/*! 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 7 Analyze the matrices \(\left[\be... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 7

Analyze the matrices \(\left[\begin{array}{lll}1 & & \\ & 0 & \\ & & 1\end{array}\right]\) and \(\left[\begin{array}{llll}1 & & & \\ & 1 & & \\ & & 0 & \\ & & & 1\end{array}\right]\) a. as affine transformations of \(\mathbb{R}^{2}\) and \(\mathbb{R}^{3}\), respectively. b. as linear transformations of homogeneous coordinate vectors. From what points can we interpret these as perspective projections?

Short Answer

Expert verified
The given matrices A and B can represent affine transformations in \(\mathbb{R}^2\) and \(\mathbb{R}^3\) respectively. Matrix A scales the x-coordinate by 1, leaves the y-coordinate unchanged, and projects onto the x-axis in \(\mathbb{R}^2\). Matrix B scales the x, y, and w coordinates by 1, leaves the z-coordinate unchanged, and projects onto the xy-plane in \(\mathbb{R}^3\). The points from which we can interpret these matrices as perspective projections are a center of projection (COP) at \((0, \infty, 1)\) for Matrix A and a COP at \((0, 0, \infty, 1)\) for Matrix B.

Step by step solution

01

Determine the Matrix Dimensions

: We will first determine the dimensions of the given matrices to better understand how to analyze them. Matrix A: \(\left[\begin{array}{lll}1 & & \\\ & 0 & \\\ & & 1\end{array}\right]\) is a 3x3 matrix. Matrix B: \(\left[\begin{array}{llll}1 & & & \\\ & 1 & & \\\ & & 0 & \\\ & & & 1\end{array}\right]\) is a 4x4 matrix.
02

Analyze the Matrices as Affine Transformations

: Affine transformations in \(\mathbb{R}^2\) and \(\mathbb{R}^3\) have specific forms: 2D Affine Transformation: \(\left[\begin{array}{ccc}a & b & c \\ d & e & f \\ 0 & 0 & 1\end{array}\right]\) 3D Affine Transformation: \(\left[\begin{array}{cccc}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ 0 & 0 & 0 & 1\end{array}\right]\) From the given matrices' dimensions and the forms of affine transformations, we can see that matrix A can represent an affine transformation in \(\mathbb{R}^2\) and matrix B can represent an affine transformation in \(\mathbb{R}^3\). Matrix A in \(\mathbb{R}^2\) represents a transformation that scales the x-coordinate by a factor of 1, leaves the y-coordinate unchanged (scaling by 0), and scales the w-coordinate by a factor of 1 (has no effect on the transformation). Matrix B in \(\mathbb{R}^3\) represents a transformation that scales the x, y, and w coordinates by a factor of 1 and leaves the z-coordinate unchanged (scaling by 0).
03

Analyze the Matrices as Linear Transformations of Homogeneous Coordinate Vectors

: Now we will analyze matrices A and B as linear transformations of homogeneous coordinate vectors. Matrix A: Homogeneous coordinates in \(\mathbb{R}^2\) are expressed as \((x, y, 1)\), so the linear transformation represented by the matrix A will not change the x-coordinate and set the y-coordinate to 0 (essentially projecting the input points onto the x-axis). Matrix B: Homogeneous coordinates in \(\mathbb{R}^3\) are expressed as \((x, y, z, 1)\). The linear transformation represented by the matrix B will not change the x, y, and w coordinates and set the z-coordinate to 0 (essentially projecting the input points onto an xy-plane).
04

Find the Points of Perspective Projections

: Perspective projections in \(\mathbb{R}^2\) and \(\mathbb{R}^3\) are created by projecting points from a center of projection (COP) onto a projection plane: a. In the case of matrix A and its transformation of vectors in \(\mathbb{R}^2\), the projection collapses points onto the x-axis. This implies that the COP is located at a point with an infinite y-coordinate value (e.g., \((0, \infty, 1)\)). b. In the case of matrix B and its transformation of vectors in \(\mathbb{R}^3\), the projection collapses points onto an xy-plane. This implies that the COP is located at a point with an infinite z-coordinate value (e.g., \((0, 0, \infty, 1)\)).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Homogeneous Coordinate Vectors
Understanding homogeneous coordinate vectors is essential for concepts ranging from computer graphics to robotics. In essence, these vectors allow for the representation of points in a coordinate system with an additional dimension, primarily to deal with transformations that include translations easily.

For example, in 2D space, a point is generally represented as \( (x, y) \), but its homogeneous coordinate would be \( (x, y, 1) \). This extra dimension makes the mathematics of affine transformations more cohesive, particularly because it enables translations to be represented as matrix multiplications, just like rotations and scaling.

When applying a matrix to a homogeneous coordinate, such as matrix A in our exercise, the transformation can either leave the point in its current location or move it to a different one. Analyzing the matrix as a linear transformation of homogeneous coordinates provides insights into the type of transformation we're dealing with. If certain elements are zero, as in matrix A's y-coordinate scaling to 0, it indicates a projection onto the x-axis, a crucial characteristic when considering perspective in graphics.
Perspective Projections
Moving on to perspective projections, they represent how we see the world 鈥� closer objects appear larger, while distant objects look smaller. Mathematically, this effect is achieved by transforming three-dimensional objects onto a two-dimensional surface, such as a computer screen or a piece of paper, in a way that mimics this visual phenomenon.

By leveraging the concepts from homogeneous coordinate vectors, perspective projections can be understood as particular types of affine transformations that emulate the way our eyes perceive depth. As seen in our exercise, matrix B is capable of representing an affine transformation leading to a projection in 3D space - that is, taking a point in 3D and figuring out where it 鈥渓ands鈥� on a 2D plane.

The 'center of projection' (COP) is a theoretical point from which all projections emanate, much like the way the lens of a camera projects light onto the film. For instance, the COP corresponding to matrix A lies at an infinitely distant point along the y-axis, flattening the 3D scene into a line. Such insights allow us to understand and create complex visual effects in graphics and design.
Linear Algebra
Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations in vector spaces and through matrices. Affine transformations, homogeneous coordinate vectors, and perspective projections are all concepts that stem from linear algebra.

Matrices, like the ones provided in our exercise (matrix A and B), are tools in linear algebra for performing transformations. By understanding the properties of these matrices鈥攕uch as their dimensions, the values they contain, and the way they operate on vectors鈥攚e can deduce their impact on geometric figures.

For example, analyzing a matrix as an affine transformation permits us to assess how an object will be moved, rotated, scaled, or sheared. Moreover, when dealing with perspective projections, linear algebra provides the framework to convert a 3D scene into a 2D image realistically.

Linear algebra not only gives us the language to describe these transformations but also the computational methods to perform them, making it a cornerstone of many fields involving geometry and transformation, including computer graphics, engineering, and physical simulations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Check that if \(A\) is an \(n \times n\) matrix and the \(n \times n\) differentiable matrix function \(E(t)\) satisfies \(\frac{d E}{d t}=A E(t)\) and \(E(0)=I\), then \(E(t)=e^{t A}\) for all \(t \in \mathbb{R}\).

In this exercise we explore the problem of displaying three-dimensional images on a two-dimensional blackboard, piece of paper, or computer screen. (A computer or good graphics calculator will be helpful for experimentation here.) a. Fix a plane \(H \subset \mathbb{R}^{3}\) containing the \(x_{3}\)-axis, say \((\cos \theta) x_{1}+(\sin \theta) x_{2}=0\), and fix \(\mathbf{a}=\left(a_{1}, a_{2}, a_{3}\right) \notin H\). Find the \(4 \times 4\) matrix that represents projection from a onto \(H\). b. Since we want to view on a standard screen, give a matrix that rotates \(H\) to the \(x_{1} x_{2}\)-plane, sending the \(x_{3}\)-axis to the \(x_{2}\)-axis. c. By multiplying the two matrices you've found and deleting the row of 0 's, show that the resulting linear transformation \(T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{3}\) is given by the matrix $$ A=\left[\begin{array}{cccc} a_{2} & -a_{1} & 0 & 0 \\ a_{3} \cos \theta & a_{3} \sin \theta & a_{1} \cos \theta+a_{2} \sin \theta & 0 \\ \cos \theta & \sin \theta & 0 & -a_{1} \cos \theta-a_{2} \sin \theta \end{array}\right] $$ d. Experiment with different values of a and \(\theta\) to obtain an effective perspective. You might compute the image of the unit cube (with one vertex at the origin and edges aligned on the coordinate axes). Pictured in Figure \(2.10\) are two images of the cube, first with \(\mathbf{a}=(5,4,3)\) and \(\theta=\pi / 6\), next with \(\mathbf{a}=(4,6,3)\) and \(\theta=\pi / 4\). e. What happens if you try usual orthogonal projection onto a plane (and then rotate that plane, as before)? Show that if we take the plane with unit normal \(\frac{1}{\sqrt{2}}(\cos \theta, \sin \theta, 1)\), the resulting projection \(\mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) is given by $$ B=\left[\begin{array}{ccc} -\sin \theta & \cos \theta & 0 \\ -\frac{1}{2} \cos \theta & -\frac{1}{2} \sin \theta & \frac{1}{2} \end{array}\right] $$ f. Experiment as before. For example, if we take \(\theta=\pi / 6\), then the image of the unit cube is as pictured in Figure 2.11. What conclusions do you reach?

Calculate \(e^{t A}\) and use your answer to solve \(\frac{d \mathbf{x}}{d t}=A \mathbf{x}, \mathbf{x}(0)=\mathbf{x}_{0}\). *a. \(A=\left[\begin{array}{ll}1 & 5 \\ 2 & 4\end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{r}6 \\ -1\end{array}\right]\) \({ }^{*} \mathrm{~d} . A=\left[\begin{array}{rr}1 & 1 \\ -1 & 3\end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{r}2 \\\ -1\end{array}\right]\) b. \(A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l}1 \\ 3\end{array}\right]\) "e. \(A=\left[\begin{array}{rrr}-1 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & -1\end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l}2 \\ 0 \\\ 4\end{array}\right]\) c. \(A=\left[\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l}5 \\ 1\end{array}\right]\) f. \(A=\left[\begin{array}{rrr}1 & -2 & 2 \\ -1 & 0 & -1 \\ 0 & 2 & -1\end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{r}3 \\ -1 \\\ -4\end{array}\right]\)

In this exercise we analyze the isometries of \(\mathbb{R}^{3}\). a. If \(A\) is an orthogonal \(3 \times 3\) matrix with det \(A=1\), show that \(A\) is a rotation matrix. (See Exercise 6.2.16.) That is, prove that there is an orthonormal basis for \(\mathbb{R}^{3}\) with respect to which the matrix takes the form $$ \left[\begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array}\right] $$ b. If \(A\) is an orthogonal \(3 \times 3\) matrix with det \(A=-1\), show that there is an orthonormal basis for \(\mathbb{R}^{3}\) with respect to which the matrix takes the form $$ \left[\begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & -1 \end{array}\right] $$ That is, \(\mu_{A}\) is the composition of a reflection across a plane with a rotation of that plane. Such a transformation is called a rotatory reflection when \(\theta \neq 0\). c. If \(A\) is an orthogonal \(3 \times 3\) matrix and \(\mathbf{a} \in \mathbb{R}^{3}\), prove that the matrix $$ \left[\begin{array}{lll|l} & A & & \mathbf{a} \\ & & & \mid \\ \hline 0 & 0 & 0 & 1 \end{array}\right] $$ is similar to a matrix of one of the following forms: $$ \left[\begin{array}{ccc|c} \cos \theta & -\sin \theta & 0 & 0 \\ \sin \theta & \cos \theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \hline 0 & 0 & 0 & 1 \end{array}\right],\left[\begin{array}{ccc|c} \cos \theta & -\sin \theta & 0 & 0 \\ \sin \theta & \cos \theta & 0 & 0 \\ 0 & 0 & -1 & 0 \\ \hline 0 & 0 & 0 & 1 \end{array}\right] $$ $$ \left[\begin{array}{rrr|r} 1 & 0 & 0 & a_{1} \\ 0 & 1 & 0 & a_{2} \\ 0 & 0 & 1 & a_{3} \\ \hline 0 & 0 & 0 & 1 \end{array}\right],\left[\begin{array}{rrr|r} 1 & 0 & 0 & a_{1} \\ 0 & 1 & 0 & a_{2} \\ 0 & 0 & -1 & 0 \\ \hline 0 & 0 & 0 & 1 \end{array}\right],\left[\begin{array}{ccc|c} \cos \theta & -\sin \theta & 0 & 0 \\ \sin \theta & \cos \theta & 0 & 0 \\ 0 & 0 & 1 & a_{3} \\ \hline 0 & 0 & 0 & 1 \end{array}\right] $$ The last such matrix corresponds to what's called a screw motion (why?). d. Conclude that any isometry of \(\mathbb{R}^{3}\) is either a rotation, a reflection, a translation, a rotatory reflection, a glide reflection, or a screw.

Prove that the set of \(n\) vectors constructed in the proof of Theorem \(1.5\) is linearly independent. (Hints: We started with a linearly independent set \(\left\\{\mathbf{w}_{1}, \ldots, \mathbf{w}_{n-d}\right\\}\). Suppose \(\sum a_{k} \mathbf{v}_{k}+\sum c_{i} \mathbf{w}_{i}+\sum d_{s} \mathbf{u}_{s}=\mathbf{0}\). Multiply by \(A-\lambda I\) and check that we get only a linear combination of the \(\mathbf{w}_{i}\), which are known to form a linearly independent set. Conclude that all the \(d_{s}\) and \(c_{i}, i \neq 1, j_{1}+1, \ldots, j_{\ell-1}+1\), must be 0 . This leaves only the terms involving the eigenvectors \(\mathbf{w}_{1}, \ldots, \mathbf{w}_{j_{l-1}+1}\), and \(\mathbf{v}_{\ell+1}, \ldots, \mathbf{v}_{d}\), but by construction these form a linearly independent set.)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.