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Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone Text: Linear Algebra With Applications, Second Edition Otto Bretscher

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Friday, Mar 14 Chapter 5.2 Page 199 Problems 14,28,40 Main Idea: Subtract off what you already have Key Words: Gram, Schmidt, Orthogonal, Orthonormal Goal: Learn to do the Gram Schmidt Process

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Previous Assignment Wednesday, Mar 12 Chapter 5.1 Page 190 Problems 14,20,26,34 Page 190 Problem 14 Leonardo da Vinci and the resolution of forces. Leonardo ( ) asked himself how the weight of a body, supported by two strings of different length, is apportioned between the two strings.

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B E A '. | / '. b | a / Longer string '. | / shorter string '. |/ ' |D | ____|___ | Weight | | |

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Three forces are acting at the point D: the tensions F1 and F2 in the strings and the weight W. Leonardo believed that | F1 | EA = ---- | F2 | EB Was he right?

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B E A '. | / '. | _ _/ '. ____ | a /| |. b | / | '. | / F2 F1 ' |D | ___|___ |Weight| | |

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Hint: Resolve F1 into a horizontal and a vertical component; do the same for F2. Since the system is at rest, the equation F1+F2+W =0 holds. Express the ratios |F1| |EA| and ---- |F2| |EB| in terms of alpha and beta, using trigonometric functions, and compare the results.

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F1 Cos[b] + F2 Cos[a] = W F1 Sin[b] - F2 Sin[a] = 0 | Cos[b] Cos[a] ||F1| = | W | | Sin[b] -Sin[a] ||F2| = | 0 | Multiply by | -Sin[a] -Cos[a] | | -Sin[b] Cos[b| | | -Sin[a] Cos[b]-Cos[a]Sin[b] 0 | = W|-Sin[a] | | 0 -Sin[b] Cos[a]-Cos[b]Sin[a]| |-Sin[b] |

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| -Sin[a] Cos[b]-Cos[a]Sin[b] 0 | = W|-Sin[a] | | 0 -Sin[b] Cos[a]-Cos[b]Sin[a]| |-Sin[b] | -W Sin[a] F1 = Sin[a+b] -W Sin[b] F2 = Sin[a+b] |F1| Sin[a] ---- = It is the torque about E that |F2| Sin[b] is balanced.

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Page 190 Problem 20 Refer to Figure 13 of this section. The least-squares line for these data is the line y = 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

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In vector notation, to minimize the sum means to find the scalar m such that |mx-y|^2 is minimal. Arguing geometrically, explain how you can find m. Use the accompanying sketch, which is not drawn to scale. / X / / /___________________________ Y

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Find m numerically, and explain the relationship between m and the correlation coefficient r. You may 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 more thorough discussion of least-squares approximations will follow in Section 5.4.)

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w = (m x1-y1)^2 + (m x2-y2)^ (m x5-y5)^2 W’= 2(mx1-y1)x1+2 (mx2-y2)x (mx5-y5)x5 W’ = 2m(x1^2 + x2^ x5^2) - 2(y1 x1+y2 x y5 x5) dw/dm = 2m |X|^2 - 2(XoY) The miximum comes when XoY m = |X|^ m = =

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| | | m| | -107/10 | | -47 1| | b | | -84/10 | | -24 1| = | -18/10 | | 57 1| | 51/10 | | 136 1| | 158/10 | Multiply by the Transpose: | | | | | SUM |xi|^2 SUM X | | m | = | SUM xi.yi | | SUM X 5 | | b | | Sum yi | | ^2 0 | | m | = | | | 0 5 | | b | | 0 | m = /198.53^2 = b = 0

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Page 190 Problem 26 |49| Find the orthogonal projection of |49| onto the |49| Subspace of R^3 spanned by | 2 | | 3 | | 3 | and |-6 | | 6 | | 2 |

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The 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 | | 49 | and each of the vectors. | 2 | | 3 | 77(1/7)| 3 | - 7(1/7)| -6 | | 6 | | 2 |

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| 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.

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Also | 2 | | 3 | | 6 | 1/7| 3 |, 1/7 |-6 |, 1/7 | 2 | | 6 | | 2 | |-3 | Are an orthonormal basis of R^3.

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Among all the unit vectors of R^n find the one for which the sum of the components is maximal. In the case n=2, explain your answer geometrically, in terms of the unit circle and the level curves of the function y = x1+x2. The maximum occurs when each component is 1/Sqrt[n].

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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[2], 1/Sqrt[2]);

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The 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 |

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| 1 | | 0 | | 0 | V1=| 0 | V2=| 1 | V3=| 0 | | 0 | | 0 | | 1 | |-1 | |-1 | |-1 | | 1 | Start W1 = V1 = | 0 | | 0 | |-1 |

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V2oW1 | 0 | 1 | 1 | | -1/2 | W2 = V W2 = | 1 | | 0 | = | 1 | W1oW1 | 0 | 2 | 0 | | 0 | |-1 | |-1 | | -1/2 | Check that this vector is perpendicular to W1.

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V3oW1 V3oW2 W3 = V W W2 W1oW1 W2oW2 | 0 | 1 | 1 | 1 | -1 | W3 = | 0 | | 0 | | 2 | | 1 | 2 | 0 | 6 | 0 | |-1 | |-1 | | -1 | | | | -2 | | 1 | 1/6| | = 1/6| -2 | use | 1 | | 6 0 0| | 6 | |-3 | |-6 3 1| | -2 | | 1 |

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This is an Orthogonal basis. To make an Orthonormal basis, divide by the lengths. | 1 | |-1| | 1 | 1/Sqrt[2] | 0 |, 1/Sqrt[6] | 2|, 1/Sqrt[12]| 1 | | 0 | | 0| |-3 | |-1 | |-1| | 1 |

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Check that they are all orthogonal, all of unit length, and in the space of the original vectors.

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Find an orthonormal basis of | 1 | | 1 | | 0 | | 1 | | 0 | | 1 | | 1 | | 1 | |-1 |

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answer 1/2 1/2 1/2 1/2 -1/2 1/2 1/2 -1/2 -1/2 1/2 1/2 -1/2

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