/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Let \(\mathbf{y}=\left[\begin{ar... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(\mathbf{y}=\left[\begin{array}{l}{7} \\ {9}\end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{r}{1 / \sqrt{10}} \\ {-3 / \sqrt{10}}\end{array}\right],\) and \(W=\operatorname{Span}\left\\{\mathbf{u}_{1}\right\\}\) a. Let \(U\) be the \(2 \times 1\) matrix whose only column is \(\mathbf{u}_{1}\) . Compute \(U^{T} U\) and \(U U^{T}\) . b. Compute proj \(_{W} \mathbf{y}\) and \(\left(U U^{T}\right) \mathbf{y}\)

Short Answer

Expert verified
Proj and \((UU^T)\mathbf{y}\) are both \(\begin{bmatrix} -2 \\ 6 \end{bmatrix}\).

Step by step solution

01

Define Matrices

First, define the matrix \(U\) from the given vector \(\mathbf{u}_1\). Here, \(U = \begin{bmatrix} \frac{1}{\sqrt{10}} \ -\frac{3}{\sqrt{10}} \end{bmatrix}\). Next, the transpose of \(U\), \(U^T = \begin{bmatrix} \frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{10}} \end{bmatrix}\).
02

Compute \(U^T U\)

Multiply \(U^T\) by \(U\):\[ U^T U = \begin{bmatrix} \frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{10}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{10}} \ -\frac{3}{\sqrt{10}} \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{\sqrt{10}}\right)^2 + \left(-\frac{3}{\sqrt{10}}\right)^2 \end{bmatrix} = \begin{bmatrix} 1 \end{bmatrix} \]
03

Compute \(U U^T\)

Multiply \(U\) by \(U^T\):\[ U U^T = \begin{bmatrix} \frac{1}{\sqrt{10}} \ -\frac{3}{\sqrt{10}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{10}} \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{\sqrt{10}}\right)^2 & \frac{1}{\sqrt{10}} \left(-\frac{3}{\sqrt{10}}\right) \ \frac{1}{\sqrt{10}} \left(-\frac{3}{\sqrt{10}}\right) & \left(-\frac{3}{\sqrt{10}}\right)^2 \end{bmatrix} = \begin{bmatrix} \frac{1}{10} & -\frac{3}{10} \ -\frac{3}{10} & \frac{9}{10} \end{bmatrix} \]
04

Calculate the Projection \(\text{proj}_W \mathbf{y}\)

The projection of \(\mathbf{y}\) onto \(W\) is given by:\[ \text{proj}_W \mathbf{y} = U(U^T U)^{-1} U^T \mathbf{y} \]Using the result from Step 2:\[ \text{proj}_W \mathbf{y} = U \cdot 1 \cdot U^T \mathbf{y} = \frac{1}{10} \begin{bmatrix} \frac{1}{\sqrt{10}} \ -\frac{3}{\sqrt{10}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{10}} \end{bmatrix} \begin{bmatrix} 7 \ 9 \end{bmatrix} \]Calculate:\[ = \frac{1}{10} \begin{bmatrix} 7 \cdot \frac{1}{\sqrt{10}} + 9 \cdot -\frac{3}{\sqrt{10}} \ (-3) \cdot \left(7 \cdot \frac{1}{\sqrt{10}} + 9 \cdot -\frac{3}{\sqrt{10}}\right) \end{bmatrix} = \frac{1}{10} \begin{bmatrix} \frac{7-27}{\sqrt{10}} \ -\frac{3(7-27)}{\sqrt{10}} \end{bmatrix} = \begin{bmatrix} -2 \ \ 6 \end{bmatrix} \]
05

Calculate \((UU^T)\mathbf{y}\)

Now compute the result of multiplying \((UU^T)\) by \(\mathbf{y}\):\[ (UU^T)\mathbf{y} = \begin{bmatrix} \frac{1}{10} & -\frac{3}{10} \ -\frac{3}{10} & \frac{9}{10} \end{bmatrix} \begin{bmatrix} 7 \ 9 \end{bmatrix} \]Calculate:\[ = \begin{bmatrix} \frac{1}{10} \cdot 7 + (-\frac{3}{10}) \cdot 9 \ (-\frac{3}{10}) \cdot 7 + \frac{9}{10} \cdot 9 \end{bmatrix} = \begin{bmatrix} -2 \ 6 \end{bmatrix} \]

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Key Concepts

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

Projection Matrix
When dealing with vectors and spaces, a projection matrix is a fundamental tool. It allows us to "project" a vector onto a particular subspace. Think of it like casting a shadow of a vector onto a flat surface—the subspace. This concept is crucial in linear algebra and has practical applications in data analysis, computer graphics, and many other fields.

A projection matrix is usually represented as \( P \) and is defined such that when it multiplies a vector, it returns that vector's projection onto the subspace. If you have a vector space represented by \( U \), and a vector \( \mathbf{y} \) you want to project, the projection matrix \( P \) is given by:
  • \( P = U(U^T U)^{-1} U^T \)
Where \( U \) is the matrix whose columns form a basis for the subspace. For our example, the calculation becomes simplified as \( U^T U \) was found to be \( 1 \), hence no inverse calculation is needed.

In this exercise, we used the projection matrix to project \( \mathbf{y} \) onto the subspace spanned by \( \mathbf{u}_1 \). By applying this matrix, you can accurately find the shadow of \( \mathbf{y} \) in the direction defined by \( \mathbf{u}_1 \). This illustrates how projection matrices make complex spaces easier to navigate.
Subspace Projection
Subspace projection is about finding how one vector can be best "represented" within another dimension—a subset of a given space. Imagine trying to shine a flashlight on a complex object, where the simpler shadow gives you an idea of its form.

In mathematical terms, when you project a vector \( \mathbf{y} \) onto a subspace \( W \), you're looking for a vector in \( W \) that is as close as possible to \( \mathbf{y} \). This operation is highly useful for simplifying complex problems by using reduced dimensions or simpler approximations.
  • A subspace is defined by its basis vectors, like \( \mathbf{u}_1 \) in our example.
  • To project \( \mathbf{y} \) onto \( W \), we use the formula \( \text{proj}_W \mathbf{y} = U(U^T U)^{-1} U^T \mathbf{y} \). This method leverages the pre-computed values of \( U^T U \) and \( U U^T \).
Projection helps us focus only on the components of \( \mathbf{y} \) that align with the direction defined by our subspace \( W \). Here, the subspace \( W \) was defined by the single vector \( \mathbf{u}_1 \), enabling a simple yet profound transformation of \( \mathbf{y} \) into its components along this direction.
Matrix Multiplication
Matrix multiplication is a key operation in linear algebra that involves a systematic combination of rows from the first matrix and columns from the second. It can be considered as the method by which transformations, like projections, are applied to vectors or other matrices.

In our exercise, matrix multiplication plays a critical role in creating the projection. Here, we used this operation to compute \( U^T U \), \( U U^T \), and the final projection onto subspace \( W \). Understanding how these matrices interact can greatly simplify many calculations in both theoretical and applied mathematics.
  • To calculate \( U^T U \), multiply the transpose of \( U \) by itself. This results in a scalar value when \( U \) is a single column vector.
  • \( U U^T \) multiplication provides us a projection matrix when \( U \) is an orthonormal matrix.
  • Finally, multiply the obtained projection matrix with the vector \( \mathbf{y} \) to get the projected vector.
These steps illustrate how elegant solutions in mathematics often stem from basic operations like multiplication. Mastery of matrix multiplication equips students to engage with more complex linear transformations, making it an essential part of one's mathematical toolkit.

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Most popular questions from this chapter

In Excrcises \(1-6,\) the given set is a basis for a subspace \(W .\) Use the Gram-Schmidt process to produce an orthogonal basis for \(W\) . $$ \left[\begin{array}{r}{1} \\ {-4} \\ {0} \\\ {1}\end{array}\right],\left[\begin{array}{r}{7} \\ {-7} \\ {-4} \\\ {1}\end{array}\right] $$

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 \(W\) $$ \mathbf{y}=\left[\begin{array}{r}{4} \\ {3} \\ {3} \\\ {-1}\end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{l}{1} \\ {1} \\\ {0} \\ {1}\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r}{-1} \\\ {3} \\ {1} \\ {-2}\end{array}\right], \mathbf{u}_{3}=\left[\begin{array}{r}{-1} \\ {0} \\ {1} \\\ {1}\end{array}\right] $$

In Exercises 17 and \(18, A\) is an \(m \times n\) matrix and \(b\) is in \(\mathbb{R}^{m} .\) Mark each statement True or False. Justify each answer. a. If \(\mathbf{b}\) is in the column space of \(A,\) then every solution of \(A \mathbf{x}=\mathbf{b}\) is a least-squares solution. b. The least-squares solution of \(A \mathbf{x}=\mathbf{b}\) is the point in the column space of \(A\) closest to \(\mathbf{b}\) . c. A least-squares solution of \(A \mathbf{x}=\mathbf{b}\) is a list of weights that, when applied to the columns of \(A,\) produces the orthogonal projection of \(\mathbf{b}\) onto \(\operatorname{Col} A .\) d. If \(\hat{\mathbf{x}}\) is a least-squares solution of \(A \mathbf{x}=\mathbf{b},\) then \(\hat{\mathbf{x}}=\left(A^{T} A\right)^{-1} A^{T} \mathbf{b}\) e. The normal equations always provide a reliable method for computing least- squares solutions. f. If \(A\) has a QR factorization, say \(A=Q R\) , then the best way to find the least-squares solution of \(A \mathbf{x}=\mathbf{b}\) is to compute \(\hat{\mathbf{x}}=R^{-1} Q^{T} \mathbf{b} .\)

[M] To measure the takeoff performance of an airplane, the horizontal position of the plane was measured every second, from \(t=0\) to \(t=12 .\) The positions (in feet) were: \(0,8.8,\) \(29.9,62.0,104.7,159.1,222.0,294.5,380.4,471.1,571.7\) \(686.8,\) and \(809.2 .\) a. Find the least-squares cubic curve \(y=\beta_{0}+\beta_{1} t+\) \(\beta_{2} t^{2}+\beta_{3} t^{3}\) for these data. b. Use the result of part (a) to estimate the velocity of the plane when \(t=4.5\) seconds.

In Exercises 5 and \(6,\) describe all least-squares solutions of the equation \(A \mathbf{x}=\mathbf{b} .\) $$ A=\left[\begin{array}{lll}{1} & {1} & {0} \\ {1} & {1} & {0} \\ {1} & {1} & {0} \\ {1} & {0} & {1} \\ {1} & {0} & {1} \\ {1} & {0} & {1}\end{array}\right], \mathbf{b}=\left[\begin{array}{l}{7} \\ {2} \\ {3} \\\ {6} \\ {5} \\ {4}\end{array}\right] $$

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