5.1#12

Hello Dr. Taylor, 

I was having trouble with the Gram-Schmidt procedure in the homework and I wanted to check the practice test too. I found that the first vector will be the vector, the second would be the second (v2) – dot product of v1 and v2 divided by v1Tv1 times v1 and the third is the same process but also have v3 -  dot product of v1 and v3 divided by v1Tv1 times v1 - dot product of the original v2 and v3 divided by v2Tv2 times v2. 

Indeed, start with v1. BUT, you're supposed to give a normalized vector, so your first vector should be u1 = v1/||v1||

 v2– (v1T v2)/( v1Tv1) v1 will indeed be perpendicular to v1 (and this is the same as v2-(u1^Tv2)u1

because u1Tu1=1) It will not typically be normalized though, except by accident. You need to normalize it: u2 = (v2– (v1T v2)/( v1Tv1) v1)/||(v2– (v1T v2)/( v1Tv1) v1)||.

Now, v3 - (u1Tv3)u1 - (u2^Tv3)u2 will be perpendicular to both v1 and v2, and to both u1 and u2. Again, it's not normalized so you need to do that.

For the first problem I got the second row right and the first wrong, the second Gram-Schmidt problem on the homework I got nothing right and on the Test I got v1 and v2 unchanged and v3 turned into [2/3,1/3,1/3,1/3,2/3,1/3] I was wondering if that was correct and what I’m doing wrong.

Again you forgot to normalize.  See my solution below

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Note ||x|| = √((-1)^2 + 2^2 + (-2)^2) = √9 = 3;   ||y||= √((-1)^2+1^2 + 0^2) = √2;
||z|| = √(1^2 + (-5)^2 +(-1)^2) = √27

SO u1 = x/||x|| = | -1/3 |
                           | 2/3  |
                           | -2/3 |
Now y - (u1Ty) u1 = | -1 |                                                        | -1/3 |
                                  |  1  |  -  ((-1/3)(-1) +(2/3)(1)+(-2/3)(0)) |  2/3 |
                                  |  0  |                                                        |-2/3  |

                                     | -1 |           | -1/3 |      | -2/3  |
                                  = |  1  |  -   1  |  2/3 |   = |  1/3  |
                                     |  0  |           |-2/3 |       | 2/3  |
and you can check that this vector is perpendicular to u1 and v1. It's also already normalized because its norm is √((-2/3)^2 + (1/3)^2 + (2/3)^2 = √(4/9 + 1/9 +4/9) = 1 so

         |-2/3|
u2 =  | 1/3|
         | 2/3|

Now we need to compute
                                              | 1 |                                                           | -1/3 | 
z - (u1Tz) u1 -(u2Tz ) u2  = | -5 |  -   ((-1/3)(1) +(2/3)(-5)+(-2/3)(-1)) |  2/3 | 
                                              |-1 |                                                          |-2/3  |
   
                                                                                                                      | -2/3 | 
                                                              -   ((-2/3)(1) +(1/3)(-5)+(2/3)(-1))  |  1/3 | 
                                                                                                                      |  2/3  |

   | 1 |               | -1/3 |           | -2/3 |
= |-5 |  -    (-3) |  2/3 |  - (-3) |  1/3 |
   |-1 |               |-2/3 |            | 2/3 |

   | 1 -  1  -  2   |       | -2  |
= |-5 -(-2)-(-1) |  =  | -2  |
   |-1 -  2 - (-2) |      | -1  |

This vector is perpendicular to u1 and u2 and also perpendicular to v1 and v2 but it's not normalized. It's norm is 3 though, so

         |-2/3|
u3 =  |-2/3|
         |-1/3|

I was also wondering if the dimension of the kernel of A was 0 because you could row reduce it to the identity matrix. 

Yes.

And is there a difference between b and g, the orthogonal basis of S and the basis of S perpendicular?

Thank you

Well yes there would have to be a difference because the subspace S and the subspace S^⟘ are perpendicular to each other, so the vectors in a basis for one would have to be perpendicular to ALL of the vectors in the other