5}\end{array}\right], \mathbf{u}=\left[\begin{array}{r}{5} \\\ {-1}\end{array}\right],\) and \(\mathbf{v}=\) \(\left[\begin{array}{r}{5} \\\ {-2}\end{array}\right] .\) Compute \(A \mathbf{u}\) and \(A \mathbf{v},\) and compare them with \(\mathbf{b}\) . Could \(\mathbf{u}\) possibly be a least- squares solution of \(A \mathbf{x}=\mathbf{b} ?\) (Answer this without computing a least-squares solution.)

Suppose \(A=Q R,\) where \(Q\) is \(m \times n\) and \(R\) is \(n \times n .\) Show that if the columns of \(A\) are linearly independent, then \(R\) must be invertible. \([\text { Hint: Study the equation } R \mathbf{x}=\mathbf{0} \text { and use the }\) fact that \(A=Q R .\)

Involve a design matrix \(X\) with two or more columns and a least-squares solution \(\hat{\beta}\) of \(\mathbf{y}=X \beta .\) Consider the following numbers. (i) \(\|X \hat{\boldsymbol{\beta}}\|^{2}-\) the sum of the squares of the "regression term." Denote this number by \(\mathrm{SS}(\mathrm{R})\). (ii) \(\|\mathbf{y}-X \hat{\boldsymbol{\beta}}\|^{2}-\) the sum of the squares for error term. Denote this number by \(\mathrm{SS}(\mathrm{E})\). (iii) \(\|\mathbf{y}\|^{2}-\) the "total" sum of the squares of the \(y\) -values. Denote this number by \(\mathrm{SS}(\mathrm{T}) .\) Every statistics text that discusses regression and the linear model \(\mathbf{y}=X \boldsymbol{\beta}+\boldsymbol{\epsilon}\) introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume that the mean of the \(y\) -values is zero. In this case, SS(T) is proportional to what is called the variance of the set of \(y\) -values. Show that \(\|X \hat{\boldsymbol{\beta}}\|^{2}=\hat{\boldsymbol{\beta}}^{T} X^{T} \mathbf{y}\) [Hint: Rewrite the left side and use the fact that \(\hat{\boldsymbol{\beta}}\) satisfies the normal equations.] This formula for \(\mathrm{SS}(\mathrm{R})\) is used in statistics. From this and from Exercise \(19,\) obtain the standard formula for \(\mathrm{SS}(\mathrm{E}) :\) \(\mathrm{SS}(\mathrm{E})=\mathbf{y}^{T} \mathbf{y}-\hat{\boldsymbol{\beta}}^{T} X^{T} \mathbf{y}\)

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