1. Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141
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

2. 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

3. Previous Assignment
Wednesday, Mar 12 Chapter 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.

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

5. 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?

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

7. 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.

8. 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] | = W|-Sin[a] |
| Sin[b] Cos[a]-Cos[b]Sin[a]| |-Sin[b] |

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.

10. 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

11. 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

12. 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.)

13. w = (m x1-y1)^2 + (m x2-y2)^2 + ... + (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|^2
4182.9
m = =

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

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

16. 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 |
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.

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

19. 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].

20. The 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[2], 1/Sqrt[2]);

21. 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 |

22. | 1 | | 0 | | 0 |
V1=| 0 | V2=| 1 | V3=| 0 |
| 0 | | 0 | | 1 |
|-1 | |-1 | |-1 |
| 1 |
Start W1 = V1 = | 0 |
| 0 |
|-1 |

23. V2oW | 0 | | 1 | | -1/2 |
W2 = V W2 = | 1 | | 0 | = | 1 |
W1oW | 0 | | 0 | | 0 |
|-1 | |-1 | | -1/2 |
Check that this vector is perpendicular to W1.

24. V3oW V3oW2
W3 = V W W2
W1oW W2oW2
| 0 | | 1 | | -1 |
W3 = | 0 | | 0 | | 2 |
| 1 | | 0 | | 0 |
|-1 | |-1 | | -1 |
| | | -2 | | 1 |
1/6| | = 1/6| -2 | use | 1 |
| | | 6 | |-3 |
| | | -2 | | 1 |

25. This is an Orthogonal basis
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 |

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

27. Find an orthonormal basis of
| 1 | | 1 | | 0 |
< | 1 |, | 0 |, | 2 | >
| 1 | | 0 | | 1 |
| 1 | | 1 | |-1 |

28. answer 1/ /2 1/2
1/ /2 1/2
1/ /2 -1/2
1/ /2 -1/2