1 / 97

Engineering Analysis

Engineering Analysis. Chapter 5 Orthogonality. Orthogonality.

kelid
Download Presentation

Engineering Analysis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Engineering Analysis Chapter 5Orthogonality

  2. Orthogonality Definition. We say that 2 vectors are orthogonal if they are perpendicular to each other. i.e. the dot product of the two vectors is zero. ... A set of vectors S is orthonormal if every vector in S has magnitude 1 and the set of vectors are mutually orthogonal If the dot product of two vectors is 0, they are orthogonal, in other words, they are perpendicular. The dot product between two vectors u⃗ ,v⃗ is given by: u⃗ ⋅v⃗ =|u⃗ ||v⃗ |cos(θ), so when u⃗ ⋅v⃗ =0 ⟹ cosθ =0 ⟹ θ =π/2(90∘).

  3. Two vectors x and y in Rnmay be regarded as n × 1 matrices. We can then form the matrix product xTy. This product is a 1×1 matrix that may be regarded as a vector in R1or, more simply, as a real number. The product xTy is called the scalar product of x and y. In particular, if x = (x1, . . . , xn)T and y = (y1, . . . , yn)T, then Example 1 If then

  4. In order to see the geometric significance of the scalar product, let us begin by restricting our attention to R2 and R3. Vectors in R2 and R3can be represented by directed line segments. Given a vector x in either R2 or R3, its Euclidean length can be defined in terms of the scalar product.

  5. Definition

  6. If x and y are nonzero vectors, then we can specify their directions by forming unit vectors If θ is the angle between x and y, then The cosine of the angle between the vectors x and y is simply the scalar product of the corresponding direction vectors u and v.

  7. EXAMPLE 3 Let x and y be the vectors in Example 2. The directions of these vectors are given by the unit vectors The cosine of the angle θ between the two vectors is

  8. If x and y are vectors in either R2or R3,then |xT y| ≤ ||x|| ||y|| with equality holding if and only if one of the vectors is 0 or one vector is a multiple of the other. If xT y= 0, it follows from Theorem 5.1.1 that either one of the vectors is the zero vector or cosθ = 0. If cosθ = 0, the angle between the vectors is a right angle.

  9. Example 4 Scalar and Vector Projections

  10. The scalar α is called the scalar projection of x onto y, and the vector p is called the vector projection of x onto y.

  11. EXAMPLE 5 The point Q in Figure 5.1.3 is the point on the line y = 1/3 x that is closest to the point (1, 4). Determine the coordinates of Q. Solution Thus, Q = (2.1, 0.7) is the closest point.

  12. Notation If P1 and P2 are two points in 3-space, we will denote the vector from P1 to P2 by If Nis a nonzero vector and P0 is a fixed point, the set of points Psuch that is orthogonal to N forms a plane πin 3-space that passes through P0. The vector Nand the plane πare said to be normal to each other. A point P = (x, y, z) will lie on π if and only if If N = (a, b, c)T and P0= (x0, y0, z0), this equation can be written in the form a (x − x0) + b (y − y0) + c (z − z0) = 0

  13. EXAMPLE 6 Find the equation of the plane passing through the point (2,−1, 3) and normal to the vector N = (2, 3, 4)T. Solution The span of two linearly independent vectors x and y in R3 corresponds to a plane through the origin in 3-space. To determine the equation of the plane we must find a vector normal to the plane. In Section 3 of Chapter 2, it was shown that the cross product of the two vectors is orthogonal to each vector. If we take N = x × y as our normal vector, then the equation of the plane is given by

  14. EXAMPLE 7 Find the equation of the plane that passes through the points Solution Let The normal vector N must be orthogonal to both x and y. If we set We can then use any one of the points to determine the equation of the plane. Using the point P1, we see that the equation of the plane is

  15. EXAMPLE 8 Find the distance from the point (2, 0, 0) to the plane x + 2y + 2z = 0. Solution The vector N = (1, 2, 2)Tis normal to the plane and the plane passes through the origin. Let v = (2, 0, 0)T . The distance d from (2, 0, 0) to the plane is simply the absolute value of the scalar projection of vonto N. Thus

  16. If x and y are nonzero vectors in R3and θis the angle between the vectors, then It then follows that and hence

  17. Thus, we have, for any nonzero vectors xand yin R3, If either x or y is the zero vector then x×y = 0 and hence the norm of x×y will be 0.

  18. 5.2 Orthogonal Subspaces Definition Thus,

  19. 5.2 Orthogonal Subspaces EXAMPLE 2 Let Xbe the subspace of R3 spanned by e1 and e2, and let Ybe the subspace spanned by e3. If x ∈ X and y ∈ Y, then

  20. 5.2 Orthogonal Subspaces Definition

  21. 5.2 Orthogonal Subspaces

  22. 5.2 Orthogonal Subspaces

  23. 5.2 Orthogonal Subspaces Example Solution According to the proposition, we need to compute the null space of the matrix. Ax=0 The free variable is x3, so the parametric form of the solution set is x1 = x3/17, x2 = − 5x3/17, and the parametric vector form is

  24. 5.2 Orthogonal Subspaces Scaling by a factor of 17, we see that We can check our work:

  25. 5.2 Orthogonal Subspaces Example Find all vectors orthogonal to Solution According to the proposition, we need to compute the null space of the matrix This matrix is in reduced-row echelon form. The parametric form for the solution set is x1 = − x2 + x3 , so the parametric vector form of the general solution is Therefore, the answer is the plane

  26. 5.2 Orthogonal Subspaces

  27. 5.2 Orthogonal Subspaces

  28. 5.2 Orthogonal Subspaces

  29. 5.2 Orthogonal Subspaces

  30. 5.2 Orthogonal Subspaces

  31. 5.2 Orthogonal Subspaces

  32. 5.2 Orthogonal Subspaces

  33. 5.2 Orthogonal Subspaces

  34. 5.2 Orthogonal Subspaces Theorem 5.2.1 Fundamental Subspaces Theorem Let us denote the range of A by R(A). Thus

  35. 5.2 Orthogonal Subspaces EXAMPLE 3 Let The column space of A consists of all vectors of the form Note that if x is any vector in R2 and b = Ax, then The null space of ATconsists of all vectors of the form β (−2, 1)T. Since (1, 2)Tand (−2, 1)T are orthogonal, it follows that every vector in R (A) will be orthogonal to every vector in N (AT).

  36. 5.2 Orthogonal Subspaces EXAMPLE 4 Let Solution We can find bases for N(A) and R(AT) by transforming A into reduced row echelon form:

  37. 5.2 Orthogonal Subspaces Since (1, 0, 1) and (0, 1, 1) form a basis for the row space of A, it follows that (1, 0, 1)T and (0, 1, 1)Tform a basis for R (AT). If x ∈ N(A), it follows from the reduced row echelon form of Athat Thus To find bases for R(A) and N(AT), transform AT to reduced row echelon form.

  38. 5.2 Orthogonal Subspaces Thus, (1, 0, 1)Tand (0, 1, 2)T form a basis for R(A). If x ∈N(AT), then x1 = −x3, x2 = −2x3. Hence, N(AT) is the subspace of R3 spanned by (−1,−2, 1)T. Note that (−1,−2, 1)Tis orthogonal to (1, 0, 1)Tand (0, 1, 2)T.

  39. 5.2 Orthogonal Subspaces Theorem5.2.2 Definition Theorem 5.2.3 If S is a subspace of Rn , then If S is a subspace of Rn , then Theorem 5.2.4

  40. 5.3 Least Squares Problems

  41. 5.3 Least Squares Problems A standard technique in mathematical and statistical modeling is to find a least squares fit to a set of data points in the plane. The least squares curve is usually the graph of a standard type of function, such as a linear function, a polynomial, or a trigonometric polynomial. Since the data may include errors in measurement or experiment-related inaccuracies, we do not require the curve to pass through all the data points. Instead, we require the curve to provide an optimal approximation in the sense that the sum of squares of errors between the y values of the data points and the corresponding y values of the approximating curve are minimized.

  42. 5.3 Least Squares Problems Least Squares Solutions of Overdetermined Systems A least squares problem can generally be formulated as an overdetermined linear system of equations. Recall that an overdetermined system is one involving more equations than unknowns. Such systems are usually inconsistent. Thus, given an m×n system Ax = b with m>n, we cannot expect in general to find a vector x ∈ Rn for which Ax equals b. Theorem 5.3.2 If A is an m × n matrix of rank n, the normal equations have a unique solution and is the unique least squares solution of the system Ax = b.

  43. 5.3 Least Squares Problems APPLICATION 2 Spring Constants Hooke’s law states that the force applied to a spring is proportional to the distance that the spring is stretched. Thus, if F is the force applied and x is the distance that the spring has been stretched, then F = kx. The proportionality constant k is called the spring constant. Some physics students want to determine the spring constant for a given spring. They apply forces of 3, 5, and 8 pounds, which have the effect of stretching the spring 4, 7, and 11 inches, respectively. Using Hooke’s law, they derive the following system of equations:

  44. 5.3 Least Squares Problems The system is clearly inconsistent, since each equation yields a different value of k. Rather than use any one of these values, the students decide to compute the least squares solution of the system.

  45. 5.3 Least Squares Problems EXAMPLE 1 Find the least squares solution of the system Solution The normal equations for this system are This simplifies to the 2 × 2 system The solution of the 2 × 2 system is

More Related