Instructor: Irvin Roy Hentzel
Office 432 Carver
Phone 515-294-8141
E-mail: hentzel@iastate.edu
http://www.math.iastate.edu/hentzel/class.307.05


Assignment:  Work the practice test for Monday

April 15    6.4       

Main Idea:  How to create vectors which are perpendicular. 

Key Words: Perpendicular, Orthogonal, Orthonormal,

                                    VoW
           Gram Schmidt       V - ------ W  etc
                                    WoW

Goal: Understand the Gram Schmidt orthonormalization process


--------------------------------------------
Previous Assignment Page 401 Problem 19.

          |  1 |       |  5 |           |  0 |
Let u1 =  |  1 |,  u2= | -1 |, and u3 = |  0 |
          | -2 |       |  2 |           |  1 |

Note that u1 and u2 are orthogonal but that u3 is 

not orthogonal to u1 or u2.  It can be shown that

u3 is not in the subspace W spanned by u1 and u2.
                                                   3
Use this fact to construct a non zero vector v in R  

that is orthogonal to u1 and u2.

      -2  |  1 |    2  |  5 |         
    ----- |  1 | + --- | -1 |  =
       5  | -2 |   30  |  2 |          

      -10 |  1 |    2  |  5 |      1   |   0 |
    ----- |  1 | + --- | -1 |  =  ---  | -12 |
       30 | -2 |   30  |  2 |     30   |  24 |


       | 0 |      1  |   0 |      1   |   0  |
Now    | 0 |  -  --- | -12 | =   ---  |  12  |
       | 1 |     30  |  24 |     30   |   6  |


     | 0 |
So   | 2 |  should be the vector perpendicular to
     | 1 |

      |  1 |  |  5 |
Both  |  1 |  | -1 |
      | -2 |  |  2 |
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New Material

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 |

                 | 1 |  
Start W1 = V1 =  | 0 |
                 | 0 |
                 |-1 |
 
          V2oW1          | 0 |     1  | 1 |   | -1/2 |          |-1 |
W2 = V2 - ----- W2  =    | 1 | -  --- | 0 | = |   1  | use W2 = | 2 |
          W1oW1          | 0 |     2  | 0 |   |   0  |          | 0 |
                         |-1 |        |-1 |   | -1/2 |          |-1 |

Check that this vector is perpendicular to W1.
      
          V3oW1       V3oW2
W3 = V3 - ------W1 - ------ W2
          W1oW1       W2oW2

      | 0 |    1   | 1 |    1  | -1 |
W3 =  | 0 | - ---- | 0 | - --- |  2 |
      | 1 |    2   | 0 |    6  |  0 |
      |-1 |        |-1 |       | -1 |


        | 0 -3  1|      | -2 |      | 1 |
     1/6| 0  0 -2| = 1/6| -2 |  use | 1 |  
        | 6  0  0|      |  6 |      |-3 |
        |-6  3  1|      | -2 |      | 1 |

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 |


Check that they are all orthogonal, all of unit length,

and in the space of the original vectors.


Find an orthonormal basis of

              | 1 |   | 1 |  | 0 |
           <  | 1 |,  | 0 |, | 2 | >
              | 1 |   | 0 |  | 1 |
              | 1 |   | 1 |  |-1 |


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