Presentation on theme: "Instructor: Irvin Roy Hentzel Office 432 Carver Phone"— Presentation transcript:
1Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 Math 307Spring, 2003HentzelTime: 1:10-2:00 MWFRoom: 1324 Howe HallInstructor: Irvin Roy HentzelOffice 432 CarverPhoneText: Linear Algebra With Applications,Second Edition Otto Bretscher
2Friday, Mar 14 Chapter 5.2 Page 199 Problems 14,28,40 Main Idea: Subtract off what you already haveKey Words: Gram, Schmidt, Orthogonal,OrthonormalGoal: Learn to do the Gram Schmidt Process
3Previous AssignmentWednesday, Mar 12 Chapter Page 190 Problems 14,20,26,34Page 190 Problem 14Leonardo da Vinci and the resolution of forces.Leonardo ( ) asked himself how theweight of a body, supported by two strings ofdifferent length, is apportioned between the twostrings.
4B E A' | /' | /' b | a /Longer string ' | / shorter string' . |/' |D|____|___| Weight || |
5Three forces are acting at the point D: the tensions F1 and F2 in the strings and the weightW. Leonardo believed that| F1 | EA= ----| F2 | EBWas he right?
6B E A' | /' | _ _/' . ____ | a /|| b | /| ' | / F2F1 ' |D|___|___|Weight|| |
7Hint: Resolve F1 into a horizontal and a vertical component; do the same for F2. Since thesystem is at rest, the equation F1+F2+W =0holds. Express the ratios|F1| |EA|and|F2| |EB|in terms of alpha and beta, using trigonometricfunctions, and compare the results.
9| -Sin[a] Cos[b]-Cos[a]Sin[b] 0 | = W|-Sin[a] | | Sin[b] Cos[a]-Cos[b]Sin[a]| |-Sin[b] |-W Sin[a]F1 =-Sin[a+b]-W Sin[b]F2 =|F1| Sin[a]---- = It is the torque about E that|F2| Sin[b] is balanced.
10Page 190 Problem 20Refer to Figure 13 of this section.The least-squares line for these data is the liney = mx that fits the data best, in that the sum of the squares of the vertical distances between the line and the data points is minimal. We want to minimize the sum( mx1-y1)^2 + (mx2-y2)^ (mx5-y5)^2
11In vector notation, to minimize the sum means to find the scalar m such that|mx-y|^2is minimal. Arguing geometrically, explain how you canfind m. Use the accompanying sketch, which is notdrawn to scale./X //___________________________Y
12Find m numerically, and explain the relationship between m and the correlation coefficient r. Youmay find the following information helpful;XoY = |X| = |Y| =To check whether your solution m is reasonable,draw the line y = mx in Figure 13. (A morethorough discussion of least-squaresapproximations will follow in Section 5.4.)
14| | | m| | -107/10 || | | b | | -84/10 || | = | -18/10 || | | 51/10 || | | 158/10 |Multiply by the Transpose:| || || SUM |xi|^2 SUM X | | m | = | SUM xi.yi || SUM X | | b | | Sum yi || ^ | | m | = | || | | b | | |m = /198.53^2 = b = 0
15Page 190 Problem 26|49|Find the orthogonal projection of |49| onto theSubspace of R^3 spanned by| 2 | | 3 || 3 | and |-6 || 6 | | 2 |
16The vectors in the subspace are orthogonal. We divide by their lengths to make them an orthonormal basis. .| 2 | | 3 |1/7| 3 |, 1/7 |-6 || 6 | | 2 || 49 |The coefficients to use are the dot product of | 49 |and each of the vectors.| 2 | | 3 |77(1/7)| 3 | - 7(1/7)| -6 || 6 | | 2 |
17| 19 || 39 | This is the projection.| 64 |The error which is| 49 | | 19 | | 30 | | 6 || 49 | - | 39 | = | 10 | = 5 | 2 || 49 | | 64 | |-15 | |-3 |Since this vector is perpendicular to the spanning set of the subspace out answer checks.
19Among all the unit vectors of R^n find the one for which the sum of the components is maximal. Inthe case n=2, explain your answer geometrically,in terms of the unit circle and the level curves ofthe function y = x1+x2.The maximum occurs when each component is 1/Sqrt[n].
20The maximum occurs when each component is 1/Sqrt[n]. For the R^2 case, you want the projection of the unit circle on the line x=y to be maximum, which occurs at (1/Sqrt, 1/Sqrt);
21The Gram-Schmidt process creates an orthonormal basis out of an ordinary basis. The process is slow and tedious, but it works.Find an orthonormal basis for the space spanned by| 1 | | 0 | | 0 |V1=| 0 | V2=| 1 | V3=| 0 || 0 | | 0 | | 1 ||-1 | |-1 | |-1 |
25This is an Orthogonal basis This is an Orthogonal basis. To make an Orthonormal basis, divide by the lengths.| 1 | |-1| | 1 |1/Sqrt | 0 |, 1/Sqrt | 2|, 1/Sqrt| 1 || 0 | | 0| |-3 ||-1 | |-1| | 1 |
26Check that they are all orthogonal, all of unit length, and in the space of the original vectors.